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3 years ago

Buddie is filing his federal income tax

Mathematics

2 answers:

nexus9112 [7]3 years ago

5 0

Answer:

$43,980

a p e x

Alexxandr [17]3 years ago

4 0

Answer: ????

Step-by-step explanation:

CAN YOU EXPLAIN WHAT THE QUESTION IS?

IS THERE ANY BACKGROUND INFORMATION ON THIS QUESTION?

DO YOU KNOW THE ANSWER?

WHERE DID THE PROBLEM COME FROM?

WITHOUT THE INFORMATION, WE ARE NOT ABLE TO ANSWER THE QUESTION.

You might be interested in

Find the volume in cubic centimeters of a cube with a side length of 5 cm

Jobisdone [24]

Answer:

125 cm3

Step-by-step explanation:

V = l * w * h

V = 5 * 5 * 5

V = 125 cm3

7 0

2 years ago

Read 2 more answers

The area of two rectangles is given by he functions: area of a rectangle A:f(x)= 4x2+6x area of rectangle B:g(x)=3x2-x Which fun

dusya [7]

Answer:

x^2+7x if you are asked to find the difference of a function f and function g

Step-by-step explanation:

We are asked to A-B or f(x)-g(x).

(4x^2+6x)-(3x^2-x)

4x^2+6x-3x^2+x

The like terms I'm going to pair up.

4x^2-3x^2+6x+x

1x^2 +7x

x^2 +7x

The answer is x^2+7x

7 0

3 years ago

Read 2 more answers

Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that

amm1812

Answer:

The value is $P(| \^ p - p| < 0.05 ) = 0.9822$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.52$

The sample size is n = 563

Generally the population mean of the sampling distribution is mathematically represented as

$\mu_{x} = p = 0.52$

Generally the standard deviation of the sampling distribution is mathematically evaluated as

$\sigma = \sqrt{\frac{ p(1- p)}{n} }$

=> $\sigma = \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }$

=> $\sigma = 0.02106$

Generally the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

$P(| \^ p - p| < 0.05 ) = P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 ))$

Here $\^ p$ is the sample proportion of persons with a college degree.

$P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma } < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )$

Here

$\frac{[\^p - p] - p}{\sigma } = Z (The\ standardized \ value \ of\ (\^ p - p))$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 } < Z < \frac{-0.47 + 0.52}{0.02106 }]$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[ -2.37 < Z < 2.37 ]$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(Z < 2.37 ) - P(Z < -2.37 )$

From the z-table the probability of (Z < 2.37 ) and (Z < -2.37 ) is

$P(Z < 2.37 ) = 0.9911$

and

$P(Z < - 2.37 ) = 0.0089$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) =0.9911-0.0089$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = 0.9822$

=> $P(| \^ p - p| < 0.05 ) = 0.9822$

3 0

3 years ago

What is the percent increase between getting a high school scholarship and bachelors degree when high school scholarship is $421

gogolik [260]

Answer:

The percent increase between getting a high school scholarship and bachelor's degree is <u>59.14%</u>.

Step-by-step explanation:

Given:

High school scholarship is $421.

Bachelor's degree is $670.

Now, to find the percent increase between a high school scholarship and bachelor's degree.

So, we get the amount of increase between a high school scholarship and bachelor's degree.

$670-421=249.$

<em>Thus, the amount of increase = $249</em>.

Now, to get the percent increase between a high school scholarship and bachelor's degree:

$\frac{249}{421}\times 100$

$=\frac{24900}{421}$

$=59.14\%.$

Therefore, the percent increase between getting a high school scholarship and bachelor's degree is 59.14%.

7 0

3 years ago

Before sending track and field athletes to the Olympics, the U.S. Holds a qualifying meet. The box plots below show the distance

sleet_krkn [62]

Question:

The options are;

A. The distances in the Olympic final were farther on average.

B. The distances in the Olympic final varied noticeably more than the US qualifier distances

C. The distances in the Olympic final were all greater than the US qualifier distances

D. none of the above

Answer:

The correct option is;

A. The distances in the Olympic final were farther on average.

Step-by-step explanation:

From the options given, we have

A. The distances in the Olympic final were farther on average.

This is true as the sum of the 5 points divided by 5 is more in the Olympic final

B. The distances in the Olympic final varied noticeably more than the US qualifier distances

This is not correct as the difference between the upper and lower quartile in the Olympic final is lesser than in the qualifier

C. The distances in the Olympic final were all greater than the US qualifier distances

This is not correct as the max of the qualifier is more than the lower quartile in the Olympic final

D. none of the above

We have seen a possible correct option in option A

7 0

3 years ago

Read 2 more answers