186.73 x (-0.0175) Please help!!

(72-y)+(26-X) when x=9,y=51

disa [49]

Answer:

38

Step-by-step explanation:

(72-51)+(26-9)

21+17

38

7 0

3 years ago

The life in hours of a 75-watt light bulb is known to be normally distributed with standard deviation σ = 25 hours. A random sam

irina [24]

Answer: a) (1008.34,1019.658) b) (1009.24,1018.76)

Step-by-step explanation:

Since we have given that

n = 75

mean = 1014 hours

Standard deviation = 25 hours

At 95% two sided , z = 1.96

So, confidence interval would be

$\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=1014\pm 1.96\dfrac{25}{\sqrt{75}}\\\\=1014\pm 5.658\\\\=(1014-5.658,1014+5.658)\\\\=(1008.34,1019.658)$

(b) Construct a 95% lower confidence bound on the mean life.

z = 1.65

So, confidence interval would be

$\bar{x}\pm z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1014\pm 1.65\times \dfrac{25}{\sqrt{75}}\\\\=1014\pm 4.76\\\\=(1014-4.76,1014+4.76)\\\\=(1009.24,1018.76)$

Hence, a) (1008.34,1019.658) b) (1009.24,1018.76)

4 0

3 years ago

A laptop producing company also produces laptop batteries, and claims that the batteries

gregori [183]

Answer:

There is enough evidence to support the claim that the batteries power the laptops for significantly less than 4 hours. (P-value = 0).

The null and alternative hypothesis are:

$H_0: \mu=4\\\\H_a:\mu< 4$

Step-by-step explanation:

The question is incomplete: To test this claim a sample or population standard deviation is needed.

We will estimate that the sample standard deviation is 2 hours, and use a t-test to test that claim.

NOTE (after solving): The difference between the sample mean and the mean of the null hypothesis is big enough to reject the null hypothesis, even when we have a sample standard deviation of 3.5 hours, which can be considered bigger than the maximum standard deviation for the sample.

This is a hypothesis test for the population mean.

The claim is that the batteries power the laptops for significantly less than 4 hours.

Then, the null and alternative hypothesis are:

$H_0: \mu=4\\\\H_a:\mu< 4$

The significance level is 0.05.

The sample has a size n=500.

The sample mean is M=3.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2}{\sqrt{500}}=0.0894$