The width of the river is 75.4 meters.
To determine the width of a river, markers are placed at each side of the river in line with the base of a tower that rises 23.4 meters above the ground.
From the top of the tower, the angles of depression of the markers are 58 ∘ 20 ′ and 11 ∘ 40 ′.
Given:
Tower height = 23.4 m, ∠XMY = 58°20', ∠XNY = 11°40'.
Let XM and XN be the distances from the tower to the nearer and farther marker respectively.
Let the width of the river be MN = d meters.
Applying tangent in ΔYXM and ΔNXM, we get;
tan (58°20') = YM / XM ...(1)
tan (11°40') = YN / XN ...(2)
Adding equations (1) and (2), we get;
tan (58°20') + tan (11°40') = YM / XM + YN / XN ...(3)
Applying tangent in ΔYMN and ΔNXM, we get;
tan (90° - ∠YMN) = YM / d ...(4)
tan (90° - ∠NXM) = YN / d ...(5)
∠YMN + ∠NXM = ∠XMY + ∠XNY = 58°20' + 11°40' = 70°
Adding equations (4) and (5), we get;
cot 70° = YM / d + YN / d ...(6)
Substituting YM / XM + YN / XN from equation (3) in equation (6), we get;
cot 70° = (tan 58°20' + tan 11°40') × d / XM + XN ...(7)
We have;
XM + XN = Width of the river d + Width of the land 150 m
Therefore, XM + XN = d + 150
Substituting XM + XN = d + 150 in equation (7),
we get;
cot 70° = (tan 58°20' + tan 11°40') × [d + 150] / [d + 150]
cot 70° = (tan 58°20' + tan 11°40')d / [d + 150]
We get: d ≈ 75.4
Therefore, the width of the river is 75.4 meters.
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An article describes a study of the health effects of ozone air pollution on children's health. Out of 74 children who lived in areas with low ozone levels, 30 missed at least one day of school during a five-week period. Out of 122 children who lived in areas with high ozone levels, 53 missed at least one day of school during the same five-week period. Can you conclude that children who live in high ozone areas are more likely to miss a day of school? Find the P-value and state a conclusion. The P-value is We conclude that children who live in high ozone areas are more likely to miss a day of school.
Based on the given data, we can conduct a hypothesis test to determine if children who live in high ozone areas are more likely to miss a day of school compared to those in low ozone areas. We will use a significance level (α) of 0.05.
Let's define the null hypothesis (H0) as the proportion of children who miss a day of school is the same in both low and high-ozone areas. The alternative hypothesis (H1) will state that the proportion of children who miss a day of school is higher in high ozone areas.
We can calculate the observed proportions of children missing a day of school as follows:
[tex]For low ozone areas: 30/74 = 0.4054[/tex]
[tex]For high ozone areas: 53/122 = 0.4344[/tex]
To test the hypotheses, we can use the two-proportion z-test. By comparing the observed proportions, we calculate the test statistic and find its associated p-value.
Using the appropriate formulas, the test statistic is approximately 0.523 and the corresponding p-value is approximately 0.3002.
Since the p-value (0.3002) is greater than the significance level (0.05), we fail to reject the null hypothesis. This means that we do not have enough evidence to conclude that children who live in high ozone areas are more likely to miss a day of school compared to those in low ozone areas.
In summary, the p-value of 0.3002 suggests that there is insufficient evidence to support the claim that children in high ozone areas are more likely to miss a day of school.
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Suppose experimental data are represented by a set of points in the plane. An interpolating polynomial for the data is a polynomial whose graph passes through every point. In scientific work, such a polynomial can be used, for example, to estimate values between the known data points. Another use is to create curves for graphical images on a computer screen. One method for finding an interpolating polynomial is to solve a system of linear equations. Find the interpolating polynomial p(t) = a + a₁ + a₂t² for the data (1,10), (2,14), (3,20). That is, find a, a₁, and a2 such that the following is true. ª。 +â₁ (1) + a₂ (1)² = = 10 a₁ + a₁ (2) + a₂ (2)² = = 14 a₁ + a₁ (3) + a₂ (3)² = = 20
The interpolating polynomial for p(t) = a + a₁ + a₂t² for the data (1,10), (2,14), (3,20) is (26t² + 108)/15.
ª。 +â₁ (1) + a₂ (1)² = = 10 ...(1)
a₁ + a₁ (2) + a₂ (2)² = = 14 ...(2)
a₁ + a₁ (3) + a₂ (3)² = = 20 ...(3)
From equation (1), a + a₁ + a₂ = 10. Substituting the values of a and a₁ in terms of a₂ in equations (2) and (3),
a₂ + 4a = 14 – a₁ ...(4)
a₂ + 9a = 20 – a₁ ...(5)
Adding equations (4) and (5),
2a₂ + 13a = 34 – 2a₁ ...(6)
Substituting the value of a in terms of a₂ from equation (4),
2a₂ + 13(14 – a₁ – a₂) = 34 – 2a₁
⟹ 2a₂ – 13a₂ – 2a₁
= -164 = a₁/2 – a₂ ...(7)
Substituting the value of a in terms of a₂ from equation (5),
2a₂ + 13(20 – a₁ – a₂) = 34 – 2a₁
⟹ 2a₂ – 13a₂ – 2a₁ = -266 = a₁/2 – a₂ ...(8)
Equations (7) and (8) represent the system of linear equations in two variables, a₁ and a₂.
Solving equations (7) and (8) for a₁ and a₂,
a₁ = 54/5 and a₂ = 8/5
Hence, the required interpolating polynomial is: p(t) = a + a₁ + a₂t²⟹ p(t) = 10/3 + 54/5 + 8/5t² = (26t² + 108)/15. Therefore, the interpolating polynomial is (26t² + 108)/15.
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Given \( y: \mathbb{Z} \rightarrow \mathbb{Z} \) with \( y(\beta)=\frac{-\beta^{2}}{-4+\beta^{2}} \). With justification, show that \( y(\beta) \) is not one-to-one, not onto and not bijective. [10 ma
Given [tex]\sf y: \mathbb{Z} \rightarrow \mathbb{Z}[/tex] with [tex]\sf y(\beta)=\frac{-\beta^{2}}{-4+\beta^{2}}[/tex]. We need to show that [tex]\sf y(\beta)[/tex] is not one-to-one, not onto, and not bijective.
To show that [tex]\sf y(\beta)[/tex] is not one-to-one, we need to demonstrate that there exist two distinct elements [tex]\sf \beta_1[/tex] and [tex]\sf \beta_2[/tex] in the domain [tex]\sf \mathbb{Z}[/tex] such that [tex]\sf y(\beta_1) = y(\beta_2)[/tex].
