For a certain company, the cost function for producing x items is C(x)=40x+150 and the revenue function for selling x items is r(x)=-0.5(x-120)^2+7200. The maximum capacity of the company is 170 items.

Answer:

[tex]\pi[/tex]([tex]r^{2}[/tex]h)

Step-by-step explanation:

r = radius

h = height

Using the z-distribution, it is found that since the p-value is greater than 0.05, there is not enough evidence to conclude that there has been a change since 1965 in the proportion of U.S. adults that have never smoked cigarettes.

At the null hypothesis, it is tested if the proportion is still of 44%, that is:

[tex]H_0: p = 0.44[/tex]

At the alternative hypothesis, it is tested if the proportion is now different of 44%, that is:

[tex]H_1: p \neq 0.44[/tex]

The test statistic is given by:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

In which:

For this problem, the parameters are:

[tex]p = 0.44, n = 1360, \overline{p} = \frac{626}{1360} = 0.4603[/tex]

Hence the test statistic is:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

[tex]z = \frac{0.4603 - 0.44}{\sqrt{\frac{0.44(0.56)}{1360}}}[/tex]

z = 1.51

Using a z-distribution calculator, for a two-tailed test, as we are testing if the proportion is different of a value, with z = 1.51, the p-value is of 0.1310.

Since the p-value is greater than 0.05, there is not enough evidence to conclude that there has been a change since 1965 in the proportion of U.S. adults that have never smoked cigarettes.

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The functions, q(x) and r(x) are defined as 2•x + 1, and -5•x - 3, respectively, therefore;

[tex] \: (q \: \circ \: r) (1)= - 15[/tex]

[tex] \: (r \: \circ \: q) (1)= -18[/tex]

The given functions are;

q(x) = 2•x + 1

r(x) = -5•x - 3

The evaluation of the composite functions can be presented as follows;

[tex](q \: \circ \: r) (x)= q(r(x))[/tex]

Therefore;

[tex](q \: \circ \: r) (1)= q(r(1))[/tex]

Which gives;

[tex](q \: \circ \: r) (1)= q( - 8)[/tex]

q(-8) = 2×(-8) + 1 = -15

Therefore;

[tex](q \: \circ \: r) (1)= - 15[/tex]

Similarly, we have;

[tex](r \: \circ \: q) (1)= r(q(1))[/tex]

q(1) = 2×1 + 1 = 3

r(3) = -5×3 - 3 = -18

Which gives;

[tex](r \: \circ \: q) (1)= -18[/tex]

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Answer:

= 5.375

Step-by-step explanation:

Use the algorithm method.

7  9 9 10

8 . 0 0 0

- 2 . 6 2 5

5 . 3 7 5

= 5.375

Answer:

5 3/8

Step-by-step explanation:

change it so that every number is a improper fraction and same denominator

64/8 - 21/8 = 43/8 = 5 3/8

Answer:

gradient = [tex]\frac{5}{3}[/tex]

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope/gradient and c the y- intercept )

given

3y = 5x - 6 ( divide through by 3 )

y = [tex]\frac{5}{3}[/tex] x - 2 ← in slope- intercept form

with gradient = [tex]\frac{5}{3}[/tex]

Answer:

13/3 or 4.333

Step-by-step explanation:

Let the number be x

the sum of x and 7 is 6 less than 4x, giving the equation: x+7+6 = 4x

Solve:

x+7+6 = 4x

x+13 = 4x

13 = 3x

x = 13/3

The times interest earned ratio for 20Y8 and 20Y9 are 11 and 12.6 respectively.

The change in the times interest earned ratio is favorable.

According to the question,

In 20Y8 and 20Y9, Averill Products Inc. disclosed the following on its income statement:

For 20Y9,

Interest expense= $440,000

Income before income tax expense= $5,544,000

Times interest earned ratio= 5544000/440000 = 12.6

For 20Y8,

Interest expense=$400,000

Income before income tax expense=$4,400,000

Times interest earned ratio= 4400000/400000 =11

As the times interest earned ratio for 20Y9 is greater than that for 20Y8,the change in the times interest earned ratio is favorable.

