True, If the null hypothesis states that there is no difference between the mean net income of retail stores in Chicago and New York City, then the test is two-tailed.
Step-by-step explanation:
1. The null hypothesis (H0) states that there is no difference between the mean net incomes of retail stores in the two cities: H0: μ1 = μ2.
2. The alternative hypothesis (H1) would be that there is a difference between the mean net incomes: H1: μ1 ≠ μ2.
3. Since the alternative hypothesis includes both possibilities of the mean net income in Chicago being either greater or less than that in New York City, it's a two-tailed test.
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Will give brainlyst
Step-by-step explanation:
The volume of the packages is 1.5³=3.38ft³
The volume of the truck is a prism: A×B×C=10.5×8×9=756ft³
The amount of basketballs that can fit is: 756/3.38=223.6 boxes.
Please show your work
Answer:
2=40 3=140 4=40
Step-by-step explanation:
the center line is equal to 180 degrees so if angle 1 is 140, 180-140=40 angle 2=40
angle 2 and 4 and the same by some process (its been awhile since geometry to remember exact terms)
angle 1 and 3 are the same (same reason)
Answer:
<2 is 40°, <3 is 140°, <4 is 40°
Step-by-step explanation:
Angle 1 and angle 3 are equal because of the vertical angles theorem. Angle 2 is supplementary to angle 3(<3 = 140°), so 180 - 140 = 40°. <2 and <4 are equal because of the vertical angles theorem.
Finding the sum of series help: ∑=1[infinity](−9)x.
To find the sum of the series ∑=1[infinity](−9)x, we first need to determine whether the series converges or diverges. We can use the ratio test to do this. Therefore, the sum of the series ∑=1[infinity](−9)x is (-9)x / (1 - x), provided that |x| < 1.
The ratio test states that for a series ∑an, if the limit of |an+1/an| as n approaches infinity is less than 1, then the series converges. If the limit is greater than 1 or does not exist, then the series diverges.
Applying the ratio test to our series, we get:
|(-9)x(n+1) / (-9)x(n)| = |x(n+1) / x(n)|
As x is a constant, this simplifies to:
|x(n+1) / x(n)| = |x|
Since |x| is a constant, the limit of |x(n+1) / x(n)| as n approaches infinity is also |x|. Therefore, if |x| < 1, the series converges, and if |x| ≥ 1, the series diverges.
Assuming that |x| < 1, we can then find the sum of the series using the formula for an infinite geometric series:
S = a / (1 - r)
where a is the first term of the series and r is the common ratio. In this case, a = (-9) x and r = x.
Substituting these values into the formula, we get:
S = (-9)x / (1 - x)
Therefore, the sum of the series ∑=1[infinity](−9)x is (-9)x / (1 - x), provided that |x| < 1.
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Problem 4
Michener Company’s standard labor cost per unit of output is $20 (2 hours * $10 per hour). During August, the company incurs 2,100 hours of direct labor at an hourly cost of $10. 50 per hour in making 1,000 units of finished product. Compute the total, price, and quantity labor variances
The total labor variance = $2,050 favorable, the price variance = $2,100 unfavorable, and the quantity variance = $1,000 favorable.
The total labor variance formula is,
Total labor variance => The actual labor cost - The standard labor cost
To use actual labor cost, first, we need to find its value
Actual labor cost = Actual hours worked x hourly rate
= 2,100hr x $10.50 per/hr
= $22,050
now,
Standard labor cost = Standard hours x Standard hourly rate
= 1,000 units x 2 hours per unit x $10
= $20,000
now,
total labor variance = $22,050 - $20,000 => $2,050(favorable)Price variance = ( Actual hourly rate - Standard hourly rate) x Actual hours worked
Price variance = ($10.50 - $10) x 2,100hr => $2,100(unfavorable)Quantity variance = (Actual hours worked - Standard hours) x Standard hourly rate
Standard hours = 1,000units x 2hr => 2,000hr
Quantity variance = (2,100 - 2,000) x $10 => $1,000(favorable)The total labor variance = $2,050 favorable, the price variance = $2,100 unfavorable, and the quantity variance = $1,000 favorable.
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Sven investigates the amount of damage to the head gaskets on the trucks in his fleet and find that the damage index depends on the ambient temperature. He develops the equation y = −23x + 14 to model the relationship. What does 14 mean?
The y-intercept of 14 may serve as a reference point for the amount of damage to the head gaskets when the temperature is relatively low.
What is an equation?
An equation is a mathematical statement that shows that two expressions are equal. It consists of two sides separated by an equals sign (=).
In the equation y = -23x + 14, y represents the amount of damage to the head gaskets on the trucks in Sven's fleet, and x represents the ambient temperature.