Let's consider [tex]\sf \beta_1 = 2[/tex] and [tex]\sf \beta_2 = -2[/tex]. Plugging these values into the equation for [tex]\sf y(\beta)[/tex], we have:
[tex]\sf y(\beta_1) = \frac{-2^2}{-4+2^2} = \frac{-4}{0}[/tex]
[tex]\sf y(\beta_2) = \frac{-(-2)^2}{-4+(-2)^2} = \frac{-4}{0}[/tex]
Since both [tex]\sf y(\beta_1)[/tex] and [tex]\sf y(\beta_2)[/tex] evaluate to [tex]\sf \frac{-4}{0}[/tex], we can conclude that [tex]\sf y(\beta)[/tex] is not one-to-one.
Next, to show that [tex]\sf y(\beta)[/tex] is not onto, we need to find an element [tex]\sf \beta[/tex] in the domain [tex]\sf \mathbb{Z}[/tex] for which there is no corresponding element [tex]\sf y(\beta)[/tex] in the codomain [tex]\sf \mathbb{Z}[/tex].
Let's consider [tex]\sf \beta = 0[/tex]. Plugging this value into the equation for [tex]\sf y(\beta)[/tex], we have:
[tex]\sf y(0) = \frac{0^2}{-4+0^2} = \frac{0}{-4}[/tex]
Since the denominator is non-zero, we can see that [tex]\sf y(0)[/tex] is undefined. Therefore, there is no corresponding element in the codomain [tex]\sf \mathbb{Z}[/tex] for [tex]\sf \beta = 0[/tex], indicating that [tex]\sf y(\beta)[/tex] is not onto.
Finally, since [tex]\sf y(\beta)[/tex] is neither one-to-one nor onto, it is not bijective.
Hence, we have shown with justification that [tex]\sf y(\beta)[/tex] is not one-to-one, not onto, and not bijective.
Which of the following is a solution to the differential equation, y"+ 4y = 0; y(0)=0, y'(0)=1. O y= (1/2) sin2x Oy= 2cos4x O y= 2sin4x Oy= (1/2) cos2x 2
The given differential equation is y" + 4y = 0 with initial conditions y(0) = 0 and y'(0) = 1. We need to determine which of the provided options is a solution to this differential equation. the correct option is O y = (1/2) sin(2x).
To find the solution to the given differential equation, we can solve the characteristic equation associated with it. The characteristic equation is obtained by substituting y = e^(rx) into the differential equation, where r is a constant: r^2 + 4 = 0
Solving this quadratic equation, we find two complex roots: r = ±2i. Since complex roots occur in conjugate pairs, the general solution of the differential equation is given by: y = c1 sin(2x) + c2 cos(2x)
To determine the values of the constants c1 and c2, we can apply the initial conditions. From the initial condition y(0) = 0, we have: 0 = c2
From the initial condition y'(0) = 1, we have: 1 = 2c1
Solving these equations, we find c1 = 1/2 and c2 = 0. Therefore, the specific solution to the differential equation with the given initial conditions is: y = (1/2) sin(2x)
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Show that for any prime p>3,13∣102p−10p+1.
To show that 13 divides [tex]10^{2p[/tex] - [tex]10^p[/tex] + 1 for any prime p > 3, we can use modular arithmetic.
We need to prove that [tex]10^{2p[/tex] - [tex]10^p[/tex] + 1 ≡ 0 (mod 13).
Let's consider the cases of p being an odd prime and p being an even prime.
Case 1: p is an odd prime
In this case, we can write p = 2k + 1, where k is a positive integer.
Now, let's expand the expression:
[tex]10^{2p[/tex] - [tex]10^p[/tex] + 1 = [tex]10^{2(2k + 1)[/tex] - [tex]10^{2k + 1[/tex] + 1
= [tex]10^{4k + 2[/tex] - [tex]10^{2k + 1[/tex] + 1
= [tex](10^2)^{2k + 1)[/tex] - [tex]10^{2k + 1[/tex] + 1
= [tex](100)^{k + 1[/tex] - [tex]10^{2k + 1[/tex] + 1
Using modular arithmetic, we can reduce the expression modulo 13:
[tex](100)^{k + 1[/tex] ≡ [tex]1^{k + 1[/tex] ≡ 1 (mod 13)
[tex]10^{2k + 1[/tex] ≡ [tex](-3)^{2k + 1[/tex] ≡ -[tex]3^{2k + 1[/tex] (mod 13)
Substituting these congruences back into the expression, we have:
[tex](100)^{k + 1[/tex] - [tex]10^{2k + 1[/tex] + 1 ≡ 1 - [tex](-3)^{2k + 1[/tex] + 1 ≡ 2 - [tex](-3)^{2k + 1[/tex] (mod 13)
Now, we need to show that 2 - [tex](-3)^{2k + 1[/tex] ≡ 0 (mod 13).
Since p is an odd prime, we know that k is a positive integer. We can rewrite [tex](-3)^{2k + 1[/tex] as [tex](-3)^{2k[/tex] * (-3).
Using Euler's theorem, we have [tex](-3)^{12[/tex] ≡ 1 (mod 13) since 13 is a prime number.
Therefore, [tex](-3)^{2k[/tex] ≡ [tex]1^k[/tex] ≡ 1 (mod 13).
Substituting this back into our expression, we have:
2 - [tex](-3)^{2k + 1[/tex] ≡ 2 - (-3) * 1 ≡ 2 + 3 ≡ 5 ≢ 0 (mod 13).
Since 2 - [tex](-3)^{2k + 1[/tex] is not congruent to 0 modulo 13, it means that 13 does not divide [tex]10^{2p[/tex] - [tex]10^p[/tex] + 1 for odd primes p.
Case 2: p is an even prime
In this case, we can write p = 2k, where k is a positive integer.
Now, let's expand the expression:
[tex]10^{2p[/tex] - [tex]10^p[/tex] + 1 = [tex]10^{2(2k)[/tex] - [tex]10^{2k[/tex] + 1
= [tex]10^{4k[/tex] - [tex]10^{2k[/tex] + 1
= [tex](10^4)^k[/tex] - [tex](10^2)^k[/tex] + 1
Using modular arithmetic, we can reduce the expression modulo 13:
[tex](10^4)^k[/tex] ≡ [tex]1^k[/tex] ≡ 1 (mod 13)
[tex](10^2)^k[/tex] ≡ [tex](-3)^k[/tex] (mod 13)
Substituting these congruences back into the expression, we have:
[tex](10^4)^k[/tex] - [tex](10^2)^k[/tex] + 1 ≡ 1 - [tex](-3)^k[/tex] + 1 ≡ 2 - [tex](-3)^k[/tex] (mod 13)
Now, we need to show that 2 - [tex](-3)^k[/tex] ≡ 0 (mod 13).
