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Brrunno [24]
3 years ago
8

How do you solve -3+r/3=-5

Mathematics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

r=-6

Step-by-step explanation:

Question: -3+r/3=-5

1) Multiply both sides of the equation by 3:

-9+r=-15

2) Add 9 to both sides:

r=-6

Footnotes:

1) In step 1 when I multiplied both sides of the equation by 3, don't forget you also have to multiply -3 by 3 because it is not part of the fraction (which is how I got -9)

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Grace drew a triangle with sides of equal length. Its measured with sides of 3 cm. What is the perimeter of the triangle?
Otrada [13]

Answer:

9 cm explained to

Step-by-step explanation:

explained that

4 0
3 years ago
20 POINTS HELPPPPP
kiruha [24]

For this case we have the following expression:

\frac {216 ^ {n-2}} {(\frac {1} {36}) ^ {3n}} = 216

We multiply both sides by: (\frac {1} {36}) ^ {3n}

216 ^ {n-2} = 216 * (\frac {1} {36}) ^ {3n}

We divide both sides by 216:

\frac {216 ^ {n-2}} {216} = (\frac {1} {36}) ^ {3n}

To divide powers of the same base, we place the same base and subtract the exponents:

216 ^ {n-2-1} = (\frac {1} {36}) ^ {3n}\\216 ^ {n-3} = (\frac {1} {36}) ^ {3n}

Rewriting:

(6 ^ 3) ^ {n-3} = (\frac {1} {6 ^ 2}) ^ {3n}\\6 ^ {3n-9} = \frac {1} {6 ^ {6n}}\\6^{ 3n-9} * 6^{ 6n} = 1

To multiply powers of the same base, we place the same base and add the exponents:

6^{ 3n-9 + 6n} = 1\\6^{ 9n-9} = 1

We know that any number raised to zero is 1, a ^ 0 = 1.

So, for equality to be true:

9n-9 = 0\\9n = 9\\n = \frac {9} {9}\\n = 1

Answer:

n = 1

3 0
3 years ago
Read 2 more answers
4/5 DIVIDED BY 16/15
Doss [256]

Answer: 3 /4

Step-by-step explanation:

Exact Form:  3 /4

Decimal Form:

0.75

6 0
4 years ago
Solve for X when the surface area is 298ft^2
jok3333 [9.3K]

Step-by-step explanation:

Total surface area of cuboid= 2(lb+bh+hl)

= 298ft²=2(28+4x+7x)

= 149=28+11x

= 121=11x

= 11=x

6 0
3 years ago
The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as t
skad [1K]

Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that \mu = 273, \sigma = 100

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that n = 30, s = \frac{100}{\sqrt{30}}

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

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

By the Central Limit Theorem

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

Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}

Z = 0.88

Z = 0.88 has a p-value of 0.8106

X = 257

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

Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}

Z = -0.88

Z = -0.88 has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that n = 50, s = \frac{100}{\sqrt{50}}

X = 289

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

By the Central Limit Theorem

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

Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}

Z = 1.13

Z = 1.13 has a p-value of 0.8708

X = 257

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

Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}

Z = -1.13

Z = -1.13 has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that n = 100, s = \frac{100}{\sqrt{100}}

X = 289

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

By the Central Limit Theorem

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

Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}

Z = 1.6

Z = 1.6 has a p-value of 0.9452

X = 257

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

Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}

Z = -1.6

Z = -1.6 has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

6 0
3 years ago
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