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Answer:

1) log7 (5*8)= log7(40)

2) log2(11) -log2(9)=log2(11/9)

3)2log9 2=log9(2^2)

The area of the shaded region is 8.1838. The area of the shaded region is calculated by subtracting the area of the triangle from the area of the sector of the circle.

The area of the sector of a circle with a radius 'r' and an angle of sector 'θ' is

A = (θ/360) πr² sq. units

The area of the triangle with measures of two sides and an angle between them is

A = 1/2 × a × b × sinC sq. units

Where a and b are the lengths of sides and ∠C is the angle between those sides.

It is given that,

The area of the sector shown in the diagram is 78.6794 cm² and the area of the triangle is 70.4956 cm².

Then to calculate the area of the shaded region, subtract the area of the sector and the area of the triangle. I.e.,

Area of the shaded region = Area of the sector - Area of the triangle

⇒ 78.6794 - 70.4956

⇒ 8.1838 cm²

Therefore, the required area of the shaded region is 8.1838 sq. cm.

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Answer:

Step-by-step explanation:

cosA×cos 2A-sin4A×sinA

=cosAcos2A-2sin2Acos2A sin A

=cos 2A(cosA-2sin2AsinA)

=cos 2A(cosA-2×2sinAcosAsin A)

=cos2A×cosA(1-4sin²A)

=cos 2AcosA(1-4(1-cos²A))

=cos2A×cosA(1-4+4cos²A)

=cos 2A(-3cosA+4cos³A)

=cos 2A(4cos³A-3cosA)

=cos 2A×cos3A

Using the addition rule of the Sine function and the Cosine function, we obtain [tex]\cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}[/tex].

Given that

[tex]\sin x + \sin y = a\\\cos x + \cos y = b[/tex]

Then using the above formulas, we get:

[tex]2\sin\frac{x+y}{2}\cos\frac{x-y}{2}=a[/tex]       (1)

[tex]2\cos\frac{x+y}{2}\cos\frac{x-y}{2}=b[/tex]       (2)

Dividing the equation (1) by (2), we get:

[tex]\dfrac{\sin\dfrac{x+y}{2}}{\cos\frac{x-y}{2}}=\dfrac{a}{b}\\\Longrightarrow \tan\dfrac{x+y}{2}=\dfrac{a}{b}[/tex]             (3)

Now, we know that  [tex]\cos\theta=\dfrac{1}{\sqrt{1+\tan^2\theta}}[/tex].

Thus, using the above formula, we get from (3):

[tex]\cos\dfrac{x+y}{2}=\dfrac{1}{\sqrt{1+\tan^2\dfrac{x+y}{2}}}\\\Longrightarrow \cos\dfrac{x+y}{2}=\dfrac{1}{\sqrt{1+\dfrac{a^2}{b^2}}}\\\Longrightarrow \cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}[/tex]

Therefore, using the addition rule of the Sine function and the Cosine function, we obtain [tex]\cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}[/tex].

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Answer:

c = -75

Step-by-step explanation:

cd + (-6c) = 150

d = 4

c(4) + (-6c) = 150

4c - 6c = 150

-2c = 150

c = -75

Answer:

  (f•g)(x) = 4x² -7x -11

Step-by-step explanation:

The product of the two functions is the product of their respective definitions.

(f•g)(x) = f(x)•g(x) = (x+1)•(4x -11)

  = x(4x -11) +1(4x -11) . . . . . use the distributive property

  = 4x² -11x +4x -11 . . . . . . . and again

(f•g)(x) = 4x² -7x -11 . . . . . collect terms

Answer:

8√2 cm²

Refer to the attached page,I've shown the complete calculation over there

Answer:

no

Step-by-step explanation:

no it is not a right triangle

two angles share the same x but no angles share the same y

you can see this clearly when graphed

the answer is no

the solutions are:

The only information of h(x) that we have is the given graph.

To find the value of x, we need to go to the vertical axis and find y = 4, then we move horizontally to the left until we meet the curve.

That intersection will give the value of x for which h(x) = 4.

