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PtichkaEL [24]
3 years ago
10

Round 6.94729776259 to 5 decimal places.

Mathematics
1 answer:
maria [59]3 years ago
8 0

Answer

6.9473

Step-by-step explanation:

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Write each equation in slope-intercept form.
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1 year ago
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A sample of 900900 computer chips revealed that 49%49% of the chips do not fail in the first 10001000 hours of their use. The co
dedylja [7]

Answer:

The proportion of chips that do not fail in the first 1000 hours of their use is 52%.

Step-by-step explanation:

The claim made by the company is that 52% of the chips do not fail in the first 1000 hours of their use.

A quality control manager wants to test the claim.

A one-proportion <em>z</em>-test can be used to determine whether the proportion of chips do not fail in the first 1000 hours of their use is 52% or not.

The hypothesis can be defined as:

<em>H₀</em>: The proportion of chips do not fail in the first 1000 hours of their use is 52%, i.e. <em>p</em> = 0.52.

<em>Hₐ</em>: The proportion of chips do not fail in the first 1000 hours of their use is different from 52%, i.e. <em>p</em> ≠ 0.52.

The information provided is:

\hat p=0.49\\n=900\\\alpha =0.02

The test statistic value is:

z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.49-0.52}{\sqrt{\frac{0.52(1-0.52)}{900}}}=-1.80

The test statistic value is -1.80.

Decision rule:

If the <em>p-</em>value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the <em>p</em>-value as follows:

p-value=2\times P(Z

*Use a <em>z</em>-table.

The <em>p</em>-value = 0.07186 > <em>α</em> = 0.02.

The null hypothesis was failed to be rejected at 2% level of significance.

<u>Conclusion</u>:

The proportion of chips that do not fail in the first 1000 hours of their use is 52%.

8 0
3 years ago
Exclude leap years from the following calculations. ​(a) Compute the probability that a randomly selected person does not have a
Scrat [10]

Answer:

a) 99.73% probability that a randomly selected person does not have a birthday on March 14.

b) 96.71% probability that a randomly selected person does not have a birthday on the 2 nd day of a month.

c) 98.08% probability that a randomly selected person does not have a birthday on the 31 st day of a month.

d) 92.33% probability that a randomly selected person was not born in February.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

A non-leap year has 365 days.

​(a) Compute the probability that a randomly selected person does not have a birthday on March 14.

There are 365-1 = 364 days that are not March 14. So

364/365 = 0.9973

99.73% probability that a randomly selected person does not have a birthday on March 14.

​(b) Compute the probability that a randomly selected person does not have a birthday on the 2 nd day of a month.

There are 12 months, so there are 12 2nds of a month.

So

(365-12)/365 = 0.9671

96.71% probability that a randomly selected person does not have a birthday on the 2 nd day of a month.

​(c) Compute the probability that a randomly selected person does not have a birthday on the 31 st day of a month.

The following months have 31 days: January, March, May, July, August, October, December.

So there are 7 31st days of a month during a year.

Then

(365-7)/365 = 0.9808

98.08% probability that a randomly selected person does not have a birthday on the 31 st day of a month.

(d) Compute the probability that a randomly selected person was not born in February.

During a non-leap year, February has 28 days. So

(365-28)/365 = 0.9233

92.33% probability that a randomly selected person was not born in February.

6 0
3 years ago
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