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SIZIF [17.4K]
3 years ago
13

1) Lorraine would like to purchase a new pair of boots that cost $80. She has a coupon for 25% off.

Mathematics
2 answers:
german3 years ago
7 0

Answer:

A) X represents the new price of the jacket after adding the coupon

B) There's a 100 in the denominator of the first ratio because it was originally a percentage, you multiply the percentage by 100 to figure out the decimal.

C) Cross multiply:

100x-2000

Then subtract 100 by both sides:

(100 cancels out each other) 100x-2000

    2000-100= 1900                                                  

X=1900      

Andrei [34K]3 years ago
3 0

Answer:

$60

Step-by-step explanation:

25/100x80=20

80-20=60

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schepotkina [342]

Answer:

t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28    

The degrees of freedom are given by

df=n-1=17-1=16  

The p value would be given by this probability:

p_v =P(t_{(16)}  

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

Step-by-step explanation:

Information given

\bar X=8.19 cm^3 represent the sample mean

s=0.8 cm^3 represent the sample deviation

n=17 sample size  

\mu_o =9.02 represent the value to verify

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic

p_v represent the p value

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We want to check if the true mean is less than the normal value of 9.02 cm^3, the system of hypothesis would be:  

Null hypothesis:\mu \geq 9.02  

Alternative hypothesis:\mu < 9.02  

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28    

The degrees of freedom are given by

df=n-1=17-1=16  

The p value would be given by this probability:

p_v =P(t_{(16)}  

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

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