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dybincka [34]
2 years ago
8

The average cost to own and operate a car in the United States for 2020 was $0.524 per mile.

Mathematics
1 answer:
sp2606 [1]2 years ago
3 0

Answer:

y = 0.524m + b is the answerrrr

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PLEASE HELP!! DUE SOON
mrs_skeptik [129]

Answer:

b=3a/8

a=8b/3

Step-by-step explanation:

b/a=k ( k is the constant proportional)

3/8=9/24=15/40

b/a=3/8

b=3a/8

a=8b/3

I hope this is right answer

7 0
3 years ago
How many 4/5 cup servings in 8 cups
Molodets [167]
To solve this question, we use simple division:
\frac{8}{ \frac{4}{5} } = 8* \frac{5}{4} = \frac{40}{4} = 10

So, there are 10 servings of 4/5 of a cup in 8 cups.
5 0
3 years ago
Read 2 more answers
Can anyone tell me what x equals in this scenario.<br><br> 3/4x=2x-5
Dvinal [7]

Hey There!

The answer to your problem is x=4

<u>Subtract 2x from both sides</u>

<u></u>3/4x-2x=2x-5-2x<u></u>

<u></u>-5/4x=-5<u></u>

<u></u>

<u>Multiply both sides by 4/(-5)</u>

<u></u>(4/-5)*(-5/4)x=(4/-5)*(-5)

3 0
3 years ago
Divide 114÷7 explain the anwser
bezimeni [28]
To get the answer you must see how many times 7 can go in 114
5 0
3 years ago
Read 2 more answers
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab
Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, H_0 : \mu = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, H_1 : \mu < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

       T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = \frac{738.44-750}{\frac{38.2}{\sqrt{50} } } ~ t_4_9

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0
3 years ago
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