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natka813 [3]
3 years ago
10

What is the surface area and volume? PLEASE EXPLAIN IN DETAIL HOW TO SOLVE.

Mathematics
1 answer:
Maslowich3 years ago
7 0

The surface area is the sum of the area of all the faces of an object.

Area of bottom face: 4×2.5=10

Area of right and left sides: 2(2.5×2)=10

Area of front and back sides: 2(4×2)=16

Area of the rectangular sides of the roof: 2(2.5×2.5)=12.5

Area of the triangle sides of the roof: 2(1/2(4×1.5))=6

Add up all of these to get 54.5 yd^2

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A 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender​ selecti
Andrews [41]

Answer:

(a) Null Hypothesis, H_0 : p = 0.50

    Alternate Hypothesis, H_A : p > 0.50

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed.

(e) The value of z test statistics is 0.96.

(f) The P-value is 0.1685.

(g) At 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.

(h) We conclude that the proportion of baby girls is equal to 0.50.

Step-by-step explanation:

We are given that a 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender​ selection, the proportion of baby girls is greater than 0.5.

Assume that sample data consists of 78 girls in 144 ​births.

Let p = <u><em>population proportion of baby girls</em></u>

(a) So, Null Hypothesis, H_0 : p = 0.50     {means that the proportion of baby girls is equal to 0.50}

Alternate Hypothesis, H_A : p > 0.50     {means that the proportion of baby girls is greater than 0.50}

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed as in the alternative hypothesis we are concerned for proportion of baby girls that is greater than 0.50.

(e) The test statistics that would be used here <u>One-sample z test for proportions</u>;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of baby girls =  \frac{78}{144} = 0.54

            n = sample of births = 144

So, <u><em>the test statistics</em></u>  =  \frac{0.54-0.50}{\sqrt{\frac{0.54(1-0.54)}{144} } }  

                                       =  0.96

The value of z test statistics is 0.96.

(f) <u>The P-value of the test statistics is given by;</u>

            P-value = P(Z > 0.96) = 1 - P(Z < 0.96)

                          = 1 - 0.8315 = 0.1685

<u></u>

(g) <u>Now, at 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.</u>

(h) Since our test statistic is less than the critical value of z as 0.96 < 1.282, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region (which was to the right of value of 1.282) due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the proportion of baby girls is equal to 0.50.

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Answer:

Relateable

Step-by-step explanation:

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Slope = -1/5
Coordinate = (1, -4)

We know, y - y1 = m(x - x1)
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y = -1/5x -19/5

When, x = 1, y = -1/5(1) - 19/5 = -20/5 = -4
x = 2, y = -1/5(2) - 19/5 = -21/5
x = 3, y = -1/5(3) - 19/5 = -22/5

Here, Your Coordinates would be: (0, -4), (1, -21/5), (2, -22/5)
Mark them & draw the lines. Graph is done!

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12/6 is 2 bottles for a dollar 
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