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3 years ago

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Arnold is planning on painting a mural. Before painting the mural, Arnold painted the picture on a paper that is 5 inches wide a

nd 11 inches tall. Arnold is using a scale in which 12 inch represents 1 foot to paint the mural.
How wide and tall will the actual mural be?

1 answer:

snow_tiger [21]3 years ago

3 0

0.42ft
0.92 ft
Hope it helps

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You are running a fuel economy study. One of the cars you find is blue. It can travel 41 1/2 on 1 1/4 of gasoline. Another car i

kogti [31]

Answer: i get point

Step-by-step explanation:

Blue car: 30.5 / 1.25 =24.4 miles per gallon

Red car: 25.6/0.80 =32 miles per gallon

You can easily see that the Red car travel farther on 1 gallon of gasoline.

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3 years ago

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The cost of a pizza with the works" is represented as a function of its

Nina [5.8K]

Answer:

diameter of the pizza = 21.88 centimeters

Step-by-step explanation:

Given:

C = d^2 – 2d + 447

where

C = cost of the pizza

d = diameter of the pizza

If the pizza costs $12.00, then what is a reasonable estimate for the diameter of the pizza?

12 = d^2 – 2d + 447

d^2 - 2d = 447 - 12

d^2 - 2d = 435

d^2 - 2d - 435 = 0

Solve the quadratic equation using formula

a = 1

b = -2

c = -435

d = -b +or- √b^2 - 4ac / 2a

= -(-2) +or- √(-2)^2 - (4)(1)(-435) / 2(1)

= 2 +or- √4 - (-1740) / 2

= 2 +or- √4 + 1740 / 2

= 2 +or- √1744 / 2

= 2 +or- 4√109 / 2

= 2/2 +or- 4√109/2

= 1 +or- 2√109

d = 1 + 2√109 or d = 1 - 2√109

= 1 + 2(10.44) or d = 1 - 2(10.44)

= 1 + 20.88 or d = 1 - 20.88

d = 21.88 or -19.88

diameter of the pizza = 21.88 centimeters

Therefore, the estimated diameter of the pizza can not be negative. So, diameter of the pizza = 21.88 cm

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2 years ago

8. In AABC, LA is a right angle and mzB = 45°. Find BC.

Margaret [11]

Answer:

BC = 11 $\sqrt{2}$

Step-by-step explanation:

using the sine ratio in the right triangle and the exact value

sin45° = $\frac{1}{\sqrt{2} }$ , then

sin 45° = $\frac{opposite}{hypotenuse}$ = $\frac{AC}{BC}$ = $\frac{11}{BC}$ = $\frac{1}{\sqrt{2} }$ ( cross- multiply )

BC = 11 $\sqrt{2}$

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2 years ago

The measures of two complementary angles are m angle 1 = (10x+7) and m angle 2=(9x-12) find the measures of both angles

USPshnik [31]

Answer:

m<1 = 57°

m<2 = 33°

Step-by-step explanation:

To find the numerical measure of both angles, let's come up with an equation to determine the value of x.

Given that m<1 = (10x +7)°, and m<2 = (9x - 12)°, where both are complementary angles, therefore, it means, both angles will add up to give us 90°.

Equation we can generate from this, is as follows:

(10x + 7)° + (9x - 12)° = 90°

Solve for x

10x + 7 + 9x - 12 = 90

Combine like terms

19x - 5 = 90

Add 5 to both sides

19x = 90 + 5 (addition property not equality)

19x = 95

Divide both sides by 19

x = 5

m<1 = (10x +7)°

Replace x with 5

m<1 = 10(5) + 7 = 50 + 7 = 57°

m<2 = (9x - 12)

Replace x with 5

m<2 = 9(5) - 12 = 45 - 12 = 33°

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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.

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2 years ago