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3 years ago

Name the values of the given digits The 4s in 4,400

Mathematics

2 answers:

liraira [26]3 years ago

8 0

Thousands and hundreds

makkiz [27]3 years ago

4 0

Hundreds and thousands

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Alybia was making 115 cookies for a Valentine’s Day party. They were each either red or purple. Forty percent of the cookies wer

NeTakaya

Answer:

69 purple cookies!

Step-by-step explanation:

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2 years ago

Help please thank you

iogann1982 [59]

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3 years ago

Cathrine was concerned about her report card grade. She had 5 tests the whole semester. If her scores on each test were 93, 82,

enyata [817]

Answer:

81.6

Step-by-step explanation:

mean/average of her grade = sum of all scores/the number of scores

= (93+82+74+92+67)/5

= 81.6

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3 years ago

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

The revenue of a company is represented by r(x) and the expenses of the company are represented by e(x). What would you do to fi

Pepsi [2]

B.) r(x) - e(x) = p(x)

Profit is how much money remains after the expenses are deducted from the total revenue

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3 years ago