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weeeeeb [17]
3 years ago
7

Persevere with Problems Suppose an online store has an item that costs $50. The item is on sale for 10% off the original price.

10% of $50 = $
Mathematics
1 answer:
ollegr [7]3 years ago
5 0

Answer:

$5

Step-by-step explanation:

To find 10% of $50 we will use 50*0.10 since 10%=.10

the result of this is 5/ $5

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The sum of 32 and what number is 105
natta225 [31]
32+n=105
32-32+n=105-32
n=73
4 0
3 years ago
The number of miles Ford trucks can go on one tank of gas is normally distributed with a mean of 350 miles and a standard deviat
kati45 [8]

Answer:

The probability that a randomly chosen Ford truck runs out of gas before it has gone 325 miles is 0.0062.

Step-by-step explanation:

Let <em>X</em> = the number of miles Ford trucks can go on one tank of gas.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 350 miles and standard deviation, <em>σ</em> = 10 miles.

If the Ford truck runs out of gas before it has gone 325 miles it implies that the truck has traveled less than 325 miles.

Compute the value of P (X < 325) as follows:

P(X

Thus, the probability that a randomly chosen Ford truck runs out of gas before it has gone 325 miles is 0.0062.

7 0
3 years ago
Find the area of the figure.
Gnoma [55]
15.5 x 12.4 = 192.2 so the answer is 192.2 square feet.
8 0
3 years ago
Read 2 more answers
Will says that in the number 8,184​, one 8 is 10 times as great as the other 8. Is he ​correct? Explain why or why not.
Vinil7 [7]
No he is not correct .
Break apart the number 8,184
You will get
8000 + 100 + 80 + 4

The numbers we are comparing are 8000 and 80

80 x 10 = 800

The number we are looking for is 8000

This makes the claim that one 8 is 10 times greater false as it is in fact 100 times as great.

4 0
3 years ago
Read 2 more answers
Assume that the national credit card interest rate is 12.83 percent. A study of 62 college students finds that their average int
meriva

Answer:

b)  95 percent confidence interval for this single-sample t test

[11.64, 16.36]

Step-by-step explanation:

Explanation:-

Given data  a study of 62 college students finds that their average interest rate is 14 percent with a standard deviation of 9.3 percent.

Sample size 'n' =62

sample mean x⁻ = 14  

sample standard deviation 'S' = 9.3

<u>95 percent confidence interval for this single-sample t test</u>

The values are (x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-}+t_{0.05}\frac{S}{\sqrt{n} })  the <u>95 percent confidence interval for the population mean  'μ'</u>

Degrees of freedom γ=n-1=62-1=61

t₀.₀₅ = 1.9996 at 61 degrees of freedom

(14 - 1.9996\frac{9.3}{\sqrt{62} } ,14+1.9996\frac{9.3}{\sqrt{62} })

(14-2.361 , 14 + 2.361)

[(11.64 , 16.36]

<u>Conclusion:-</u>

95 percent confidence interval for this single-sample t test

[11.64, 16.36]

<u></u>

6 0
3 years ago
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