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ELEN [110]
3 years ago
11

kim rented skates for h hours.The rental free was $2.50 per hour. she paid a total a 2.5h=20 b 1.5h=10 c 1.5h=20 d 2.5h=10l of $

20
Mathematics
1 answer:
Genrish500 [490]3 years ago
3 0

Answer:

2.5h=20

Step-by-step explanation:

:D

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A recipe calls for mixed nuts with 50% peanuts. 1/2 pound of 15% peanuts has already been used. How many pounds of 75% peanuts n
Galina-37 [17]
Now you have
<u />
<u>(.15)0.5 lb peanuts</u>     =     <u>0.075 lb peanuts</u>
0.5 lb  nuts                              0.5 lb nuts

for every pound of 75% mixed nuts you add, you get .75 lb peanuts, how many pounds, x, do you need to add to get the ratio 50 to 100?

<u>0.075 lb peanuts + 0.75x lb peanuts</u><em>       =    <u /></em><u>   50   </u><u>
</u>0.5 lb nuts + x lb nuts                                      100

cross multiply

(0.075 + .75x) 100 = (0.5 + x) 50
7.5 + 75x = 25 + 50x
25x = 17.5
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6 0
3 years ago
What is 1 over 4 as a percent​
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1/4×100

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Midpoint of 4 – 3i and –2 + 7i please help!!!!!
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Hello : 
 let  : z = 4-3ii       z' = -2+7i
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4 0
3 years ago
A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she
Nataly_w [17]

Answer:

The minimum sample size needed is n = (\frac{1.96\sqrt{\sigma}}{4})^2. If n is a decimal number, it is rounded up to the next integer. \sigma is the standard deviation of the population.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.9}{2} = 0.05

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.645.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large a sample must she select if she desires to be 90% confident that her estimate is within 4 ounces of the true mean?

A sample of n is needed, and n is found when M = 4. So

M = z\frac{\sigma}{\sqrt{n}}

4 = 1.96\frac{\sigma}{\sqrt{n}}

4\sqrt{n} = 1.96\sqrt{\sigma}

\sqrt{n} = \frac{1.96\sqrt{\sigma}}{4}

(\sqrt{n})^2 = (\frac{1.96\sqrt{\sigma}}{4})^2

n = (\frac{1.96\sqrt{\sigma}}{4})^2

The minimum sample size needed is n = (\frac{1.96\sqrt{\sigma}}{4})^2. If n is a decimal number, it is rounded up to the next integer. \sigma is the standard deviation of the population.

4 0
3 years ago
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