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3 years ago

7. Matthew drove 350 miles on 10 gallons of gasoline. How many miles per gallon does

Mathematics

1 answer:

FinnZ [79.3K]3 years ago

6 0

Answer:

Step-by-step explanation:

Just divide 350 by 10 since it asks, "How many miles per gallon".

= 35

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A license plate is to have the following form: 4 letters followed by 4 numbers. How many different license plates can be made, a

Darina [25.2K]

Answer ≈ 4 and a half billion

If we assume your letters range from A to Z and your digits from 0 to 9, you have 26 possibilities per letter and 10 per digit. This gives you 26^4*10^4 which is roughly 4 billion and a half.

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3 years ago

Please help!

Kobotan [32]

Answer:

You have three card : 3, 4, and 5.

You first pick one card randomly, the probability of picking a 3 is:

P = 1/3

You then pick a second card without putting the first card back, the probability of picking a 5 is:

P = 1/2

=> The probability of picking a 3 and then picking a 5:

P = (1/3) x (1/2) = 1/6

Hope this helps!

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3 years ago

I need help you solve this by using elimination and substitution thank you

AnnZ [28]

Answer: Just use MathPapa.

Step-by-step explanation:

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3 years ago

Read 2 more answers

Jack usually mows his lawn in 6 hours. marilyn can mow the same yard in 4 hours. How much time would it take for them to mow the

Anuta_ua [19.1K]

Answer:

2.4 hours or 2 hours and 24 minutes

Step-by-step explanation:

Jack mows 1/6 of the lawn per hour while Marilyn mows 1/4 of the lawn together. Adding up those rates yields in their combined mowing rate:

$R = \frac{1}{6}+ \frac{1}{4}\\R=\frac{10}{24}$

The time required for them to mow the entire lawn is:

$t = \frac{1}{R}\\t = \frac{1*24}{10}\\ t=2.4\ hours$

It would take them both 2.4 hours to mow the lawn together.

5 0

3 years ago

Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.

zysi [14]

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step explanation:

Part a

For this case we have the following data given

$\bar X = 78.25$ represent the sample mean for the customer order totals

$\sigma =22.50$ represent the population deviation

$n= 40$ represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be $\alpha=0.1$ and $\alpha/2 = 0.05$ and the critical value from the normal standard distirbution would be given by:

$z_{\alpha/2}=1.64$

And the confidence interval is given by:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

3 0

3 years ago