The constant term, 14, is the y-intercept of the equation. It represents the value of y when x is equal to zero.
In this context, the y-intercept of 14 means that when the ambient temperature is zero (which is unlikely in most cases), the amount of damage to the head gaskets on the trucks in Sven's fleet is equal to 14. However, this value may not be meaningful in practical terms since it is unlikely that the ambient temperature will ever be exactly zero.
Therefore, in practical terms, the y-intercept of 14 may serve as a reference point for the amount of damage to the head gaskets when the temperature is relatively low.
It is important to note that the value of y will decrease by 23 for every increase of one unit in the ambient temperature (x).
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It’s says find the area of this answer I have feeling it’s between choices B and D but I am not sure please help if you can!
Answer:
B
Step-by-step explanation:
Most triangle part to the other side.
11*14
Assume H0: μ ≤ 6 and Ha: μ > 6. Is this a left-tailed, right-tailed, or two-tailed test?
A. left-tail
B. right-tail
C. two-tail
D. none of the above
Given the hypotheses H0: μ ≤ 6 and Ha: μ > 6, this is a right-tailed test. So, the correct answer is B. right-tail.
The hypotheses given are about the population mean μ being either less than or equal to 6 (null hypothesis, H0) or greater than 6 (alternative hypothesis, Ha). The alternative hypothesis Ha indicates a one-sided or directional hypothesis because it specifies a particular direction of change (i.e., increase) in the population mean.
In this case, the test is a right-tailed test because the alternative hypothesis indicates that the population mean is greater than the null hypothesis value of 6. A right-tailed test is used when the alternative hypothesis suggests that the population parameter of interest is greater than the null hypothesis value.
Therefore, the answer is B. right-tail.
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Find the rectangular equation for the surface by eliminating the parameters from the vector-valued function. r(u, v) = 5 cos(v) cos(u)i + 5 cos(v) sin(u)j + 3 sin(v)k Identify the surface A. spheroid B. plane C. cylinder D. ellipsoid
The rectangular equation for the surface is (x²)/25 + (y²)/25 = (z²)/9 thus the surface is an ellipsoid (option D).
To find the rectangular equation for the surface, we will eliminate the parameters u and v from the vector-valued function r(u, v) = 5 cos(v) cos(u)i + 5 cos(v) sin(u)j + 3 sin(v)k.
1. Break down the vector-valued function into its components:
x = 5 cos(v) cos(u)
y = 5 cos(v) sin(u)
z = 3 sin(v)
2. Divide the first two equations to eliminate u:
y/x = sin(u)/cos(u)
y/x = tan(u)
u = arctan(y/x)
3. Now, let's square and add the first two equations to eliminate v:
x² + y² = (5 cos(v) cos(u))² + (5 cos(v) sin(u))²
x² + y² = 25(cos²(v))(cos²(u) + sin²(u))
4. Use the trigonometric identity cos²(u) + sin²(u) = 1:
x² + y² = 25 cos²(v)
5. Square the third equation:
z² = 9 sin²(v)
6. Divide the fourth equation by the fifth equation to eliminate v:
(x² + y²)/z² = 25 cos²(v) / 9 sin²(v)
7. Use the trigonometric identity sin²(v) + cos²(v) = 1 to eliminate v:
(x² + y²)/z² = 25(1 - sin²(v))/9 sin²(v)
8. Simplify and rearrange the equation:
(x²)/25 + (y²)/25 = (z²)/9
9. Recognize the equation as that of an ellipsoid:
x²/a² + y²/b² + z²/c² = 1 (where a = 5, b = 5, and c = 3)
Thus, the surface is an ellipsoid (option D).
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A rectangular classroom seats 120 students. When the seats are rearranged with three more seats in each row, the classroom has two fewer rows. Find the original number of seats in each row.
seats per row
By solving the equations, we can determine that each row originally had 12 seats.
What are equations?A mathematical assertion that proves the equality of two mathematical expressions is what an equation in algebra means.
For instance, the expressions 3x + 5 and 14 make up the equation 3x + 5 = 14, with the 'equal' sign separating them.
So, S is the original number of seats:
Total number of rows at first = 120/S
Rows were rearranged = 120/S+3
Then, we got:
120/S - 120/S+3 = 2
Now take LCM and calculate as follows:
120(S+3)−120(S)/S(S+3)=2
S(S+3) multiplied by both sides:
S(S+3)
120(S)+360−120(S)=2S(S+3)
360=2S²+6S
2S2+6S−360=0
Using the factorization approach, solve the quadratic problem shown above:
2S²+6S−360=0
2(S²+3S−180)=0
By 2 divide both sides:
S²+3S−180=0
S²+15S−12S−180=0
S(S+15)−12(S+15)=0
(S−12)(S+15)=0
Get S as:
S = 12, -15
The number of seats is never negative.
