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poizon [28]
4 years ago
15

A shop has one-pound bags of peanuts for $2 and three-pound bags of peanuts for $5.50. If you buy 8 bags and spend $37, how many

of each size bag did you buy?
Mathematics
1 answer:
Andrew [12]4 years ago
4 0

Answer:

  • 6 3-lb bags
  • 2 1-lb bags

Step-by-step explanation:

Let x represent the number of 3-lb bags purchased. Then the total purchase was ...

  $2(8 -x) +$5.50(x) = $37

  16 +3.50x = 37 . . . . . . . . . divide by $, collect terms

  3.50x = 21 . . . . . . . . . . . . . subtract 16

  21/3.50 = x = 6 . . . . . . . . divide by the coefficient of x

You bought 6 3-lb bags of peanuts and 2 1-lb bags.

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JulsSmile [24]

Answer:

1) n=114

2) n=59

3) On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.

Step-by-step explanation:

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".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by \alpha=1-0.80=0.2 and \alpha/2 =0.1. And the critical value would be given by:

z_{\alpha/2}=-1.28, z_{1-\alpha/2}=1.28

Part 1

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.06 or 6% points, and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

Since we don't have a prior estimate of \het p we can use 0.5 as a good estimate, replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.06}{1.28})^2}=113.77  

And rounded up we have that n=114

Part 2

On this case we have a prior estimate for the population proportion and is \hat p =0.85 so replacing the values into equation (b) we got:

n=\frac{0.85(1-0.85)}{(\frac{0.06}{1.28})^2}=58.027

And rounded up we have that n=59

Part 3

On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.

5 0
4 years ago
Solve the volume of the box! Help ASAP also you get extra points <3.
olga2289 [7]

Answer:

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Step-by-step explanation:

V=bwh

V=5(3)(1.5)

V=15(1.5)

V=22.5

8 0
3 years ago
SOMEONE PLEASE HELP!!
german

-12=\dfrac{-12}{1}=\dfrac{12}{-1}=\dfrac{-24}{2}=\dfrac{24}{-2}=\dfrac{-36}{3}=\dfrac{36}{-3}=\dfrac{-48}{4}=\dfrac{48}{-4}=\dots

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It would take both Hannah and destiny 28 hours to paint the room together
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3 years ago
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