Beth wants to buy a shirt that costs $40. She has a 20% discount. The sales tax is 8%.

2 answers:

iogann1982 [59]2 years ago

8 0

Correct answer:
$22.84

Explanation:
The tax is calculated by 0.075 * 21.25 = 1.59375 or $1.59. So the total bill is the cost of items purchased plus tax, which is $21.25 + $1.59 = $22.84.

This doesn't make sense I know but I tried ok

Aleks04 [339]2 years ago

6 0

Answer:

$34.56

Step-by-step explanation:

She will save 8 dollars from the discount.

The tax will be $0.56.

The final price of the shirt will be $34.56

You might be interested in

Solve - 9/2y=27/2 for y.

dedylja [7]

Answer:

It’s y = 3

Step-by-step explanation:

Solve for y by simplifying both sides of the equation, then isolating the variable.

5 0

3 years ago

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95