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Ket [755]
2 years ago
6

The price of a notebook in 2012 is $2000. In 2013, its value decreased by 30%. In 2014, its value decreased by 30% of its value

in 2013. Find the value of the notebook at the end of 2013
Mathematics
1 answer:
jeyben [28]2 years ago
8 0

Answer:

The value of the notebook at the end of 2013 is $980

Step-by-step explanation:

First, find 30% of 2,000. To do so you need to convert 30% into a decimal by dividing it by 100.

30% ⇒ 0.30

Now multiply 0.30 by 2,000 to find 30% of 2,000.

2,000 × 0.30 = 600

Subtract 600 from 2,000 to find the price of the notebook in 2013.

2,000 - 600 = 1,400

Now find 30% of 1,400 using the same process.

1,400 × 0.30 = 420

Subtract 420 from 1,400 to find the price of the notebook in 2014.

1,400 - 420 = 980

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Answer:

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

df=n-1=15-1=14  

Since is a right tailed test the p value would be:  

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation  

\bar X=5.63 represent the sample mean  

s=1.61 represent the standard deviation for the sample  

n=15 sample size  

\mu_o =15/tex] represent the value that we want to test  &#10;[tex]\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:  

Null hypothesis:\mu \leq 5.63  

Alternative hypothesis:\mu > 5.63  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

Now we need to find the degrees of freedom for the t distribution given by:

df=n-1=15-1=14  

P value

Since is a right tailed test the p value would be:  

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

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3 years ago
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7 0
3 years ago
Read 2 more answers
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Harrizon [31]

Hello!

f(g(x)) = 4 - <u>2</u><u> </u><u>×</u><u> </u><u>(</u><u>3</u><u>x</u><u>²</u><u>)</u> <=>

<=> f(g(x)) = 4 - 6x²

Answer: B. f(g(x)) = 4 - 6x²

Good luck! :)

5 0
2 years ago
Read 2 more answers
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