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1 year ago

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Find the answer The weight of an elephant is 10 to the 3rd power times the weight of a cat .If the the elephant weighs 14,000 po

unds,how many pounds does the cat weigh? How did you find the answer?

1 answer:

Hitman42 [59]1 year ago

3 0

Answer:

$14\ pounds$

Below is the procedure that was used to find the answer.

Step-by-step explanation:

Let be "e" the weight in pounds of the elephant and "c" the weight in pounds of the cat.

According to the information provided in the exercise, we know that The weight of an elephant is $10^3$ times the weight of a cat. Based on this we can write the following equation:

$e=10^3c$

If the weight in pounds of the elephant is:

$e=14,000$

We must substitute this value into the equation and then solve for "c" in order to find the weight in pounds of the cat.

Then we get:

$14,000=10^3c\\\\\frac{14,000}{10^3}=c\\\\c=14$

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Answer:

a) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

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d) Null and alternative hypothesis

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Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

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

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Then, we can calculate the t-statistic as:

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The degrees of freedom for this sample size are:

$df=n-1=65-1=64$

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

<u>Critical value approach</u>

<u></u>

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

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1 year ago