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Alex_Xolod [135]
2 years ago
5

We expect a car’s highway gas mileage to be related to its city gas mileage (in mpg). Data for all 1209 vehicles in the governme

nt’s 2016 Fuel Economy Guide give the regression linehighway mpg=7.903+(0.993×city mpg)for predicting highway mileage from city mileage.(a) What is the slope of this line? (Enter your answer rounded to three decimal places.)slope:
Mathematics
1 answer:
seropon [69]2 years ago
8 0

Step-by-step explanation:

thejwksma naanansniewkna

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What is the maximum number of pages the report can have so that it can be completely stored on a USB drive that holds 4 \cdot 10
lana [24]

Answer:

<em>2,300,000pages  report.</em>

Step-by-step explanation:

Given a USB drive with memory of 4.6*10⁶KB, in order to know the maximum number of pages a report can have so that it can be completely stored on the drive, the conversion factor must be used.

We must first understand that 1 page of a document uses up approximately 2kB of the storage.

Since 2kB = 1page

4.6*10⁶KB = x page

Cross multiply

2kB * x = 4.6*10⁶KB * 1

2x =  4.6*10⁶

x =  4.6*10⁶/2

x =  2.3*10⁶

x = 2,300,000

<em>Hence the maximum number of pages that the report can have so that it can be completely stored on a USB drive that holds 4.6*10⁶KB is approximately 2,300,000pages  report.</em>

4 0
2 years ago
Identify the 35th term of an arithmetic sequence where a1 = −7 and a18 = 95.
borishaifa [10]
Th term of an arithmetic sequence:

We have to find the difference (d)

an=a₁+(n-1)d

Data:
a₁=-7
a₁₈=95

95=-7+(18-1)d
95+7=17d
17d=102
d=102/17
d=6

Now, we can calculate the 35 th term of an arithmetic sequence:
a₃₅=-7+(35-1)*6
a₃₅=-7+34*6
a₃₅=-7+204
a₃₅=197

Answer: the 35th term of this arithmetic sequence is 197. (a₃₅=197)

4 0
3 years ago
Rate this pic lol. I will give u a brainlist
nika2105 [10]
It is a good pic, good quality and all that
7 0
2 years ago
Let X represent the amount of gasoline (gallons) purchased by a randomly selected customer at a gas station. Suppose that the me
Alexus [3.1K]

Answer:

a) 18.94% probability that the sample mean amount purchased is at least 12 gallons

b) 81.06% probability that the total amount of gasoline purchased is at most 600 gallons.

c) The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums, we can apply the theorem, with mean \mu and standard deviation s = \sqrt{n}*\sigma

In this problem, we have that:

\mu = 11.5, \sigma = 4

a. In a sample of 50 randomly selected customers, what is the approximate probability that the sample mean amount purchased is at least 12 gallons?

Here we have n = 50, s = \frac{4}{\sqrt{50}} = 0.5657

This probability is 1 subtracted by the pvalue of Z when X = 12.

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

By the Central Limit theorem

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

Z = \frac{12 - 11.5}{0.5657}

Z = 0.88

Z = 0.88 has a pvalue of 0.8106.

1 - 0.8106 = 0.1894

18.94% probability that the sample mean amount purchased is at least 12 gallons

b. In a sample of 50 randomly selected customers, what is the approximate probability that the total amount of gasoline purchased is at most 600 gallons.

For sums, so mu = 50*11.5 = 575, s = \sqrt{50}*4 = 28.28

This probability is the pvalue of Z when X = 600. So

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

Z = \frac{600 - 575}{28.28}

Z = 0.88

Z = 0.88 has a pvalue of 0.8106.

81.06% probability that the total amount of gasoline purchased is at most 600 gallons.

c. What is the approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers.

This is X when Z has a pvalue of 0.95. So it is X when Z = 1.645.

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

1.645 = \frac{X- 575}{28.28}

X - 575 = 28.28*1.645

X = 621.5

The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.

5 0
2 years ago
Help me with this i only have 10 mis left
Tpy6a [65]

Answer:

It's definitely C

Step-by-step explanation:

Boston is 35 degrees as well but the lowest temperature of Seattle is 45

4 0
2 years ago
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