Therefore, by solving the equations, we can determine that each row originally had 12 seats.
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For each function below, choose the correct description of its graph.
Answer:
A) Parabola opening up
B) Line with a negative slope
c) horizontal line
Step-by-step explanation:
A) Parabola opening up - because function f(x) contains positive [tex]x^{2}[/tex]
B) Line with a negative slope - because function g(x) has a negative x
c) horizontal line - because function k(x) is a constant with no variables
a. Create your "Grade" column A~F b. Create your "P(Grade)" column corresponding to the intended grade distribution referenced in the question c. In a cell to the right, create the "P(Grade)" percentiles. Think back to the definition of a percentile: a score below which a given percentage of scores in its frequency distribution fall. Develop these through the formula: Percentile A = 1 - P(A) = 0.9, Percentile B = 1 – [P(A)+P(B)] = 0.6, ..., Percentile F = 1- [P(A)+P(B) +P(C) +P(D) +P(F)] = 0. Think of these as the lowest percentiles that will still qualify as the intended grade. If only ten percent of students will receive an A, anyone with a score between the 100th and 90th percentile will earn an A. Likewise, the lowest five percent of scores will earn F's. This means that scores falling between the 5th and oth percentiles will earn an F. d. Create your cutoffs by the formula "=NORMINV(PercentileA:PercentileF, 70, 10) i. Remember that the scores are normally distributed with a mean of 70 and a standard deviation of 10. e. Once you determine your cutoff scores (the lowest score that will earn a given letter grade), communicate your grade distribution in a text box.
a. Grade column: A, B, C, D, F
b. P(Grade) column: 0.1, 0.2, 0.3, 0.2, 0.2
c. P(Grade) percentiles: Percentile A = 1 - P(A) = 0.9, Percentile B = 1 – [P(A)+P(B)] = 0.6, Percentile C = 1 - [P(A)+P(B)+P(C)] = 0.3, Percentile D = 1 - [P(A)+P(B)+P(C)+P(D)] = 0.1, Percentile F = 0.
d. Cutoffs: Using the NORMINV function in Excel, we can calculate the score cutoffs for each grade level. The formula is: "=NORMINV(percentile, 70, 10)", where percentile is the corresponding percentile for the grade level.
Cutoff for Grade A: =NORMINV(0.9, 70, 10) = 85.11
Cutoff for Grade B: =NORMINV(0.6, 70, 10) = 77.69
Cutoff for Grade C: =NORMINV(0.3, 70, 10) = 70.88
Cutoff for Grade D: =NORMINV(0.1, 70, 10) = 64.26
Cutoff for Grade F: Anything below 64.26
e. Grade distribution: We can communicate the grade distribution in a text box using the cutoff scores we calculated. For example:
A: Scores between 85.11 and 100
B: Scores between 77.69 and 85.11
C: Scores between 70.88 and 77.69
D: Scores between 64.26 and 70.88
F: Scores below 64.26
Note that these are approximate cutoffs and may need to be adjusted based on the specific distribution of scores in the class. Additionally, the instructor may choose to round up or down in borderline cases.
Overall, creating a grade distribution involves determining the cutoff scores for each percentile, assigning grades to score ranges, and communicating the distribution to stakeholders. It's a useful tool for assessing student performance and providing feedback on their progress.
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Determine the open t-intervals on which the curve is concave downward or concave upward. x=5+3t2, y=3t2 + t3 Concave upward: Ot>o Ot<0 O all reals O none of these Concave downward: otso Otso O all reals O none of these
The curve is concave upward on the open t-interval (-1, ∞) and concave downward on the open t-interval (-∞, -1).
To determine the open t-intervals on which the curve is concave downward or concave upward, we need to find the second derivative of y with respect to t and analyze its sign.
First, let's find the second derivative of y:
y'' = 6t
To determine the concavity of the curve, we need to analyze the sign of y''.
- If y'' > 0, then the curve is concave upward.
- If y'' < 0, then the curve is concave downward.
So,
- The curve is concave upward for all values of t since y'' = 6t > 0 for all real values of t.
- The curve is concave downward when t < 0 since y'' = 6t < 0 for t < 0.
Therefore, the open t-interval on which the curve is concave downward is (negative infinity, 0), and the open t-intervals on which the curve is concave upward are (0, infinity).
To determine the open t-intervals on which the curve is concave upward or downward, we need to find the second derivative of y with respect to t.
