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Norma-Jean [14]
3 years ago
6

What is the unit rate of $1.12 for 8.2 ounces?

Mathematics
1 answer:
pashok25 [27]3 years ago
3 0

That is around 13 or 14 cents. The specific answer is 13.65853659 cents, but you can either round up to 14 or round down to 13 (rounding up to 14 is the correct way to do it.)

First you divide 1.12 by 8.2 to get the unit rate because It takes $1.12 to buy 8.2 ounces. 1.12/8.2=13.65853659

Hope this helped!

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A measurement that closely agrees with an accepted value is best described as a numerical

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A librarian knows that 45% of the books currently checked out will NOT be returned on the due date. The library has a
malfutka [58]

Answer:

79 books

Step-by-step explanation:

45% books not returned = 55% books will be returned

145 books currently checked out

55% of 145 = 79.75

since you can't return 3/4 of a book, i would round down to 79.

7 0
3 years ago
NEED HELP SOON!!!!<br> for any numbers a, b, and c, a(b + c)=ab+ac
navik [9.2K]

The property displayed here is the distributive property.

If you have a variable or unknown number inside or outside of parentheses, you can distribute it to each term and add the terms together, and it will remain true.

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6 0
3 years ago
I need the answer as fast as i can get it!
blondinia [14]

Answer:

The answer should be choice D.

HOPE THIS HELPS! :)

7 0
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
2 years ago
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