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vfiekz [6]
3 years ago
10

hanna buys 3 packs of 36 exposure film and 2 packs of 24 exposure film. she uses 8 rolls of film. how many rolls does she have l

eft?
Mathematics
2 answers:
RSB [31]3 years ago
7 0
To solve this problem, first we have to figure out how many total rolls Hanna buys.

3 packs*36 rolls in each pack = 108 total rolls
2 packs*24 rolls in each pack = 48 total rolls

108 rolls + 48 rolls = 156 rolls.

Next, it says that Hanna uses 8 rolls of film.  To model this in our expression, we need to subtract 8 rolls from our total amount.  

156-8 = 146 rolls

Hanna has 146 rolls left.


Arturiano [62]3 years ago
3 0
The question can't be answered with the given information . . . we don't know how many rolls there are in a pack.
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Answer:

6(2x - 5)

Step-by-step explanation:

The greatest common factor of 12x and 30 is 6.

12x - 30 = 6(2x - 5)

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Answer:

∠ HGL = 73°

Step-by-step explanation:

KL is a midsegment of the triangle and is parallel to HG, then

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9x - 62 = 5x - 2 ( subtract 5x from both sides )

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3 years ago
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Step-by-step explanation:

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3 years ago
The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books check
Verdich [7]

Answer:

n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11

So the answer for this case would be n=11 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma =150 represent the population standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The confidence interval for this case is given by: (740, 920)

We can find the estimate for the mean and we got:

\bar X = \frac{740+920}{2} = 830

and the margin of error is given by :

ME = \frac{920-740}{2}= 90

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

And on this case we have that ME =90 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.960, replacing into formula (b) we got:

n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11

So the answer for this case would be n=11 rounded up to the nearest integer

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