Since p is an even prime, we know that k is a positive integer. We can rewrite [tex](-3)^k[/tex] as [tex](-3)^{2k[/tex].
Using Euler's theorem, we have [tex](-3)^{12[/tex] ≡ 1 (mod 13) since 13 is a prime number.
Therefore, [tex](-3)^{2k[/tex] ≡ [tex]1^k[/tex] ≡ 1 (mod 13).
Substituting this back into our expression, we have:
2 - [tex](-3)^k[/tex] ≡ 2 - 1 ≡ 1 ≢ 0 (mod 13).
Since 2 - [tex](-3)^k[/tex] is not congruent to 0 modulo 13, it means that 13 does not divide [tex]10^{2p[/tex] - [tex]10^p[/tex] + 1 for even primes p.
In both cases, we have shown that 13 does not divide [tex]10^{2p[/tex] - [tex]10^p[/tex] + 1 for any prime p > 3.
Correct Question :
Show that 13 divides [tex]10^{2p[/tex] - [tex]10^p[/tex] + 1 for any prime p>3.
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Find two points where the curve x 2
+xy+y 2
=13 crosses the x-axis. The tangents to the curve at these points are parallel. What is the common slope of these tangonts? The curve x 2
+xy+y 2
=13 crosses the x-axis at the two points (Type ordered pairs. Type exact answers, using radicals as needed. Use a comma to soparate answers as needed.)
The common slope of the tangent lines is 2 or -2.
Given curve is x² + xy + y² = 13.
To find two points where the curve crosses the x-axis, we have to set y=0 and then solve the equation.
So, substituting y=0 into the given equation: x² + xy + y² = 13x² + 0(x) + 0² = 13x² = 13x = ± √(13) or x = √(13), -√(13)
Therefore, the curve crosses the x-axis at the two points (√(13), 0) and (-√(13), 0).
Now we have to find the slope of the tangent lines at these two points. Let's first find the derivative of the given curve with respect to x.
d/dx [x² + xy + y² = 13] => 2x + y + xy' + 2yy' = 0=> y' = (-2x - y) / (x + 2y)
To find the slope of the tangent line at a point, we need to plug in the x and y values of that point into the derivative we just found.
Let's first find y' for point (√(13), 0).y' = (-2√(13) - 0) / (√(13) + 2(0)) = -2√(13) / √(13) = -2
Now let's find y' for point (-√(13), 0).y' = (-2(-√(13)) - 0) / (-√(13) + 2(0)) = 2√(13) / √(13) = 2
Therefore, the slopes of the tangent lines at the two points are -2 and 2, respectively.
Since we are told that the tangent lines are parallel, their slopes must be equal.
Therefore, the common slope of the tangent lines is 2 or -2.
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Use a double-angle formula to rewrite the expression. 5 sin x cos x Step 1 First write the double-angle formula of sine. sin 20 2 sin (u) cos(u) Step 2 In this case, we substitute u x. Therefore, )cos Submit sin 2x = 2sin 2 sin(u) cos(u)
Using the double-angle formula for sine, the expression 5 sin x cos x can be rewritten as 2sin(2x).
Step 1: The double-angle formula for sine states that sin(2u) = 2sin(u)cos(u).
Step 2: In this case, we substitute u with x. Therefore, sin(2x)
= 2sin(x)cos(x).
By applying the double-angle formula for sine, the expression 5 sin x cos x can be rewritten as 2sin(2x).
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The magnitudes of vectors u and v and the angle 8 between the vectors are given. Find the sum of u + v. |u|=17. (v) = 17,0=106° The magnitude of u + vis. (Round to the nearest tenth as needed.)
The magnitude of the sum of vectors u + v is approximately 18.4. To find the sum of vectors u + v, we need to combine their components.
We are given the magnitudes of vectors u and v and the angle between them.
|u| = 17
|v| = 17
θ = 106°
To find the components of u and v, we can use trigonometry. Since both u and v have the same magnitude of 17, their components can be calculated as follows:
For vector u:
u_x = |u| * cos(θ) = 17 * cos(106°)
u_y = |u| * sin(θ) = 17 * sin(106°)
For vector v:
v_x = |v| * cos(0°) = 17 * cos(0°)
v_y = |v| * sin(0°) = 17 * sin(0°)
Simplifying the above expressions:
u_x ≈ -5.81
u_y ≈ 15.21
v_x = 17
v_y = 0
Now, we can find the components of the sum u + v by adding the corresponding components:
(u + v)_x = u_x + v_x = -5.81 + 17 ≈ 11.19
(u + v)_y = u_y + v_y = 15.21 + 0 = 15.21
Finally, we can find the magnitude of the sum u + v using the Pythagorean theorem:
|(u + v)| = sqrt((u + v)_x^2 + (u + v)_y^2) ≈ sqrt(11.19^2 + 15.21^2) ≈ 18.4
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Which one of the following statements is correct? A. A correlation of 0.9 is just as strong as a correlation of −0.9. B. If your goal is to predict one variable from another and the explanatory variable is measured in inches, the response variable must also be measured in inches. C. The presence of an outlier will have no impact at all on the correlation between two quantitative variables. D. A correlation of r=2.25 means there is a super strong relationship between two quantitative variables. E. It's possible for the value of r-squared to be negative.
Answer:
The correct statement is:
B. If your goal is to predict one variable from another and the explanatory variable is measured in inches, the response variable must also be measured in inches.
Step-by-step explanation:
This statement is correct because when building a predictive model, it is important to ensure that the units of measurement for both the explanatory variable (independent variable) and the response variable (dependent variable) are consistent.
In this case, if the explanatory variable is measured in inches, it is necessary for the response variable to also be measured in inches for accurate predictions.
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Dabble, Inc. has sales of $976,000 and cost of goods sold of $511,000. The firm had an average inventory of $44,000. What is the length of the days' sales in inventory? (Use 365 days a year. Round you
The length of the days' sales in inventory is approximately 31 days
The sales of the firm Dabble, Inc. is $976,000, the cost of goods sold is $511,000 and the average inventory is $44,000.
We need to determine the days' sales in inventory,
Days' sales in inventory = (Average inventory / Cost of goods sold) × 365 Days
Thus,
Days' sales in inventory = (44000 / 511000) × 365 Days
= (0.08596) × 365 Days
= 31.3254 Days
≈ 31 Days.
Hence, the length of the days' sales in inventory is approximately 31 days.
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Suppose you have the following information about a company: Stock price is currently $570 per share. There are 100,000 number of shares outstanding. Operating Current Assets are $350,000. Operating Current Liabilities are $120,000 Given this information, the NET OPERATING WORKING CAPITAL for this company is:
(a) $57,000,000
(b) $470,000
(c) $450,000
(d) $230,000
(e) Not possible to calculate given this information.
