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astraxan [27]
3 years ago
11

Jonathan's car used 3 gallons to travel 45 miles. How many miles can the car go for each gallon of gas?

Mathematics
2 answers:
vredina [299]3 years ago
6 0

Answer:

Jonathans car can go 15 miles per gallon of gas

Step-by-step explanation:

Do 45 divided by 3

Wittaler [7]3 years ago
6 0

Answer:

15 miles

Step-by-step explanation:

Take the number of miles and divide by the number of gallons

45 miles / 3 gallons

15 miles per gallon

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Answer:

b

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The table below shows the cube roots of different numbers:
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part A)

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part B)

f(x) = 10 + 20x

so if you rent the bike for a few hours that is

1 hr.............................10 + 20(1)

2 hrs..........................10 + 20(2)

3 hrs..........................10 + 20(3)

so the cost is really some fixed 10 + 20 bucks per hour, usually the 10 bucks is for some paperwork fee, so you go to the bike shop, and they'd say, ok is 10 bucks to set up a membership and 20 bucks per hour for using it, thereabouts.

f(100) = 10 + 20(100) => f(100) = 2010.

f(100), the cost of renting the bike for 100 hours.

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500 meters divided by 10 days
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Step-by-step explanation:

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Read 2 more answers
PLEASE HELP ME I BEG I WILL GIVE BRAINLIEST
ohaa [14]

Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

Step-by-step explanation:

7 0
3 years ago
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