What is 7/12 foot long

julia-pushkina [17]

B because the amount of money they earn per hour multiplied by how many hours they worked is equal to what they earned in total

7 0

2 years ago

What is 90% of 20????

pickupchik [31]

Answer:

18

Step-by-step explanation:

Multiply 20 by 0.90 (that is 90%)

18 is the answer

Hope this helps :)

6 0

3 years ago

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PLEASE HELP ASAP IF YOU RIGHT YOU GET BRAINLLEST

ratelena [41]

Answer:

its number 3

Step-by-step explanation:

8 0

3 years ago

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A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

4 0

3 years ago

Kim wants to know how many families in her small neighborhood of 70 homes would participate in a neighborhood sports event. She

tatuchka [14]

Answer:

Yes, this is a valid inference because she took a random sample of the neighborhood.

Step-by-step explanation:

As her survey was non bias and completely random and she found that 15% of the families would participate in the event. She claims that 15% of the neighborhood families would be expected to participate in the sports event. So, yes, this is a valid inference because she took a random sample of the neighborhood.

7 0

3 years ago

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