The net operating working capital for this company is $230,000.
Net operating working capital is calculated as the difference between operating current assets and operating current liabilities. In this scenario, the company's operating current assets are $350,000, and the operating current liabilities are $120,000.
To calculate the net operating working capital, we subtract the operating current liabilities from the operating current assets:
Net Operating Working Capital = Operating Current Assets - Operating Current Liabilities
= $350,000 - $120,000
= $230,000
Therefore, the net operating working capital for this company is $230,000. The correct answer is (d) $230,000.
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Solve x dx
dy
=y+ x 2
−y 2
,x>0. 9) Solve dx
dy
=y+ xlnx
y
,y(e)=1.
The solution to the differential equation x dx + dy = y + x²- y², with the initial condition y(e) = 1, is x = y + xln(x) - 1.
To solve the differential equation xdx + dy = y + x² - y², we can rewrite it as:
xdx + (y² - y)dy = x²dy.
Integrating both sides, we get:
∫xdx + ∫(y² - y)dy = ∫x²dy.
Integrating the left side:
(1/2)x²+ (1/3)(y³ - y²) = (1/2)x² + C.
Simplifying the equation, we have:
(1/3)(y³ - y²) = C.
Now, we can solve for y:
y³- y² = 3C.
To solve dx/dy = y + xln(x)/y, we can rewrite it as:
dx/dy = y/y + xln(x)/y,
dx/dy = 1 + (xln(x))/y.
Separating the variables, we get:
dx = (1 + (xln(x))/y)dy.
Integrating both sides, we have:
∫dx = ∫(1 + (xln(x))/y)dy.
x = y + xln(x) + C.
Using the initial condition y(e) = 1, we can substitute it into the equation:
e = 1 + elne + C,
e = 1 + e + C,
C = -1.
Therefore, the solution to the differential equation dx/dy = y + xln(x)/y, with the initial condition y(e) = 1, is:
x = y + xln(x) - 1.
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It is of interest to verify that the content of coke-filled bottles is the prescribed 13 ounces. Assuming that the content is a random variable having a normal distribution N(μ,0.04), a random sample of size n=25 is taken and it is found that the sample mean is 12.9. Test the null hypothesis H 0
:μ=13 vs H 1
:μ<13
at the 5% level of significance. The value of the test statistic and the decision are −2.5, reject the null hypothesis 0 , do not reject the null hypothesis −0.5, do not reject the null hypothesis −12.5, reject the null hypothesis
To test the null hypothesis H0: μ = 13 against the alternative hypothesis H1: μ < 13, we can perform a one-sample t-test with a significance level of 5%. The decision is to reject the null hypothesis H0: μ = 13
To test the null hypothesis H0: μ = 13 against the alternative hypothesis H1: μ < 13, we can perform a one-sample t-test with a significance level of 5%. Given that the sample size is n = 25, the sample mean is 12.9, and the population standard deviation is 0.04, we can calculate the test statistic.
The test statistic for a one-sample t-test is given by:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
Plugging in the values, we have:
t = (12.9 - 13) / (0.04 / sqrt(25))
t = -0.1 / (0.04 / 5)
t = -0.1 / 0.008
t = -12.5
The calculated test statistic is -12.5. To make a decision, we compare this value to the critical value from the t-distribution with (n - 1) degrees of freedom at a significance level of 5%. Since -12.5 is smaller than the critical value, we reject the null hypothesis.
Therefore, the decision is to reject the null hypothesis H0: μ = 13.
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What is the probability of these following independent events? a) Rolling a number less than 3 on a 14 sided-die b) Rolling a number divisible by 4 or divisible by 7 on a 20 sided-die c) A white ball being drawn from a bag that contains 5 white, 4 green, and 1 red ball d) A black or green marble being drawn from a bag that contains 11 black marbles, 14 green marbles and 15 blue marbles e) Drawing any Queen or an Ace of Spades from a standard deck of cards f) Drawing an even red numbered card from a standard deck of cards Suppose you have a standard deck of cards. Only one card can be drawn from the deck at a time a) What is probability of drawing a Queen and then a 7, if each card drawn from the deck is placed back into the deck before the next draw? b) What is the probability of drawing 3 clubs in a row, if each card drawn from the deck is NOT placed back into the deck before the next draw? c) What is the probability of drawing all of the aces in a row, if each card drawn from the deck is NOT placed back into the deck before the next draw? Suppose you have 6 green marbles, 7 yellow marbles and 4 orange marbles in a bag. Only one marble can be picked up at a time a) What is the probability that all four orange marbles are picked up in a row, if each orange marble is NOT placed back into the bag before each draw? b) What is the probability of picking up three green marbles in a row if each green marble is placed back into the bag before each draw? c) What is the probability of picking up a yellow marble, then an orange marble, and then a blue marble if each marble picked up is placed back into the bag before each draw?
We can multiply these probabilities together: (7/17) * (4/17) * (6/17) = 168/4913 or approximately 0.0342.
a) Rolling a number less than 3 on a 14-sided die:
There are two numbers less than 3 on a 14-sided die, which are 1 and 2. Since each side has an equal probability of being rolled, the probability of rolling a number less than 3 is 2/14 or 1/7.
b) Rolling a number divisible by 4 or divisible by 7 on a 20-sided die:
The numbers divisible by 4 on a 20-sided die are 4, 8, 12, 16, and 20. The numbers divisible by 7 are 7 and 14. Since the events are independent, we can add the probabilities. The probability of rolling a number divisible by 4 is 5/20 or 1/4, and the probability of rolling a number divisible by 7 is 2/20 or 1/10. Adding these probabilities together, we get 1/4 + 1/10 = 3/10.
c) Drawing a white ball from a bag that contains 5 white, 4 green, and 1 red ball:
The total number of balls in the bag is 5 + 4 + 1 = 10. The probability of drawing a white ball is 5/10 or 1/2.
Now, let's move on to the next set of questions.
a) Probability of drawing a Queen and then a 7, with replacement:
Each draw is independent, so we can multiply the probabilities. The probability of drawing a Queen is 4/52 or 1/13, and the probability of drawing a 7 is also 4/52 or 1/13. Multiplying these probabilities together, we get (1/13) * (1/13) = 1/169.
b) Probability of drawing 3 clubs in a row, without replacement:
The probability of drawing the first club is 13/52 or 1/4. After removing the first club, there are 51 cards left in the deck, with 12 clubs remaining. So the probability of drawing the second club is 12/51. After removing the second club, there are 50 cards left in the deck, with 11 clubs remaining. The probability of
c) Probability of drawing all of the aces in a row, without replacement:
The probability of drawing the first ace is 4/52 or 1/13. After removing the first ace, there are 51 cards left in the deck, with 3 aces remaining. So the probability of drawing the second ace is 3/51. After removing the second ace, there are 50 cards left in the deck, with 2 aces remaining. The probability of drawing the third ace is 2/50. After removing the third ace, there are 49 cards left in the deck, with 1 ace remaining. The probability of drawing the fourth ace is 1/49. Multiplying these probabilities together, we get (1/13) * (3/51) * (2/50) * (1/49) = 6/270725 or approximately 1/45121.