Given x = 5 + 3t^2 and y = 3t^2 + t^3, we first find dy/dt.
y'(t) = d(3t^2 + t^3)/dt = 6t + 3t^2
Now, let's find the second derivative, y''(t):
y''(t) = d(6t + 3t^2)/dt = 6 + 6t
Now, we'll find the intervals for concavity:
Concave upward: y''(t) > 0
6 + 6t > 0
t > -1
Concave downward: y''(t) < 0
6 + 6t < 0
t < -1
So, the curve is concave upward on the open t-interval (-1, ∞) and concave downward on the open t-interval (-∞, -1).
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Prove that if A and B are idempotent and AB = BA then AB is idempotent.
It has been proved that if A and B are idempotent and AB = BA then AB is idempotent.
To prove that if A and B are idempotent and AB = BA, then AB is idempotent, follow these steps:
1. Given that A and B are idempotent matrices, we have:
A² = A
B² = B
2. Also given that AB = BA (commutative property).
3. Now we need to prove that (AB)² = AB to show that AB is idempotent.
4. Calculate (AB)²:
(AB)² = (AB)(AB)
5. Using the commutative property (AB = BA), rewrite the expression:
(AB)² = (AB)(BA)
6. Now use the associative property to rearrange the expression:
(AB)² = A(BA)B
7. Substitute BA with AB (since AB = BA):
(AB)² = A(AB)B
8. Now substitute A² and B² with A and B, respectively (since A² = A and B² = B):
(AB)² = A(AB)B
= (A²)(AB)(B²)
= A(AB)B
9. Thus, we have proven that (AB)² = AB, which means that AB is idempotent.
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if a letter is selected at random from the word Mississippi find the probability that it is an i
Calculate the partial derivatives ∂U∂T and ∂T∂U using implicit differentiation of (TU−V)2ln(W−UV)=ln13 at (T,U,V,W)=(4,3,13,52).
To calculate the partial derivatives ∂U∂T and ∂T∂U using implicit differentiation of (TU−V)2ln(W−UV)=ln13 at (T,U,V,W)=(4,3,13,52), we can follow these steps:
1. Take the natural logarithm of both sides of the equation:
ln[(TU-V)2ln(W-UV)] = ln13
2. Use the chain rule to differentiate both sides with respect to T:
(2ln(TU-V) + (TU-V) * 2/(W-UV) * (-U)) * (U/T) = 0
3. Simplify the expression and solve for ∂U/∂T:
∂U/∂T = -2Uln(TU-V)/(TU-V) * (W-UV)/2U(TU-V) = -(W-UV)ln(12)/9
4. Use the chain rule to differentiate both sides with respect to U:
(2ln(TU-V) + (TU-V) * 2/(W-UV) * T) * (1/U) + (TU-V) * ln(W-UV) * (-1/U2) = 0
5. Simplify the expression and solve for ∂T/∂U:
∂T/∂U = -2Tln(TU-V)/(TU-V) * (W-UV)/2U(TU-V) - ln(W-UV)/3U = -(W-UV)ln(12)/27U - ln(W-UV)/9
Therefore, at (T,U,V,W)=(4,3,13,52), we have:
∂U/∂T = -(52-4*3)ln(12)/9 = -8ln(12)/3
∂T/∂U = -(52-4*3)ln(12)/27*3 - ln(52-4*3)/9 = -8ln(12)/27 - ln(40)/9
To find the partial derivatives ∂U/∂T and ∂T/∂U, we'll first implicitly differentiate the given equation with respect to T and U, and then evaluate the derivatives at the given point (T, U, V, W) = (4, 3, 13, 52).
Given equation: (TU - V)^2 * ln(W - UV) = ln(13)
1. Differentiate with respect to T:
Using product and chain rules, we get:
2(TU - V)(U + T * ∂U/∂T) * ln(W - UV) + (TU - V)^2 * (1/(W - UV)) * (-U * ∂U/∂T) = 0
2. Differentiate with respect to U:
Using product and chain rules, we get:
2(TU - V)(T + T * ∂T/∂U) * ln(W - UV) + (TU - V)^2 * (1/(W - UV)) * (-T * ∂T/∂U - V) = 0
Now, evaluate the derivatives at (T, U, V, W) = (4, 3, 13, 52):
1. For ∂U/∂T:
2(4*3 - 13)(3 + 4 * ∂U/∂T) * ln(52 - 4*3*13) + (4*3 - 13)^2 * (1/(-40)) * (-3 * ∂U/∂T) = 0
2. For ∂T/∂U:
2(4*3 - 13)(4 + 4 * ∂T/∂U) * ln(52 - 4*3*13) + (4*3 - 13)^2 * (1/(-40)) * (-4 * ∂T/∂U - 13) = 0
Now, you can solve these equations to find the values of the partial derivatives ∂U/∂T and ∂T/∂U.
