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2 years ago

during a basketball game justin attempted 40 shots and made 18. He says he made 35% of the shots is justin correct

Mathematics

1 answer:

Vikentia [17]2 years ago

5 0

Answer:

No, Justin is incorrect.

Step-by-step explanation:

1. To find how much 18/40 as a percent is, we just multiply 18/40 by 100:

$\frac{18}{40} * 100$
$\frac{9}{20} * 100$
$\frac{900}{20}$
$45$

2. As you can see, 18/40 as a percent is 45%, not 35%. Therefore, Justin is incorrect.

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The difference of nine times a number and the quotient of 6 and the same number.

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2 years ago

Ken's fitness member ship cost $100 to join plus a $25 monthly fee. Ken also used a coupon to get 10% off the monthly fee. What

GREYUIT [131]

The equation that models the total cost is y = 100 + 22.5m

Step-by-step explanation:

Fitness member ship cost = $100

Monthly fee = $25

Discount = 10% on monthly fee

To calculate new monthly fee after discount,

25 - (25*10%) --> 25 - 2.5 --> $22.5

Number of months = m

Fee in m months = 22.5*m = 22.5m

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(x) is the monthly fee $22.5

(m) is the number of months

(y) is the total cost

So,

y = mx + c

y = m(22.5) + 100

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2 years ago

The realtor and her clients do not know the average home sale price for all of Guelph (500 was actually just a guess). However,

strojnjashka [21]

Answer:

a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=400$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=80$ represent the population standard deviation

n=15 represent the sample size

2) Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

Now we have everything in order to replace into formula (1):

$400-2.58\frac{80}{\sqrt{15}}=346.708$

$400+2.58\frac{80}{\sqrt{15}}=453.292$

So on this case the 99% confidence interval would be given by (346.708;453.292)

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

We need to find the degrees of freedom first df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that $t_{\alpha/2}=2.98$