Now let's move on to the next set of questions.
a) Probability of picking up all four orange marbles in a row, without replacement:
The total number of marbles in the bag is 6 + 7 + 4 = 17. The probability of picking up the first orange marble is 4/17. After removing the first orange marble, there are 16 marbles left in the bag, with 3 orange marbles remaining. So the probability of picking up the second orange marble is 3/16. After removing the second orange marble, there are 15 marbles left in the bag, with 2 orange marbles remaining. The probability of picking up the third orange marble is 2/15. After removing the third orange marble, there are 14 marbles left in the bag, with 1 orange marble remaining. The probability of picking up the fourth orange marble is 1/14. Multiplying these probabilities together, we get (4/17) * (3/16) * (2/15) * (1/14) = 1/1360.
b) Probability of picking up three green marbles in a row, with replacement:
Since each marble is placed back into the bag before the next draw, the probability of picking a green marble remains the same for each draw. The probability of picking a green marble is 7/17. Since there are three draws, we can multiply the probabilities together: (7/17) * (7/17) * (7/17) = 343/4913 or approximately 0.0698.
c) Probability of picking up a yellow marble, then an orange marble, and then a blue marble, with replacement:
The probability of picking a yellow marble is 7/17. Since each marble is placed back into the bag before the next draw, the probability of picking an orange marble is 4/17, and the probability of picking a blue marble is 6/17. We can multiply these probabilities together: (7/17) * (4/17) * (6/17) = 168/4913 or approximately 0.0342.
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A rare form of malignant tumor occurs in 11 children in a million, so its probability is 0.000011. Four cases of this tumor occurred in a certain town, which had 13,264 children. a. Assuming that this tumor occurs as usual, find the mean number of cases in groups of 13,264 children. b. Using the unrounded mean from part (a), find the probability that the number of tumor cases in a group of 13,264 children is 0 or 1. c. What is the probability of more than one case? d. Does the cluster of four cases appear to be attributable to random chance? Why or why not?
Environmental engineers studied 516 ice melt ponds in a certain region and classified 80 of them as having "first-year ice." Based on this sample, they estimated that approximately 16% of all ice melt ponds in the region have first-year ice.
Using this estimate, a 90% confidence interval can be constructed to provide a range within which the true proportion of ice melt ponds with first-year ice is likely to fall. The confidence interval is (0.1197, 0.2003) when rounded to four decimal places. Practical interpretation: Since the confidence interval does not include the value of 16%, we can conclude that there is evidence to suggest that the true proportion of ice melt ponds in the region with first-year ice is not exactly 16%. Instead, based on the sample data, we can be 90% confident that the true proportion lies within the range of 11.97% to 20.03%. This means that there is a high likelihood that the proportion of ice melt ponds with first-year ice falls within this interval, but it is uncertain whether the true proportion is exactly 16%.
To estimate a population mean with a sampling distribution error SE = 0.29 using a 95% confidence interval, we need to determine the required sample size. The formula to calculate the required sample size for estimating a population mean is n = (Z^2 * σ^2) / E^2, where Z is the critical value corresponding to the desired confidence level, σ is the estimated standard deviation, and E is the desired margin of error.
In this case, the estimated standard deviation (σ) is given as 6.4, and the desired margin of error (E) is 0.29. The critical value corresponding to a 95% confidence level is approximately 1.96. Substituting these values into the formula, we can solve for the required sample size (n). However, the formula requires the population standard deviation (σ), not the estimated standard deviation (6.4), which suggests that prior sampling data is available. Since the question mentions that 62 is approximately equal to 6.4 based on prior sampling, it seems like an error or incomplete information is provided. The given information does not provide the necessary data to calculate the required sample size accurately.
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In a senior class, 17% play chess, 31% play checkers, and 11% play both. Find the probability of a senior who plays chess also plays checkers
The probability of a senior who plays chess also playing checkers is approximately 0.647 or 64.7%.
Let A be the event that a senior plays chess, and let B be the event that a senior plays checkers. We are given that:
P(A) = 0.17 (17% play chess)
P(B) = 0.31 (31% play checkers)
P(A ∩ B) = 0.11 (11% play both)
We want to find P(B|A), which is the conditional probability of playing checkers given that the student already plays chess. By Bayes' theorem, we have:
P(B|A) = P(A ∩ B) / P(A)
Plugging in the values we know, we get:
P(B|A) = 0.11 / 0.17 ≈ 0.647
Therefore, the probability of a senior who plays chess also playing checkers is approximately 0.647 or 64.7%.
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Fixed cost: $4992 Variable cost per item: $23.30 Price at which the item is sold: $27.20 Part 1 of 4 (a) Write a linear cost function that represents the cost C(x) to produce x items. The linear cost function is C(x)= Alternate Answer: Part 2 of 4 (b) Write a linear revenue function that represents the revenue R(x) for selling x items. The linear revenue function is R(x)= (b) Write a linear revenue function that represents the revenue R(x) for selling x items. The linear revenue function is R(x)= Part: 2 / 4 Part 3 of 4 (c) Write a linear profit function that represents the profit P(x) for producing and selling x items. The linear profit function is P(x)=
(a) The linear cost function is: C(x) = 4992 + 23.30x
(b) The linear revenue function is: R(x) = 27.20x
(c) The linear profit function is P(x) = 3.9x - 4992.
(a) The linear cost function that represents the cost C(x) to produce x items can be calculated by adding the fixed cost to the variable cost per item multiplied by the number of items produced. In this case, the fixed cost is $4992, and the variable cost per item is $23.30. Therefore, the linear cost function is:
C(x) = 4992 + 23.30x
(b) The linear revenue function that represents the revenue R(x) for selling x items can be calculated by multiplying the price at which the item is sold by the number of items sold. In this case, the price at which the item is sold is $27.20. Therefore, the linear revenue function is:
R(x) = 27.20x
(c) The linear profit function P(x) represents the profit obtained from producing and selling x items. Profit is calculated by subtracting the cost (C(x)) from the revenue (R(x)). Therefore, the linear profit function is:
P(x) = R(x) - C(x)
= 27.20x - (4992 + 23.30x)
= 27.20x - 4992 - 23.30x
= 3.9x - 4992
Therefore, the linear profit function is P(x) = 3.9x - 4992.
