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Kitty [74]
3 years ago
15

The dishwasher Marlin wants costs $720. This week only, the dishwasher is discounted by 25%. Marlin must pay 6% sales tax. Which

saves the most money? A Adding on the sales tax and then taking the 25% discount. B Taking the discount first and then adding on the sales tax. C Both methods save the same amount of money.
Mathematics
1 answer:
Mkey [24]3 years ago
7 0

Answer:

C, both methods save the same amount of money

Step-by-step explanation:

First, we need to calculate what each answer choice will equal.

For A, it asks us to add the sales tax and then take the discount off. So, to find out what the sales tax is equal to, we multiply 720 x 0.06 (because 6% is equal to 0.06).

720 x 0.06 = 43.2

Add the sales tax to the original amount and you get 763.2.

Now we take the discount, which is 25% or 1/4.

1/4 of 763.2 is 190.8.

Subtract the discount from the price and you get 572.4.

For B, it says to take the discount first and then add the sales tax.

25% = 1/4, so 1/4 of 720 is equal to 180.

Subtract 180 from 720, and you get 540.

Now add the sales tax.

540 x 0.06 = 32.4.

540 + 32.4 = 572.4.

The answer is C, both methods save the same amount of money!

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The quality-control manager at a compact flourescent light bulb factory wants to test the claim that the mean life of a large sh
MAXImum [283]

Answer:

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. CI [6583.336 ,6816.336]

d.  The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

Step-by-step explanation:

Population mean = u= 6500 hours.

Population standard deviation = σ=500 hours.

Sample size =n= 50

Sample mean =x`=  6,700 hours

Sample standard deviation=s=  600 hours.

Critical values, where P(Z > Z) =∝ and P(t >) =∝

Z(0.10)=1.282  

Z(0.05)=1.645  

Z(0.025)=1.960  

t(0.01)(49)= 1.299

t(0.05)= 1.677  

t(0.025,49)=2.010

Let the null and alternate hypotheses be

H0: u = 6500 against the claim Ha: u ≠ 6500

Applying Z test

Z= x`- u/ s/√n

z= 6700-6500/500/√50

Z= 200/70.7113

z= 2.82=2.82

Applying  t test

t= x`- u /s/√n

t= 6700-6500/600/√50

t= 2.82

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

Yes we reject H0  for z- test as it falls in the critical region,at the 0.05 level of significance, z=2.82 > z∝=1.645

For t test  we reject H0   as it falls in the critical region,at the 0.05 level of significance, t=2.82 > t∝=1.677 with n-1 = 50-1 = 49 degrees of freedom.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. The 95 % confidence interval of the population mean life is estimated by

x` ±  z∝/2  (σ/√n )

6700± 1.645 (500/√50)

6700±116.336

6583.336 ,6816.336

d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

6 0
3 years ago
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