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Julio is doing an experiment in which he test how high, in centimeters, different balls bounce when dropped from a hight of 1 meter. 85,83,57,89,62,68,91,22,28,58,37,41,36,59,88,54,76,90,
Answer the question which needed to be considered when choosing the intervals for a histogram to repeat the data. What is most reasonable range of the intervals
The most reasonable range of intervals will depend on the specific characteristics of your data set and the level of granularity you want in your histogram.
The range of data values and the number of data points should be considered when choosing intervals for a histogram to represent your data.
In this particular dataset, the data values range from 22 to 91. You can divide this range into equal intervals. 10, 5, or 2.5 centimeters, depending on the granularity you want in your histogram.
The number of information focuses is generally little at 18 focuses. A great run show of thumb for choosing the number of interims is to require the square root of the number of information focuses.
In this case, it is approximately 4.24. This can be rounded to 5 or 6 intervals depending on the range selected.
For illustration, in case you select an interim of 10 centimeters, you'll be able to make the taking after the histogram.
20 30 40 50 60 70 80 90
1 shot 1 shot
In this histogram, the first interval contains the values 22 and 28, so there is a ball in that interval. The second interval contains the values 36, 37, 41, and 54, so there are 4 balls in this interval.
The third interval contains the values 57, 58, and 59, so there are 3 balls in this interval. The fourth interval contains the values 62, 68, and 76, so there are 3 balls in this interval.
The fifth interval contains the values 83, 85, 88, and 89, so there are 4 balls in this interval. The sixth interval contains the values 90 and 91, so there are two balls in this interval.
Overall, the most reasonable range of intervals will depend on the specific characteristics of your data set and the level of granularity you want in your histogram.
In general, it is important to choose equally sized intervals to capture a range of data values.
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help me get the answers
The coordinate of point D after rotating the line 90 degrees is (-2, 4).
What is the coordinate of point D?When a line segment is rotated 90 degrees clockwise, its endpoints will be rotated 90 degrees in a clockwise direction around the center of rotation, which is typically the origin (0,0) on a standard coordinate plane.
This means that the x-coordinate of each endpoint will become the negative of its original y-coordinate, and the y-coordinate of each endpoint will become the positive of its original x-coordinate.
In other words, if the line segment has endpoints (x₁, y₁) and (x₂, y₂), after a 90 degree clockwise rotation, the new endpoints will be (-y₁, x₁) and (-y₂, x₂).
The resulting line segment will be perpendicular to the original line segment, with the same length and in the opposite direction.
The coordinate of point D after the rotating the line segment AB is calculated as follows;
For endpoint (x₁, y₁):
Swap x₁ and y₁ to get (3, 0)Negate the new x-coordinate to get (-3, 0)For endpoint (x₂, y₂):
Swap x₂ and y₂ to get (2, 4)Negate the new x-coordinate to get (-2, 4)Therefore, the new coordinates of the endpoints after the 90 degree clockwise rotation are (-3, 0) and (-2, 4).
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I roll a fair number cube. What is the probability that it lands on 3?
The probability that the dice rolled lands 3 is 1/6 respectively.
What is probability?Probability is simply the possibility that something will happen.
When we don't know how something will turn out, we can talk about the possibility of one outcome or the likelihood of several.
The study of events that fit into a probability distribution is known as statistics.
So, find the probability as follows:
Probability formula: P(E) = Favourable events/Total events
Favorable events = 1 which is 3
Total events = 6 which are (1, 2, 3, 4, 5, 6)
Now, insert values as follows:
P(E) = Favourable events/Total events
P(E) = 1/6
Therefore, the probability that the dice rolled lands 3 is 1/6 respectively.
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caculate the account balance if you invest 15000 for 15 years at 1.5% interest compounded quarterly
The account balance after 15 years with an initial investment of $15,000, compounded quarterly at an interest rate of 1.5%, is $18,899.82.
Formula for compound interestTo calculate the account balance after 15 years with an initial investment of $15,000, compounded quarterly at an interest rate of 1.5%, we can use the formula for compound interest:
A = P(1 + r/n)ⁿᵇ
where:
A = final amount (account balance)
P = principal amount (initial investment)
r = annual interest rate (as a decimal)
n = number of times the interest is compounded per year
b = time (in years)
Plugging in the given values, we get:
A = 15000(1 + 0.015/4)⁴×¹⁵
A = 15000(1 + 0.00375)⁶⁰
A = 15000(1.00375)⁶⁰
A = 15000(1.260655)
A = $18,899.82
Therefore, the account balance after 15 years with an initial investment of $15,000, compounded quarterly at an interest rate of 1.5%, is approximately $18,899.82.
