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Inessa05 [86]
2 years ago
8

4.

Mathematics
1 answer:
solong [7]2 years ago
3 0
The answer is D). 125 = 1.28f. This accounts for both sales tax and the cost of food.
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Convert \frac{5}{9} to a decimal.
Dmitry [639]

Answer:0.55 i think

Step-by-step explanation:

5 0
2 years ago
While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who en
AVprozaik [17]

Answer:

z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}   (1)

z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057  

p_v =P(Z>0.0057)=0.4977  

The p value is a very high value and using the significance level \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.  

Step-by-step explanation:

1) Data given and notation  

X_{M}=23 represent the number of men that said they enjoyed the activity of Saturday afternoon shopping

X_{W}=8 represent the number of women that said they enjoyed the activity of Saturday afternoon shopping

n_{M}=66 sample of male selected

n_{W}=23 sample of demale selected

p_{M}=\frac{23}{66}=0.34848 represent the proportion of men that said they enjoyed the activity of Saturday afternoon shopping

p_{W}=\frac{8}{23}=0.34782 represent the proportion of women with red/green color blindness  

z would represent the statistic (variable of interest)  

p_v represent the value for the test (variable of interest)

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment , the system of hypothesis would be:  

Null hypothesis:p_{M} \leq p_{W}  

Alternative hypothesis:p_{M} > p_{W}  

We need to apply a z test to compare proportions, and the statistic is given by:  

z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}   (1)

Where \hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{23+8}{66+23}=0.34831

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057  

4) Statistical decision

Using the significance level provided \alpha=0.05, the next step would be calculate the p value for this test.  

Since is a one side right tail test the p value would be:  

p_v =P(Z>0.0057)=0.4977  

So the p value is a very high value and using the significance level \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.  

4 0
3 years ago
The heights of a certain type of tree are approximately normally distributed with a mean height p = 5 ft and a standard
arsen [322]

Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean

Step-by-step explanation:

⇒ 1^s^t statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)

an X value is found Z standard deviations from the mean mu if:

\frac{X-\mu}{\sigma} = Z

In this case we have:  \mu=5\ ft\sigma=0.4\ ft

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒2^n^d statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft  

Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒3^r^d statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean (FALSE)

For X =5.8 ft

Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒4^t^h statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

6 0
3 years ago
310=-6.2f<br> f=??? so confused
Mariulka [41]
F=-50 as multiplying it by -6.2 would give you a positive 310
6 0
3 years ago
Use the fact that the bacteria is doubling every five
victus00 [196]

Answer:

I think it's 1/16

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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