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WARRIOR [948]
3 years ago
5

HELP TIMED TEST!!!

Mathematics
1 answer:
sdas [7]3 years ago
7 0
B is the correct answer
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Consider this right triangle. Determine whether each equation is correct. Select yes or no using the dropdown for each equation.
Romashka [77]

Answer:

Yes

Yes

NO    CosD = 5/13

Yes

5 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
3 years ago
7n + 6(n+4) -15 <br><br> how do i solve this with a step by step example
GrogVix [38]

Answer:

  13n +9

Step-by-step explanation:

There is nothing to solve. We can simplify the expression by eliminating parentheses and combining like terms.

  7n +6(n +4) -15 . . . . . . given

  7n +6n +6(4) -15 . . . . use the distributive property

  (7 +6)n +(24 -15) . . . .identify and group like terms

  13n +9 . . . . . . . . . . combine like terms

8 0
2 years ago
Solve the system 6x -2y+z= -2 2x+ 3y - 3z =11 x+ 6y=31
weqwewe [10]

Answer:

x = 1

y = 5

z = 2

Step-by-step explanation:

System of equations:

6x - 2y + z = -2

2x + 3y - 3z = 11

x + 6y = 31

Isolate one variable in any of the equations:

x + 6y = 31

x = 31 - 6y

Plug in this value for x in another equation:

6(31 - 6y) - 2y + z = -2

186 - 36y - 2y + z = -2

186 - 38y + z = -2

-38y + z = -188

z = -188 + 38y

Plug in these values in the remaining equation:

2(31 - 6y) + 3y - 3(-188 + 38y) = 11

62 - 12y + 3y + 564 - 114y = 11

626 - 12y + 3y - 114y = 11

626 - 9y - 114y = 11

626 - 123y = 11

-123y = -615

y = 5

Plug in value of y into our other answers to solve for x and z:

x = 31 - 6(5)

x = 31 - 30

x = 1

z = -188 + 38(5)

z = -188 + 190

z = 2

Check your work:

6x - 2y + z = -2

6(1) - 2(5) + 2 = -2

6 - 10 + 2 = -2

-4 + 2 = -2

-2 = -2

Correct!

*Note there are several ways to solve for these types of problems. I used substitution*

8 0
3 years ago
-43x - 121 = 0. Type ONLY THE NUMBER for the answer!!
Elza [17]

Answer:

Step-by-step explanation:

The answer is -2.81

Plz mark me brainliest.

7 0
2 years ago
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