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Triangle ABC is similar to triangle DEF. What is AC?
Answer:
AC = 12
Step-by-step explanation:
similar = same shape but different measures, the sides are in proportion.
18 : x = 12 : 8
x = 18 × 8 : 12
x = 144 : 12
x = 12
What are the three main phases of audit sampling? Are these phases the same for statistical and nonstatistical samplingmethods?
The three main phases for statistical sampling methods are:(List the three steps in the proper order.)
1.
2.
3.
The three main phases of audit sampling are planning, performing, and evaluating. These phases are the same for both statistical and nonstatistical sampling methods.
The three steps in the proper order for statistical sampling methods are:
1. Planning: This involves determining the objectives of the audit, selecting the sample size, and choosing the sampling method.
2. Performing: This involves actually selecting and examining the items in the sample.
3. Evaluating: This involves evaluating the results obtained from the sample and drawing conclusions about the population being audited based on those results.
The three main phases of audit sampling are:
1. Planning: This phase involves determining the audit objectives, defining the population and sampling unit, and selecting the sampling method, whether statistical or nonstatistical.
2. Sample Selection: In this phase, auditors select the sample items from the population using the chosen sampling method.
3. Evaluation: The final phase involves analyzing the sample results, projecting them to the entire population, and forming conclusions based on these projections.
These phases are the same for both statistical and nonstatistical sampling methods. However, the techniques used for sample selection and evaluation may differ between the two methods.
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Fiona draws a circle with a diameter of 14 meters. what is the area of fiona's circle? 7 14 28 49
The required area of Fiona's circle is49π. Option D is correct.
What is a circle?The circle is the locus of a point whose distance from a fixed point is constant i.e center (h, k). The equation of the circle is given by
(x - h)² + (y - k)² = r²
Where h, k is the coordinate of the center of the circle on a coordinate plane and r is the radius of the circle.
The radius of Fiona's circle is half of the diameter, which is 14/2 = 7 meters.
The formula for the area of a circle is A = πr², where A is the area and r is the radius.
Substituting the value of the radius in the formula, we get:
A = π(7)² = 49π
So, the area of Fiona's circle is 154 square meters, if we use the value of π as 3.14.
Therefore, the answer is 49.
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what is the probability that in a random sample of 10 students, more than 4 students have midterm test mark that is less than 80
Summing the probabilities for k=5 to k=10 will give you the probability that more than 4 students in a random sample of 10 have a midterm test mark less than 80.
To calculate the probability that in a random sample of 10 students, more than 4 students have a midterm test mark less than 80, we'll use the following terms: random sample, probability, and binomial distribution.
Random sample: We're selecting a random sample of 10 students from a larger population. This means that each student has an equal chance of being selected for the sample.
Probability: We want to find the probability that more than 4 students (i.e., 5, 6, 7, 8, 9, or 10 students) have a midterm test mark less than 80.
Binomial distribution: To solve this problem, we'll use the binomial distribution formula, which calculates the probability of a specific number of successes (students with marks less than 80) in a fixed number of trials (the 10 students in the sample).
First, we need to know the probability of success (p) - the probability that a single student has a midterm test mark less than 80. Let's assume this probability is given as p. The probability of failure (q) is then 1-p, as it represents the probability that a student has a midterm test mark equal to or greater than 80.
Now, we can apply the binomial distribution formula to find the probability of more than 4 students having a midterm test mark less than 80:
P(X > 4) = P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10)
To calculate each probability term, we use the formula:
P(X=k) = C(n, k) * p^k * q^(n-k)
Where C(n, k) is the number of combinations of n items taken k at a time, n is the total number of trials (10 students), k is the number of successes (students with marks less than 80), p is the probability of success, and q is the probability of failure.
Summing the probabilities for k=5 to k=10 will give you the probability that more than 4 students in a random sample of 10 have a midterm test mark less than 80.
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approximate the area under the function f(x)=4/x on the interval [1,5] using 8 right-sided rectangles. give your answer as a fraction
The estimated area under the curve of f(x) = 4 on the interval (1,5) using 8 right-sided rectangles has the fraction form 16/1.
The area under the function f(x) = 4 on the interval (1,5) can be approximated using 8 right-sided rectangles.
To do this, first divide the interval (1,5) into 8 equal subintervals. Each subinterval will have a width of 0.5. The x-coordinates of the 8 subintervals will be 1, 1.5, 2, 2.5, 3, 3.5, 4, and 4.5.
