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antoniya [11.8K]
2 years ago
14

The 99 percent confidence interval for the average weight of a product is from 71.36 lbs. to 78.64 lbs. Can we conclude that μ i

s equal to 71, using a 99 percent confidence interval? Briefly explain.
Mathematics
1 answer:
Ilya [14]2 years ago
6 0

Answer:

Step-by-step explanation:

Hello!

To decide over a hypothesis test using a confidence interval the following conditions are to be met:

1. The hypotheses should be two-tailed.

2. The hypotheses and the confidence interval should be made for the same parameter.

3. The confidence level and the significance level should be complementary, so if the CI is 1-α: 0.99 the level of significance should be α: 0.01

If all conditions are met, the decision rule is the following:

If the value stated in the null hypothesis is contained by the CI, the decision is to not reject the null hypothesis.

If the value stated in the null hypothesis is not contained by the CI, the decision is to reject the null hypothesis.

Let's say in this example the hypotheses are:

H₀: μ = 71

H₁: μ ≠ 71

α: 0.01

99% Confidence level is (71.36;78.64).

As you can see, 71 is not contained in the CI so the decision to take is to reject the null hypothesis, in other words, the population mean is not 71.

I hope it helps!

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Answer:

1) Juans claim is incorrect. The correct experimental probablilty is 2/9

Step-by-step explanation:

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jose is choosing a 3-letter password from the letters a,b,c,d. the password cannot have the same letter repeated in it. how many
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Alright, so we are dealing with permutations. Permutations are the number of combinations in a specific order possible for the set. 

There's 4 letters, each one can be paired with 3 other letters, but those letters could be in different orders. To figure out how many variations of each combination there are (aka the number of permutations) use this formula:

_{n}P_{r}= \frac{n!}{(n-r)!}

r= number of elements in the subset

n= number of elements in the set

P= permutations of the set

There are only 3 elements in the subset because there is 1 that will not be repeated in each set, and there are 4 elements in the set.

Here's the math:

_{n}P_{r}= \frac{n!}{(n-r)!}

_{4}P_{3}= \frac{4!}{(4-3)!}

_{4}P_{3}= \frac{4!}{(1)!}

_{4}P_{3}= \frac{4(3)(2)(1)}{1}

_{4}P_{3}=24

There are 24 permutations. I can prove this by showing you the model:

ABCD, ABDC, ACBD, ACDB, ADBC, ADCB are the 6 arrangements possible of the set starting with the letter A. Because there are 4 letters, the total amount of permutations without repeated letters is 4 (letters) times 6 (possible arrangements), which equals 24.

Hope this helps!

Answer: 24 passwords are possible
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2 years ago
Gabriel is saving money by mowing lawns on the weekends. Last month, he saved $150. This month, saved 20% more than he saved las
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Answer:

180

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so last month he saved 150 right first you find 20% of 150. that is 30. you add 150 and 30 and get 180.

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Dilation refers to a non rigid motion where a figure is transform and its image has the same form but a different size measure.

On this exercise is asked to find the scale factor by which the triangle ABC was 
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As can be see on the picture the dilation produce was an enlargement meaning that the image is larger that the preimage.
Of this form you can discard the choices A and B as possible solutions.

Lets try 5/2 as the possible scale factor:
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A(0,2)--(5/2(0),5/2(2))=A'(0,5)
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Lets try 5/1 or 5 as the scale factor:
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As said at the beginning of the question the triangle was not only dilated.
After a dilation and a translation, the scale factor of the dilation is letter C or 5/2.  

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