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docker41 [41]
2 years ago
6

A carpenter is making doors that are 2058 millimeters tall. If the doors are too long they must be trimmed, and if they are too

short they cannot be used. A sample of 17 doors is taken, and it is found that they have a mean of 2069 millimeters with a standard deviation of 24. Assume the population is normally distributed. Is there evidence at the 0.02 level that the doors are either too long or too short
Mathematics
1 answer:
marysya [2.9K]2 years ago
3 0

There is not enough evidence at the 0.02 level that the doors are either too long or too short. So, the null hypothesis is accepted.

<h3>How to decide whether the null hypothesis is rejected or accepted?</h3>

Using the p-value approach for the hypothesis test, whether the null hypothesis is rejected or accepted.

  • If the p-value is greater than the significance level then the null hypothesis is accepted.
  • If the p-value is less than the significance level then the null hypothesis is rejected.

<h3>Calculation:</h3>

It is given that,

Sample size n = 17

Sample mean X = 2069

Standard deviation σ = 24

Population mean  μ = 2058

Significance level α = 0.02

Hypothesis:

The null hypothesis: H0: μ = 2058

The alternative hypothesis: Ha: μ ≠ 2058

So, the z-test score for the given distribution is,

z-score = (X - μ)/(σ/√n)

On substituting,

z-score = (2069 - 2058)/(24/√17)

            = 11/5.82

            = 1.89

Thus, the p-value for the z-score of 1.89 is 0.0587 ≅ 0.06.

Since 0.06 > 0.02(significance level), the null hypothesis is accepted. So, there is no sufficient evidence that the doors are either too long or too short.

Learn more about p-value approaches for hypothesis testing here:

brainly.com/question/14805123

#SPJ4

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Answer:

Part 1) The measure of arc EHL is 108\°

Part 2) The measure of angle LVE is 54\°

Step-by-step explanation:

step 1

Let

x-----> the measure of arc EHL

y----> the measure of arc EVL

we know that

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m

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y=144\°+x ------> equation A

Remember that

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substitute equation A in equation B and solve for x

x+(144\°+x)=360\°

2x=360\°-144\°

x=216\°/2=108\°

Find the value of y

y=144\°+x

y=144\°+108\°=252\°

therefore

The measure of arc EHL is 108\°

The measure of arc EVL is 252\°

step 2

Find the measure of angle LVE

we know that

The inscribed angle measures half that of the arc comprising

Let

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3 years ago
I need help i need the statement and reason
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Arrange the cones in order from lease volume to greatest volume
bearhunter [10]

Answer:

Volume of the cone in ascending order.

V_{2}=270\pi\ units^{3}

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

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Step-by-step explanation:

Let V_{2}. V_{3}. and\  V_{4}. be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

d_{1} = 20\ and\ h_{1}=12

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Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

V=\pi r^{2} \frac{h}{3}----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for d_{1} = 20\ and\ h_{1}=12

r_{1} = \frac{d_{1}}{2}

r_{1} = \frac{20}{2}=10\ units

V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}

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r_{2} = \frac{d_{2}}{2}

r_{2} = \frac{18}{2}=9\ units

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V_{2}=\pi (9)^{2} \frac{10}{3}

V_{2}=\pi\times 81\times \frac{10}{3}

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Similarly, for volume of the cone for r_{3} = 10\ and\ h_{3}=9

V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}

V_{3}=\pi (10)^{2} \frac{9}{3}

V_{3}=\pi\times 100\times 3

V_{3}=\pi\times 300

V_{3}=300\pi\ units^{3}

Similarly, for volume of the cone for r_{4} = 11\ and\ h_{4}=9

V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}

V_{4}=\pi (11)^{2} \frac{9}{3}

V_{4}=\pi\times 121\times 3

V_{4}=\pi\times 363

V_{4}=363\pi\ units^{3}

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V_{2}=270\pi\ units^{3}

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see the attached figure to better understand the problem

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