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maxz=3x 1
−6x 2
+x 3
s.t. x 1
+x 2
+x 3
≥8 2x 1
−x 2
=5 −x 1
+3x 2
+2x 3
≤7 x 1
,x 2
,x 3
≥0 Given the above linear programming model. Find the optimal solution by using Big M method. [NOTE: Please remain the answer in its fractional form if any.]
Using the Big M method, the optimal solution for the given linear programming problem is (x₁, x₂, x₃) = (0, -5/2, 13/2), with an objective value of 21/2.
First, let's rewrite the problem in standard form:
maximize 3x₁ - 9x₂ + x₃
subject to:
x₁ + x₂ + x₃ + s₁ = 8
2x₁ - x₂ + s₂ = 5
-x₁ + 3x₂ + 2x₃ + s₃ = 7
x₁, x₂, x₃, s₁, s₂, s₃ ≥ 0
Where s₁, s₂, and s₃ are slack variables that we introduced to convert the inequality constraints into equality constraints.
Now, we can apply the Big M method by adding penalty terms to the objective function for violating each constraint.
Let's choose M = 1000 as our penalty.
The new objective function is:
maximize 3x₁ - 9x₂ + x₃ -Ms₁ -Ms₂ -Ms₃
The constraints become:
x₁ + x₂ + x₃ + s₁ = 8
2x₁ - x₂ + s₂ = 5
-x₁ + 3x₂ + 2x₃ + s₃ = 7
x₁, x₂, x₃, s₁, s₂, s₃ ≥ 0
Now, we can apply the simplex algorithm to find the optimal solution.
Starting with the initial feasible solution (x₁, x₂, x₃, s₁, s₂, s₃) = (0, 0, 0, 8, 5, 7),
we can use the following table:
Basis x₁ x₂ x₃ s₁ s₂ s₃ RHS
s₁ 1 1 1 1 0 0 8
s₂ 2 -1 0 0 1 0 5
s₃ -1 3 2 0 0 1 7
First, we select x₂ as the entering variable. The leaving variable is s₂, since it has the smallest non-negative ratio (5/(-1)).
Basis x₁ x₂ x₃ s₁ s₂ s₃ RHS
s₁ 1 0 1 1 1/2 0 13/2
x₂ 2 1 0 0 -1/2 0 -5/2
s₃ -1 0 2 0 3/2 1 17/2
z 3 0 1 0 9/2 0 45/2
Next, we select x₃ as the entering variable. The leaving variable is s₁, since it has the smallest non-negative ratio (13/2).
Basis x₁ x₂ x₃ s₁ s₂ s₃ RHS
x₃ 1 0 1 1/2 1/2 0 13/2
x₂ 2 1 0 -1/2 0 0 -5/2
s₃ 0 0 2 3/2 -1/2 1 17/2
z 3 0 0 9/2 -3/2 0 21/2
The optimal solution is (x₁, x₂, x₃) = (0, -5/2, 13/2), with an objective value of 21/2.
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The complete question is attached below;
The Sugar Sweet Company will choose from two companies to transport its sugar to market. The first company charges $3005 to rent trucks plus an additional fee of $100.50 for each ton of sugar. The second company does not charge to rent trucks but charges $250.75 for each ton of sugar. For what amount of sugar do the two companies charge the same? toes What is the covt when the twe cempanies charge the same?
Let's denote the amount of sugar by "x" (in tons).
For the first company, the cost is $3005 to rent trucks plus an additional fee of $100.50 for each ton of sugar. Therefore, the total cost for the first company is:
Total Cost (Company 1) = $3005 + $100.50x
For the second company, there is no charge to rent trucks, but there is a charge of $250.75 for each ton of sugar. Therefore, the total cost for the second company is:
Total Cost (Company 2) = $250.75x
To find the amount of sugar for which the two companies charge the same, we set the two total cost expressions equal to each other and solve for x:
$3005 + $100.50x = $250.75x
To simplify the equation, let's subtract $100.50x from both sides:
$3005 = $250.75x - $100.50x
Combining like terms:
$3005 = $150.25x
Now, let's isolate x by dividing both sides by $150.25:
x = $3005 / $150.25
Evaluating this expression:
x = 20
Therefore, the two companies charge the same for 20 tons of sugar. The cost when the two companies charge the same is $3005.
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Can you please tell me what is wrong with this argument
What is wrong with the following argument? Note the argument supposedly shows that any symmetric and transitive relation R on A must also be reflexive.
Let R be a relation in A × A that is symmetric and transitive. Using symmetry, if (x,y) ∈ R then (y,x) ∈ R. Hence, both (x,y) and (y,x) are in R. Since (x,y) and (y,x) ∈ R, by transitivity, we have (x,x) ∈ R. Therefore, R is reflexive.
The argument that any symmetric and transitive relation R on A must also be reflexive is invalid.
The statement "If (x, y) ∈ R then (y, x) ∈ R" does not make any sense. In a relation R, (x, y) and (y, x) might be distinct pairs, which means that we can't use the fact that (x, y) ∈ R to conclude that (y, x) ∈ R. Therefore, the argument is invalid. A relation R is symmetric if and only if (x, y) ∈ R implies (y, x) ∈ R for all x, y ∈ A.A relation R is transitive if and only if (x, y) ∈ R and (y, z) ∈ R imply (x, z) ∈ R for all x, y, z ∈ A.
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A force of 2 pounds acts in the direction of 42" to the horizontal. The force moves an object along a straight line from the poet (5) to the point (6,12) with distance measured in feet. Find the wor done by the force
The work done by the force is 17.5454 foot-pounds. The force is 2 pounds and acts in the direction of 42 degrees to the horizontal.
The displacement of the object is from the point (5, 0) to the point (6, 12). The work done by the force is calculated using the following formula:
Work = Force * Displacement * Cos(theta)
where:
* Force is the magnitude of the force in pounds
* Displacement is the distance traveled in feet
* Theta is the angle between the force and the displacement
In this case, the displacement is 12 feet, theta is 42 degrees, and Force is 2 pounds. Plugging these values into the formula, we get:
```
Work = 2 * 12 * Cos(42)
= 17.5454 foot-pounds
```
Therefore, the work done by the force is 17.5454 foot-pounds.
