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Mrac [35]
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
Mathematics
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
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