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MatroZZZ [7]
3 years ago
12

Me Harris can drive an average of 33.1 miles per hour if his gas tank only holds 23 gallons how many miles can he drive

Mathematics
1 answer:
pishuonlain [190]3 years ago
8 0
Well if it is 33.1 miles per gallon in his tank, and his tank holds 23 gallons, what is 33.1*23
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Order the numbers from least to greatest. 9/5, -2.5,-1.1, -4/5, 0.8
prohojiy [21]

Answer:

-4/5, -2.5, -1.1, 0.8, 9/5

Step-by-step explanation:

7 0
3 years ago
A loan for $1,200 has an annual interest rate of 5.2%. There is a $15 processing fee to receive the loan. The loan’s APR is .
Charra [1.4K]

<u>Answer: </u>

A loan for $1,200 has an annual interest rate of 5.2%. There is a $15 processing fee to receive the loan. The loan’s APR is 5.2%

<u>Solution: </u>

Since full form of APR is Annual Percentage Rate and it is defined as rate of interest for whole year which means it is same as annual interest rate. Also processing fee is totally different component.

So APR = annual interest rate ----- eqn 1

Given that a loan of $1200 has an annual interest rate of 5.2% .

So by using eqn 1 we can say APR = annual interest rate = 5.2%

Hence APR for a loan of $1200 is 5.2%

6 0
3 years ago
at farm the ratio of cows to horses was 10:3if there were 20 cows at the farm how many horses were there
Lubov Fominskaja [6]
If you think the cows are doubled so if you double the amount of horses  so the ratio is 10:3 so 20:6 is the new ratio the answer would be 6 horses
7 0
3 years ago
In a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50. Determ
shepuryov [24]

Answer:

t-distribution should be used to construct a confidence interval.

Step-by-step explanation:

We are given that a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50.

We have to determine whether a normal distribution (Z values) or a t- distribution should be used or whether neither of these can be used to construct a confidence interval.

<em><u>Since in this question we are provided with;</u></em>

Sample average weekly food expense, \bar X = $95.60

Sample standard deviation, s = $22.50

Sample of families, n = 18

The distribution that we will use here to construct a confidence interval will be <u>t-distribution</u> because in the question we don't know anything about population standard deviation (\sigma) .

Normal distribution is used when we know population standard deviation (\sigma).

So, the pivotal quantity for confidence interval that will be used is One-sample t-test statistics;

                P.Q. = \frac{\bar X -\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

Therefore, t-distribution should be used to construct a confidence interval.

3 0
3 years ago
What is 100/7 divided by 4 1/3 simplified if you answer thank you and happy holidays
SVETLANKA909090 [29]

Answer: 300/91

Step-by-step explanation:

1. Convert 4 1/3 to an improper fraction

13/3

2. Divide, using the keep, change, flip method to divide fractions

- Keep the first fraction the same (100/7)

- Change the division sign to a multiplication sign

- Flip the second fraction (13/3 -> 3/13)

3. Plug-in and evaluate

100/7 * 3/13

300/91

If you are looking for the solution as a mixed number, it is 3 27/91

4. Simplify, if applicable. This fraction cannot be simplified any further.

7 0
3 years ago
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