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3 years ago

PLEASE HELP ME WITH THIS THANKS

Mathematics

1 answer:

kotykmax [81]3 years ago

7 0

Answer:

SA = 1244.64 square centimeters

Step-by-step explanation:

From the attached figure

The formula of the surface area of the prism is SA = 2B + PH, where

B is the area of its base
P is the perimeter of its base
H is the distance between its bases

The base of the prism is a regular hexagon with side 8 cm

If you join each vertex of the hexagon with its center you will form 6 congruent triangles with base 8 cm and height 6.93 cm

The area of the hexagon = 6 × area of a triangle

∵ The base of the triangle = 8 cm

∵ Its height = 6.93 cm

- The formula of the area of a triangle is A = $\frac{1}{2}$ × base × height

∴ Area of the triangle = $\frac{1}{2}$ × 8 × 6.93 = 27.72 cm²

- Lets find the area of the hexagon

∴ The area of the hexagon = 6 × 27 .72 = 166.32 cm²

∴ B = 166.32 cm²

The formula of the perimeter of the regular hexagon is P = 6 × s, where s is the length of its side

∵ The side of the hexagon is 8 cm

∴ P = 6 × 8

∴ P = 48 cm

∵ The distance between the two bases is 19 cm

∴ H = 19 cm

Substitute the values of B, P and H in the formula of the surface area above

∵ SA = 2(166.32) + (48)(19)

∴ SA = 332.64 + 912

∴ SA = 1244.64 square centimeters

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4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B

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Answer:

$z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123$

$p_v =P(Z>2.123)=0.0169$

The p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .

Step-by-step explanation:

1) Data given and notation

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

$\hat p=\frac{63}{200}=0.315$ estimated proportion of lawyers had used some form of advertising for their business

$p_o=0.25$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that more than 25% of all the lawyers in Boston advertise for their business:

Null hypothesis: $p\leq 0.25$

Alternative hypothesis: $p > 0.25$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(Z>2.123)=0.0169$

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