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3 years ago

14

Anabel uses 10 gallons of gas to drive 220 miles on the highway. If she travels at the same rate, how many miles can she drive o

n 7 gallons of gas

1 answer:

NeTakaya3 years ago

5 0

Answer:

196 miles

Explanation:

You can use the unitary method by finding how far she can drive on ONE gallon.

You might be interested in

In the parallelogram shown, find x.

Olin [163]

The side with all the 9s and x is 12 just like its opposite side because this is a parallelogram. We can write the equation to add up the three segments; the ones on the left and right are 9-x:

9-x + x + 9-x = 12

18 - x = 12

x = 6

4 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So $\mu = 373, \sigma = 67$

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So $n = 5, s = \frac{67}{\sqrt{5}} = 29.96$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 373}{29.96}$

$Z = 1.23$

$Z = 1.23$ has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So $\mu = 526, \sigma = 107$

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So $n = 5, s = \frac{107}{\sqrt{5}} = 47.86$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 526}{47.86}$

$Z = -2.42$

$Z = -2.42$ has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

7 0

3 years ago

100 POINTS PRE-CALCULUS

Sonja [21]

Answer:

not sure for the 1st one but pretty sure 2nd one is B

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Samuel surveys 15 students at his school and asks them: "How many candies per day do you usually eat?"

shepuryov [24]

Answer:

Minimum number of candies:

0

Maximum number of candies:

13

Range:

13

Median number of candies:

4

Interquartile range:

5

Step-by-step explanation: There u go

6 0

3 years ago

What is the solution to x^6 – 6x ^5 + 15x ^4 – 20x ^3 + 15x^ 2 – 6x + 1 ≥ 0? x = 0 x = 1 all real numbers all real numbers excep

g100num [7]

Answer:

Answer: C all real numbers

8 0

2 years ago