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Gekata [30.6K]
3 years ago
15

Mr. Anderson paid $19.18 for a CD cover. This amount included a tax of 2%. What was the cost of the CD cover before tax

Mathematics
1 answer:
uysha [10]3 years ago
7 0
The price is around 18.80
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Due for 10:00am need help​
Sveta_85 [38]

Answer:

y=9b to the second power t over m

Step-by-step explanation:

should look like this

y=9b^2t/m

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16 is 4% of what number?
IrinaK [193]
16 is 4% of 400 

Change the percentage into a decimal by dividing the decimal over 100:
\frac{4}{100} = 0.04

Divide 16 by the decimal (0.04):
16 \div 0.04 = 400
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The wills tower has a height of 1,451feet.what is the estimated height of the building in meters​
Ivan
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What is the length of the hypothenuse of the triangle?
BlackZzzverrR [31]

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26ft

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2 years ago
Read 2 more answers
For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df
rjkz [21]

Answer:

a) For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

b) For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

Part b

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

3 0
3 years ago
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