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cupoosta [38]
1 year ago
8

Mr. Hollins determines that he gives away $800 each month. If he gives away 16% of his budget, then how much is his overall budg

et
Mathematics
1 answer:
Fynjy0 [20]1 year ago
6 0

Answer:

5000 dollars

Step-by-step explanation:

Let his budget be x

16% of x is 800 dollars

.16 * x = 800

Divide each side by .16

.16x/.16 = 800/.16

x=5000

His budget is 5000 dollars

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What does it mean to eliminate a variable?
uranmaximum [27]

Answer:

The elimination method for solving systems of linear equations uses the addition property of equality. You can add the same value to each side of an equation.

So if you have a system: x – 6 = −6 and x + y = 8, you can add x + y to the left side of the first equation and add 8 to the right side of the equation. And since x + y = 8, you are adding the same value to each side of the first equation.

7 0
2 years ago
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

To solve this question, we need to understand 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 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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
3 years ago
PLEASE HELP ASAP!!!! BRAINLIEST ANSWER AWAITS YOU +30 POINTS!!
tester [92]
C is the correct answer



Hope this helps :D
8 0
3 years ago
Read 2 more answers
What is the surface area of the prism in square inches?
Kay [80]

<u>Given</u>:

Given that the triangular prism with height 10 inches.

The side lengths of the base of the triangle are 12 inches, 13 inches and 5 inches.

We need to determine the surface area of the prism.

<u>Surface area of the prism:</u>

The surface area of the prism can be determined using the formula,

SA=bh+(s_1+s_2+s_3)H

where b is the base and h is the height of the triangle.

s₁, s₂, s₃ are the side lengths of the triangle and

H is the height of the prism.

Substituting b = 12, h = 5, s₁ = 12, s₂ = 5, s₃ = 13 and H = 10 in the above formula, we get;

SA=(12)(5)+(12+5+13)(10)

SA=60+(30)(10)

SA=60+300

SA=360 \ in^2

Thus, the surface area of the triangular prism is 360 square inches.

Hence, Option b is the correct answer.

8 0
3 years ago
Quality control is an important issue at ACME Company which manufacturers light bulbs. In order to conduct testing of the life h
mrs_skeptik [129]

Answer: 3424.28

Step-by-step explanation:

Given data : 378, 361, 350, 375, 200, 391, 375, 368,  321

Number of data values = 9

Mean :\overline{x}=\dfrac{\sum^n_{i=1} x_i}{n}

\\\\=\dfrac{378+361+350+375+200+391+375+368+321}{9}\\\\=\dfrac{3119}{9}\approx346.56

Variance = \dfrac{\sum^n_{i=1} (x_i-\overline{x})^2}{n-1}

\sum^n_{i=1} (x_i-\overline{x})^2 = (31.44)^2+(14.44)^2+(3.44)^2+(28.44)^2+(-146.56)^2+(44.44)^2+(28.44)^2+(21.44)^2+(-25.56)^2\\\\=27394.2224

Variance = \dfrac{27394.2224}{8}=3424.2775\approx3424.28

4 0
3 years ago
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