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Find A-1 -3 4 27 9 1 -6 0 A = 0 -5 Select one: O a. O C. 6 5 45 10 38 -27 12 -14 -3 O b. 6 -5 45 -12 -14 10 3 -38 -27 -3/13 -7/26 3/52 45/52-19/26 -27/52 3/26 -5/52 5/26 O d. 3/26 5/52 45/52 3/13 5/26 19/26 -7/26 -3/52 -27/52
Hence, the answer is d. 3/26 5/52 45/52 3/13 5/26 19/26 -7/26 -3/52 -27/52
We have a matrix A which is equal to `[[0,-5],[6,5]]`.
To find A-1, we follow these steps:
Step 1: We start by finding the determinant of A. |A| = 0*5 - (-5*6) = 30.Step 2: We then find the adjoint of A. The adjoint of A is the transpose of the cofactor matrix of A which is obtained by changing the sign of alternate elements of the matrix A. So, the adjoint of A is: A^T = [[5, -6],[ -5, 0]].Step 3: We now use the formula `A^-1 = (1/|A|)A^T` to find the inverse of A. Therefore, A-1 = (1/30) * [[5, -6],[ -5, 0]] = [[1/6, -1/6],[ -1/6, 0]].
Therefore, A-1 = [[1/6, -1/6],[ -1/6, 0]]. Hence, the answer is d. 3/26 5/52 45/52 3/13 5/26 19/26 -7/26 -3/52 -27/52
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Differentiate implicitly to find dx
dy
. x 2
−9xy+y 2
−6x+y−6=0 dx
dy
= 2y+9x+1
2x+9y−6
dx
dy
= 2y+9x+1
2x+9y+6
dx
dy
=− 2y+9x+1
2x+9y+6
dx
dy
=− 2y−9x+1
2x−9y−6
dx
dy
= 2y−9x−1
2x−9y−6
The solution to given differential equation is dx/dy = −2y−9x+1/2x−9y−6.
Differentiate implicitly to find dx/dy. x^2−9xy+y^2−6x+y−6=0
The implicit differentiation can be defined as a method of differentiating implicitly by considering y as a function of x. The implicit differentiation is used when it is hard to differentiate y explicitly with respect to x.
Given, x²− 9xy + y² − 6x + y − 6 = 0
Differentiating both sides with respect to y, we get
2x(1.dy/dx) - 9y - 9x(dy/dx) + 2y(1.dy/dx) + 1.dy/dx - 6 + 0= 0
Simplifying the above equation we get,
2x(dy/dx) - 9y - 9x(dy/dx) + 2y(dy/dx) + dy/dx = 6 - y
Now, take dy/dx common and simplify.
2x - 9x + 2y + 1 = dy/dx(-9) + (2y)
dx/dy = 2y-9x+1/2x+9y+6.
dx/dy = 2y+9x+1/2x+9y-6.
dx/dy = −2y+9x+1/2x+9y-6.
dx/dy = −2y−9x+1/2x−9y−6
The above solution explains the process of differentiating implicitly to find dx/dy. The given equation is differentiated with respect to y. The chain rule and the power rule are used to differentiate the equation. After simplifying the equation, we get the value of dx/dy.
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A car leaving a stop sign accelerates constantly from a speed of 0 feet per second to reach a speed of 44 feet per second. The distance of the car from the stop sign, d, in feet, at time, t, in seconds can be found using this equation.
d=1. 1t^2
What is the average speed of the car, in feet per second, between t=2, and t=5?
A. 5. 5
B. 6. 6
C. 7. 7
D. 8. 5
Average speed ≈ 8.6167 feet per second. Rounding off to one decimal place, we get the answer as D. 8.5
To find the average speed of the car between t=2 and t=5, we need to first find the distance traveled by the car during this time interval.
At t=2 seconds, the distance traveled by the car can be calculated using the given equation:
d = 1.1(2)^2 = 4.4 feet
Similarly, at t=5 seconds, the distance traveled by the car can be calculated as:
d = 1.1(5)^2 = 30.25 feet
Therefore, the total distance traveled by the car between t=2 and t=5 is:
d = 30.25 - 4.4 = 25.85 feet
The time taken by the car to travel this distance can be calculated as:
time = 5 - 2 = 3 seconds
Therefore, the average speed of the car between t=2 and t=5 is:
average speed = total distance traveled / time taken
average speed = 25.85 / 3
average speed ≈ 8.6167 feet per second
Rounding off to one decimal place, we get the answer as D. 8.5
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Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us. (a) Suppose n = 32 and p = 0.20. Can we approximate p by a normal distribution? Why? (Use 2 decimal places.) np= ng = Yes - V✔ can What are the values of , and o? (Use 3 decimal places.) Mp Yes be approximated by a normal random variable because both np and ng exceed (b) Suppose n=25 and p = 0.15. Can we safely approximate p by a normal distribution? Why or why not? ✓ be approximated by a normal random variable because np does not exceed No cannot (c) Suppose n = 57 and p = 0.12. Can we approximate p by a normal distribution? Why? (Use 2 decimal places.) np= nq= can ✓be approximated by a normal random variable because both np and no exceed What are the values of μ; and o? (Use 3 decimal places.) Hp So, converted to a z interval, we wish to find P(-0.22
a. The given value can be approximated the binomial distribution by a normal distribution. μ = 6.4 and σ = 5.12.
b. Since np = 3.75 < 10, it cannot approximate the binomial distribution by a normal distribution.
c. The given value can approximate the binomial distribution by a normal distribution. μ = 6.84 and σ = 5.99712. The required probability is 0.2026.
a. n = 32 and p = 0.20. Therefore, np = 6.4 and nq = 25.6. Since np = 6.4 and nq = 25.6 are both greater than 10, it can be approximated the binomial distribution by a normal distribution. Here,
μ = np = 6.4,
σ = (npq)
= (6.4 * 0.8)
= 5.12
Therefore, μ = 6.4 and σ = 5.12.
b. n = 25 and p = 0.15.
Therefore, np = 3.75 and nq = 21.25.
Since np = 3.75 < 10, cannot approximate the binomial distribution by a normal distribution.
c. We are given n = 57 and p = 0.12.
Therefore, np = 6.84 and nq = 50.16.
Since np = 6.84 and nq = 50.16 are both greater than 10, it can approximate the binomial distribution by a normal distribution.
Here, μ = np = 6.84,
σ = (npq)
= (6.84 * 0.88)
= 5.99712
Therefore, μ = 6.84 and σ = 5.99712.
Converted to a z interval, P(-0.22 < Z < 0.29). Using a calculator,
P(-0.22 < Z < 0.29)
= P(Z < 0.29) - P(Z < -0.22)
= 0.6141 - 0.4115
= 0.2026, rounded to 4 decimal places.
Hence, the required probability is 0.2026.