Doing that, we conclude that h(x) = 4 when x = 0 (on the vertical axis).

Now we want to find h(-2), and we can do that using the graph.

By finding x = -2 on the horizontal axis and then moving up until we intercept the graph, we can see that:

h(-2) = 2

Concluding, the solutions are:

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Answer:

15. 4400 yards

16. 47.4 seconds

Step-by-step explanation:

15. 1 mile = 1760 yards

2.5 miles:

2.5(1760) = 4400 yards

16. t = 2.5/190 = 0.013150 h

1 hour = (60)(60) = 3600 seconds

Then 1 lap take

=(0.013150)(3600) = 47.4 seconds

Hope this helps

Using proportions, the estimate of the population percentage of seniors that will say they are going to the graduation party is of 62.5%.

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct or inverse proportional, can be built to find the desired measures in the problem.

The estimate of a population proportion is the sample proportion. In this problem, the sample proportion is of 75 out of 120 students, hence:

p = 75/120 = 0.625 = 62.5%.

The estimate of the population percentage of seniors that will say they are going to the graduation party is of 62.5%.

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Let the function be (3x+15)/(6-x) then the value of x exists at -5.

Given: Rational Expression (3x+15)/(6-x)

To find the value of x when given a rational expression equivalent to 0.

To estimate the value of x, convey the variable to the left side and convey all the remaining values to the right side. Simplify the values to estimate the result.

Consider, (3x+15)/(6-x) = 0

3x + 15 = 0(6-x)

3x + 15 = 0

Subtract 15 from both sides of the equation, e get

3x + 15 - 15 = 0 - 15

simplifying the above equation, we get

3x = 0 - 15

3x = -15

Divide both sides by 3, then we get

x/3 = -15/3

x = -5

Therefore, the value of x exists at -5.

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Answer:

-1

Step-by-step explanation:

Rise=2

Run=-2

2/-2=

-1

Answer:

36Q

Step-by-step explanation:

multiply all together we have 36Q

The sample proportion [tex]$\hat{p}=0.22+0.033=0.253$[/tex].

We know that the confidence interval for sample proportion exists estimated as;

90% confidence interval = Sample proportion  Margin of Error

Here, let [tex]$\hat{p}[/tex] = sample proportion

Level of significance = 1 - 0.90 = 0.[tex]$(0.22,0.28)=\hat{p} \pm 1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$[/tex]10 or 10% Critical value of z at 5% (two-sided) level of significance exists 1.645.

So, 90% confidence interval [tex]$=\hat{p} \pm 1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$[/tex]

[tex]$0.22=\hat{p}-1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \ldots(1)$[/tex]

[tex]$0.28=\hat{p}+1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \ldots (2)[/tex]

From (1) and (2) , we get;

[tex]$&0.22+1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.28-1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\[/tex]

Simplifying the equation, we get

[tex]$&1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}+1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.28-0.22 \\[/tex]

[tex]$&2 \times 1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.06 \\[/tex]

[tex]$&\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\frac{0.06}{2 \times 1.645} \\[/tex]

[tex]$&\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.02[/tex]

Now, squaring both sides, we get;

[tex]$\frac{\hat{p}(1-\hat{p})}{n}=0.0004 \\[/tex]

[tex]$n=\frac{\hat{p}(1-\hat{p})}{0.0004}[/tex]

Now, putting value of n in (1), we get;

[tex]$0.22=\hat{p}-1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$[/tex]

[tex]$0.22=\hat{p}-1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{\hat{p}(1-\hat{p})} \times 0.0004}$[/tex]

Simplifying the equation, we get

[tex]$0.22=\hat{p}-1.645 \times \sqrt{0.0004}$[/tex]

[tex]$0.22=\hat{p}-(1.645 \times 0.02)$[/tex]

[tex]$0.22=\hat{p}-0.033$[/tex]

[tex]$\hat{p}=0.22+0.033=0.253$[/tex].

The sample proportion [tex]$\hat{p}=0.22+0.033=0.253$[/tex].

Therefore, the correct answer is option D. 0.25.

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