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Luda [366]
3 years ago
11

The price of the jeans was reduced $8 per week for 6 weeks. By how much did the price of the jeans change over the 6 weeks?

Mathematics
1 answer:
Damm [24]3 years ago
7 0

Answer:

48$

Step-by-step explanation: The jeans were reduced by 8$ per week for 6 weeks (8 times 6= 48)

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5k^3-8-4k^2+5k^2-2+3k^3
lara31 [8.8K]
<h3>I'll teach you how to solve 5k^3-8-4k^2+5k^2-2+3k^3</h3>

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5k^3-8-4k^2+5k^2-2+3k^3

Group like terms:

5k^3+3k^3-4k^2+5k^2-8-2

Add similar elements:

5k^3+3k^3+k^2-8-2

Add similar elements:

8k^3+k^2-8-2

Subtract the numbers:

8k^3+k^2-10

Your Answer Is 8k^3+k^2-10

Plz mark me as brainliest :)

7 0
3 years ago
A toy store’s percent of markup is 30%. A model train costs the store $50. Find the markup.
Over [174]

Answer:  $15

Step-by-step explanation:

To find the markup price just multiply 30% by the amount.

30% * 50  

0.3 * 50 = 15

4 0
2 years ago
Read 2 more answers
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
Is it preferable to pay for these items with tax dollars or be charged ad each is used? What do you think?
Nastasia [14]

Answer:

In my opinion I would say Tax Dollars

Hope This Helps!     Have A Nice Day!!      

7 0
3 years ago
You have been hired as a marketing consultant to Johannesburg Burger Supply, Inc., and you wish to come up with a unit price for
kotegsom [21]

Answer:

q(p)= -3000p+12000

Step-by-step explanation:

For the function to be linear,

q(p)= mp + c

where

q(p): number of hamburgers sold

p: price per hamburger

m: gradient of the function

c: constant of the function

q(p)=6000 when p=2

6000=2m+c .................... equation I

0=4m+c

c=-4m........................ equation II

Substitute value of c in equation I

6000=2m-4m

m= -3000

c=12000

q(p)= -3000p+12000

6 0
2 years ago
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