Now calculate the height of each of the 8 rectangles. Given that f(x) = 4 for all x in the range (1,5), each rectangle will have a height of 4.
Next, multiply the width of each rectangle (0.5) with its height (4). We will then know how big each of the 8 rectangles is. Hence, each rectangle's area will be 2.
Finally, add up the area of the 8 rectangles to get the total area under the curve. So, the approximate area under the curve f(x) = 4 on the interval (1,5) using 8 right-sided rectangles is 16.
Therefore, the fractional form of the approximate area under the curve f(x) = 4 on the interval (1,5) using 8 right-sided rectangles is 16/1.
Complete Question:
Approximate the area under the function f(x) = 4 on the interval (1,5) using 8 right-sided rectangles. Give your answer as a fraction. Provide your answer below:
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m/5+9=11 please help
Answer:
m/5+9=11m/5=11-9m/5=2m=2×5m=10Step-by-step explanation:
Answer:m=10
Step-by-step explanation:some number (m) divided by 5 plus 9 = 11
10 divided by 5 = 2
2+9 = 11
Imagine you are a developer for a large construction firm, and your company will receive a huge sum
of money to build the school if they choose to do so. You are about to give a presentation before the School Board. Do you present them with predictions from a linear, logarithmic, or exponential regression model?
Explain and justify your choice.
Include information that supports your stance from your trends and /or real-world factors.
Therefore, based on the trends in construction costs and the real-world factors that affect construction projects, I believe that an exponential regression model would be the most appropriate and accurate choice for predicting the cost of building a school.
As a developer for a large construction firm, I would present the School Board with predictions from an exponential regression model.
Exponential regression models are appropriate when data points are increasing or decreasing at an accelerating rate. In the context of building a school, this means that the cost of construction may increase at an accelerating rate due to inflation, increased demand for construction materials, and other factors.
Additionally, exponential regression models are often used in financial forecasting, as they can account for compounding growth or interest rates. This is relevant in the context of school construction because the cost of construction may increase significantly over time if the project is delayed or if there are unforeseen issues during construction.
Furthermore, exponential regression models have been shown to be effective in predicting costs for construction projects. A study by Pande and Bavikar (2014) found that an exponential regression model was more accurate than other models in predicting construction costs for residential buildings.
Therefore, based on the trends in construction costs and the real-world factors that affect construction projects, I believe that an exponential regression model would be the most appropriate and accurate choice for predicting the cost of building a school.
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this exercise refers to a standard deck of playing cards. assume that 5 cards are randomly chosen from the deck. how many hands contain exactly 3 kings?
To have exactly 3 kings in a hand, we need to choose 3 kings out of 4 and 2 non-kings out of 48. This can be done in 4512 ways.
To calculate the number of hands containing exactly 3 kings, we need to use the concept of combinations. We know that there are 4 kings in a deck of 52 cards. To choose 3 kings out of 4, we can use the combination formula, also known as "n choose k," which is written as nCk.
So, the number of ways to choose 3 kings out of 4 is 4C3 = 4.
Next, we need to choose 2 non-kings out of the remaining 48 cards. This can be done using the combination formula again. The number of ways to choose 2 non-kings out of 48 is 48C2 = 1128.
Now, we can use the multiplication principle to find the total number of hands containing exactly 3 kings. We multiply the number of ways to choose 3 kings by the number of ways to choose 2 non-kings:
Total number of hands = 4C3 * 48C2 = 4 * 1128 = 4512.
Therefore, there are 4512 hands that contain exactly 3 kings when 5 cards are randomly chosen from a standard deck.
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1. What is the standard form of eight hundred two
thousand, eight hundred three and 18 thousandths?
A. 802,803.080
B. 802,803.018
C. 802,803.18
D. 802,803.80
The standard form of the number can be written as follows: 802,803.018
Writing the the standard form of the numberThe standard form of a number is a way of writing it using digits and place value.
In the given number "eight hundred two thousand, eight hundred three and 18 thousandths", the digits 8, 0, 2, 8, 0, 3 represent the whole number part, and 1, 8 represent the decimal part.
The place value positions in a whole number are ones, tens, hundreds, thousands, ten thousands, hundred thousands, millions, and so on. The place value positions in a decimal number are tenths, hundredths, thousandths, ten-thousandths, and so on.
So, the standard form of the given number can be written as follows: 802,803.018
Therefore, the correct answer is B. 802,803.018.