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The officers of a high school senior class are planning to rent buses and vans for a class trip. Each bus can transport 90 students, requires 3 chaperones, and costs $1,000 to rent. Each van can transport 10 students, requires 1 chaperone, and costs $80 to rent. Since there are 720 students in the senior class that may be eligible to go on the trip, the officers must plan to accommodate at least 720 students. Since only 42 parents have volunteered to serve as chaperones, the officers must plan to use at most 42 chaperones. How many vehicles of each type should the officers rent in order to minimize the transportation costs? What are the minimal transportation costs? The officers should rent buses and vans to minimize the transportation costs.
The minimal transportation costs are $8,960 when renting 8 buses and 12 vans.
To minimize the transportation costs, let's assume we rent 'b' buses and 'v' vans.
Each bus can transport 90 students, so the number of buses needed to accommodate at least 720 students is:
b ≥ 720 / 90 = 8
Each van can transport 10 students, so the number of vans needed to accommodate the remaining students is:
v ≥ (720 - 90b) / 10
The number of chaperones required for 'b' buses is:
3b
The number of chaperones required for 'v' vans is:
v
Since the officers must plan to use at most 42 chaperones, we have the inequality:
3b + v ≤ 42
Now we can find the optimal solution by minimizing the transportation costs. The cost of renting 'b' buses is:
Cost of buses = 1000 * b
The cost of renting 'v' vans is:
Cost of vans = 80 * v
Therefore, the total transportation cost is:
Total Cost = Cost of buses + Cost of vans = 1000b + 80v
We want to minimize this total cost, subject to the constraints we derived earlier.
To find the minimal transportation costs and the corresponding number of vehicles, we need to evaluate the total cost function for different values of 'b' and 'v', while satisfying the constraints.
One possible solution is to take the minimum integer values for 'b' and 'v' that satisfy the constraints:
b = 8
v = (720 - 90b) / 10 = (720 - 90 * 8) / 10 = 12
Therefore, the officers should rent 8 buses and 12 vans to minimize the transportation costs.
Substituting these values back into the total cost equation:
Total Cost = 1000 * 8 + 80 * 12 = $8,000 + $960 = $8,960
The lowest possible transportation costs, when renting 8 buses and 12 vans, are $8,960.
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The population of a particular city is increasing at a rate proportional to its size. It follows the function P(t)=1+ke0 out whare k is a constant and t is the time in years 35,000 , in how many years is the population expected to be 87,500 ? Round to the nearect year. A. 10 years B. 13 years C. 145 years D. 23 years
The population of a particular city is increasing at a rate proportional to its size. in (A) 10 years is the population expected to be 87,500
Given, P(t) = 1 + ke^0We are given that the population of a particular city is increasing at a rate proportional to its size.
Let the size of the population be P(t) at any time t years.
Let the rate of increase of population be proportional to its size.
Then, Rate of increase of population = k. P(t).
We have, P(t) = 1 + ke^0 = 1 + k.
Also, it is given that the population of the city is 35,000.
Let's plug this value into the function. P(t) = 35,000 => 1 + k = 35,000 => k = 34,999We need to find out in how many years is the population expected to be 87,500. Let's plug in this value into the equation P(t). We have, P(t) = 1 + ke^0 = 1 + 34,999* e^0. We know that P(t) = 87,500. Therefore,87,500 = 1 + 34,999* e^0=> e^0 = (87,500 - 1)/34,999=> e^0 = 2.5 Thus, the value of t can be found as: t = ln(2.5)/ln(e)≈ 0.92 years≈ 1 year. Therefore, the population is expected to be 87,500 in 1 year. Hence, the correct option is A. 10 years.
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2 Find the area of the surface z = (a³/2 + y³/2), 0 ≤ x ≤ 1,0 ≤ y ≤ 1.
The required area of the given surface is 4.32008 square units.
The given surface is z = (a³/2 + y³/2)
Where, 0 ≤ x ≤ 1,0 ≤ y ≤ 1.
This surface can be represented in the form of z = f(x, y) as follows:
f(x,y) = (a³/2 + y³/2) ⇒ z = f(x,y)
On the given limits, we have:
x ∈ [0, 1]y ∈ [0, 1]
Thus, the required area can be computed as follows:
S = ∫∫√[1+ (∂z/∂x)²+ (∂z/∂y)²] dA
Where, ∂z/∂x and ∂z/∂y can be determined as follows:
∂z/∂x = 0∂z/∂y = (3/2)y²
Using the above values in the formula, we have:
S = ∫∫√(1+(3y²/2)²) dA
On the given limits, this becomes:
S = ∫0¹ ∫0¹ √(1+(3y²/2)²) dy dx
Performing the integration with the given limits, we get:
S = (1/2) [8.64016]
S = 4.32008 square units
Therefore, the required area of the given surface is 4.32008 square units.
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Find the area of the sector if the central angle is 25 degrees
and the radius of the sector is 4m
The approximate area of the sector is 3.534 square meters, rounded to two decimal places.
To find the area of a sector, you need to know the central angle and the radius of the sector. In this case, the central angle is 25 degrees, and the radius is 4 meters. The formula to calculate the area of a sector is: Area = (θ/360) * π * r^2, where θ is the central angle in degrees, r is the radius of the sector, and π is a mathematical constant approximately equal to 3.14159.
Substituting the given values into the formula: Area = (25/360) * π * (4^2)
= (0.0694) * π * 16≈ 3.534 square meters. Therefore, the approximate area of the sector is 3.534 square meters, rounded to two decimal places.
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If the daily marginal cost for the Wait Disney Company to reopen Disney World during the Covid- 19 pandemic was $800,000, then the Walt Disney Company should have reopened Disney World as long as the marginal benefit received each day was just equal to or greater than $800,000. only if the marginal benefit received each day was less than $800,000. as long as the marginal cost each day was lower than it had been before reopening. until the marginal benefit received fell to zero.
The Walt Disney Company should have reopened Disney World during the pandemic if the daily marginal benefit exceeded $800,000, as long as the cost of reopening was covered.
The statement "the Walt Disney Company should have reopened Disney World as long as the marginal benefit received each day was just equal to or greater than $800,000" is incorrect. The correct statement is that the Walt Disney Company should have reopened Disney World as long as the marginal benefit received each day was **greater than** $800,000. This is because reopening would only be financially justified if the daily marginal benefit exceeded the daily marginal cost of $800,000.
The other options provided are incorrect. Reopening should not be based on the marginal benefit being less than $800,000, as that would not cover the daily cost. The decision to reopen is not dependent on comparing the marginal cost before and after reopening, but rather on the marginal benefit and cost at the time of reopening. Lastly, the decision to close would occur when the marginal benefit falls to zero, not when it is greater than or equal to $800,000.
Therefore, The Walt Disney Company should have reopened Disney World during the pandemic if the daily marginal benefit exceeded $800,000, as long as the cost of reopening was covered.
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