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Umnica [9.8K]
3 years ago
8

2/5(5k+35) -8 =12 please helpppppppp

Mathematics
2 answers:
balu736 [363]3 years ago
8 0

See attachment for math work and answer.

Mrac [35]3 years ago
3 0

Answer:

k =3

Step-by-step explanation:

2/5(5k+35) -8 =12

Add 8 to each side

2/5(5k+35) -8+8 =12+8

2/5(5k+35)  =20

Multiply each side by 5/2

5/2 *2/5(5k+35)  =20*5/2

5k+35 = 50

Subtract 35 from each side

5k+35-35 = 50-35

5k = 15

Divide by 5

5k/5 = 15/5

k =3

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P(X = 0) = 0.9891

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Given

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Required [This completes the question]

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8 0
3 years ago
An automobile manufacturer has given its car a 52.6 miles/gallon (MPG) rating. An independent testing firm has been contracted t
Arte-miy333 [17]

Answer:

Null hypothesis:\mu = 52.6  

Alternative hypothesis:\mu \neq 52.6  

z=\frac{52.8-52.6}{\frac{1.6}{\sqrt{250}}}=1.976    

p_v =2*P(z>1.976)=0.0482  

Since the p value is lower than the significance level wedon't have enough evidence to conclude that the true mean is significantly different from 52.6 MPG.

Step-by-step explanation:

Information provided

\bar X=52.8 represent the sample mean  for the MPG of the cars

\sigma=1.6 represent the population standard deviation

n=250 sample size  of cars

\mu_o =52.6 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test

Hypothesis

We need to conduct a hypothesis in order to check if the true mean of MPG is different from 52.6 MPG, the system of hypothesis would be:  

Null hypothesis:\mu = 52.6  

Alternative hypothesis:\mu \neq 52.6  

Since we know the population deviation the statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Calculate the statistic

Replacing we have this:

z=\frac{52.8-52.6}{\frac{1.6}{\sqrt{250}}}=1.976    

Decision

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p_v =2*P(z>1.976)=0.0482  

Since the p value is lower than the significance level wedon't have enough evidence to conclude that the true mean is significantly different from 52.6 MPG.

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7 0
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