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Prove the following statement by mathematical induction. 1 For every integer n > 1,- 1 2.3 1 - 3.4 + ... +- 1.2 n(n + 1) n + 1 Proof (by mathematical induction): Let P(n) be the equation 1 1.2 1 2.3 - + 1 - 3.4 + ... + 1 n(n + 1) n n +1 We will show that P(n) is true for every integer n 2 1. Show that P(1) is true: Select P(1) from the choices below. • 172 171 .2 1+1 1 + + 1.2 1(1 + 1) 1 + 1 • P(1) - 1+1 + + 1 1+1 1 1.2 1 1.2 1 - 2.3 1 1.2 1 3.4 1 1.2 1 1.2 1 1 + 1 - + The selected statement is true because both sides of the equation equal the same quantity. Show that for each integer k > 1, if P(k) is true, then P(k + 1) is true: Let k be any integer with k > 1, and suppose that P(k) is true. We identify the expression on the left-hand side of P(k) by selecting from the choices below. + • 2 wck+1) + + 2 3 3.4 kk + 1) w 1:2 1 1.2 1 1 2.3 + 1 3.4 + ... + - kk + 1) The right-hand side of P(k) is [The inductive hypothesis states that the two sides of P(k) are equal.] 1 We must show that Pk + 1) is true.
To prove the statement using mathematical induction, we start by assuming that the equation P(n) is true for every integer n > 1, where P(n) is defined as:
1 - 1.2 + 2.3 - 3.4 + ... + (-1)^(n-1) * (n-1).n + 1 / n(n + 1)
We need to prove that P(1/2) = -1/2 is also true.
First, we show that P(1) is true by substituting n = 1 into the equation:
P(1) = 1 / (1*(1+1)) = 1/2
This matches the left-hand side of the equation, so P(1) is true.
Next, we assume that P(k) is true for some integer k > 1. We need to show that P(k+1) is also true.
To do this, we first simplify the left-hand side of P(k+1) using the definition of P(n):
1 - 1.2 + 2.3 - 3.4 + ... + (-1)^(k-1) * (k-1).k + 1 / k(k + 1) + (-1)^k * k.(k+1) + 1 / (k+1)(k+2)
= P(k) + (-1)^k * k.(k+1) + 1 / (k+1)(k+2)
= P(k) - (k+1).(k+2) / (k+1)(k+2) + k.(k+1) / (k+1)(k+2)
= P(k) - 1 / (k+1)
The last step follows from the fact that (k+1).(k+2) - k.(k+1) = k+1.
Since we assumed that P(k) is true, we can substitute P(k) with its value from the equation:
P(k+1) = P(k) - 1 / (k+1)
= 1 - 1.2 + 2.3 - 3.4 + ... + (-1)^(k-1) * (k-1).k + 1 / k(k + 1) - 1 / (k+1)
= (k+1).(1 - 1/(k+1)) / k(k+1) + (-1)^(k-1) * (k-1).k + 1 / k(k + 1)
= (-1)^(k-1) * (k-1).k + 1 / k(k + 1)
This matches the right-hand side of the equation for P(k+1), so P(k+1) is true.
Therefore, by mathematical induction, we have proven that the statement is true for every integer n > 1.
To prove the statement by mathematical induction, we will follow these steps:
1. Base Case: Show that P(1) is true
2. Inductive Step: Assume P(k) is true for some integer k > 1, and show that P(k+1) is also true.
Let P(n) be the equation: 1 - 1(1+1) + 1(2)(3) - 1(3)(4) + ... + (-1)^n 1(2n)(n+1) = n/(n+1)
Base Case (n=1):
P(1) = 1 - 1(1+1) = 1 - 2 = -1
The right-hand side of the equation is: 1/(1+1) = 1/2
Since -1 ≠ 1/2, the statement is false for n=1, and induction cannot be used to prove the given statement.
However, if the question intended to prove the statement for n > 2, the base case would be n=2 and we could proceed with the induction steps. But as the question is stated, induction cannot be used for this specific case.
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Find the relative rate of change of f(x)=100x−0.4x2 The relative rate of change of f(x) is ___
The relative rate of change of f(x) is [100(1 - 0.008x)] / [x(250 - x)].
To find the relative rate of change of the function f(x) = 100x - 0.4x^2, we need to take the derivative of the function and then divide it by the function itself.
First, let's find the derivative of f(x):
f'(x) = 100 - 0.8x
Next, we can find the relative rate of change of f(x) by dividing f'(x) by f(x):
[f'(x) / f(x)] = [100 - 0.8x] / [100x - 0.4x^2]
Simplifying this expression, we get:
[f'(x) / f(x)] = [100(1 - 0.008x)] / [x(250 - x)]
Therefore, the relative rate of change of f(x) is [100(1 - 0.008x)] / [x(250 - x)].
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