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Lerok [7]
3 years ago
12

Sammy usually takes a 18 minute shower in the morning. If a 4 minute shower uses an average of 20-40 gallons of water, what is t

he maximum gallons of water that Sammy could be using in her shower?
a.
72 gallons of water
b.
360 gallons of water
c.
90 gallons of water
d.
180 gallons of water
Mathematics
2 answers:
taurus [48]3 years ago
7 0

Answer: Option 'D' is correct.

Step-by-step explanation:

Since we have given that

In a 4 minute, the shower uses an average of 20-40 gallons of water,

If we consider the maximum gallons of water,

In 4 minutes, the maximum gallons of water = 40

In 1 minute, the maximum gallons of water is given by

\frac{40}{4}=10

In 18 minutes, the maximum gallons of water is given by

18\times 10=180\ gallons

Hence, Option 'D' is correct.

mylen [45]3 years ago
5 0
D 180 gallons of water
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Suppose ABCD is a parallelogram. Find the value of x and y.
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answer:

x = 4

y = 19

step-by-step explanation:

  • we know that this is a parallelogram, so opposite sides are equal
  • this is one of the properties of a parallelogram

therefore,

segment AB = segment DC <em>and</em> segment AD = segment BC

  • now plug in the expressions according to what we know (& solve)

AB = DC

5x - 13 = 9x - 29

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16 = 4x

x = 4

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3y - 11 = 2y + 8

y - 11 = 8

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A magazine reports that women trust recommendations from a particular social networking site more than recommendations from any
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Answer:

a) p_A represent the real population proportion for women

\hat p_A =\frac{123}{150}=0.82 represent the estimated proportion for women

n_A=150 is the sample size selected for women

b) p_B represent the real population proportion for male

\hat p_B =\frac{102}{170}=0.6 represent the estimated proportion for male

n_B=170 is the sample size selected for male

c) (0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124  

(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316  

And the 95% confidence interval would be given (0.124;0.316).  

We are confident at 95% that the difference between the two proportions is between 0.124 \leq p_A -p_B \leq 0.316

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".  

Part a

p_A represent the real population proportion for women

\hat p_A =\frac{123}{150}=0.82 represent the estimated proportion for women

n_A=150 is the sample size selected for women

Part b

p_B represent the real population proportion for male

\hat p_B =\frac{102}{170}=0.6 represent the estimated proportion for male

n_B=170 is the sample size elected for male

z represent the critical value for the margin of error  

Part c

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96  

And replacing into the confidence interval formula we got:  

(0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124  

(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316  

And the 95% confidence interval would be given (0.124;0.316).  

We are confident at 95% that the difference between the two proportions is between 0.124 \leq p_A -p_B \leq 0.316

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