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2 years ago

5

The gross income of Maurice Vaughn is $785 per week. His deductions are $42.25, FICA tax; $90.33, income tax; 2% state tax; 1% c

ity tax; and 3% retirement fund. What is his net income for one week?

1 answer:

vivado [14]2 years ago

3 0

State tax = 2% of 785 = 0.02 x 785 = $15.70
city tax = 1% of 785 = 0.01 x 785 = $7.85
retirement fund = 3% of 785 = 0.03 x 785 = $23.55

Total deductions = $42.25 + $90.33 + $15.70 + $7.85 + $23.55 = $179.68

Net income = $785 - $179.68 = $605.32

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A bakery sells rolls in units of a dozen. The demand X (in 1000 units) for rolls has a gamma distribution with parameters α = 3,

bearhunter [10]

Answer:

The value is $E(X) = \$ 1.7067$

Step-by-step explanation:

From the question we are told that

The parameters are α = 3, θ = 0.5

The cost of making a unit on the first day is c = $2

The selling price of a unit on the first day is s = $5

The selling price of a leftover unit on the second day is v = $ 1

Generally the profit of a unit on the first day is

$p_1 = 5 - 2$

$p_1 = \$3$

The profit of a unit on the second day is

$p_2 = 1 - 2$

=> $p_2 = - \$1$

Generally the probability of making profit greater than $ 1 is mathematically represented as

$P(X > 1 ) = Gamma (X ,\alpha , \theta)$

=> $P(X > 1 ) = Gamma (1 ,3 , 0.5)$

Now from the gamma distribution table we have that

$P(X > 1 ) = 0.67668$

Generally the probability of making profit less than or equal to $ 1 is mathematically represented as

$P(X \le 1 ) = 1 - P(X > 1 )$

=> $P(X \le 1 ) = 1 - 0.67668$

=> $P(X \le 1 ) = 0.32332$

So the probability of making $3 is $P(X > 1 ) = 0.67668$

and the probability of making -$1 is $P(X \le 1 ) = 0.32332$

Generally the value of profit per day is mathematically represented as

$E(X) = 3 * P(X > 1 ) + (-1 * P(X \le 1 ) )$

=> $E(X) = 3 * 0.67668 + (-1 * 0.32332 )$

=> $E(X) = \$ 1.7067$

4 0

1 year ago

A sample of 1714 cultures from individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiot

Damm [24]

Answer:

a) $p$ represent the real population proportion of people who showed partial or complete resistance to the antibiotic

b) $\hat p=\frac{973}{1714}=0.568$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

c) We are confident (95%) that the true proportion of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic is between 54.5% to 59.1%.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

Description in words of the parameter p

$p$ represent the real population proportion of people who showed partial or complete resistance to the antibiotic

$\hat p$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part b

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

$\hat p=\frac{973}{1714}=0.568$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

The confidence interval for a proportion is given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.568 - 1.96 \sqrt{\frac{0.568(1-0.568)}{1714}}=0.545$

$0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591$

And the 95% confidence interval would be given (0.545;0.591).

We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.

5 0

2 years ago

Two sides of a triangle have lengths 12 m and 14 m. The angle between them is increasing at a rate of 2°/min. How fast is the le

Dvinal [7]

Answer:

1.692 m/min

Step-by-step explanation:

Let $\theta$ be the angle between the two sides and x be the length of the third side. By cosine rule,

$x^2 = 12^2+14^2-2\times12\times14\cos\theta = 340 - 336\cos\theta$

$x= \sqrt{340 - 336\cos\theta}$

We differentiate x with respect to $\theta$ by applying chain rule.

$\dfrac{dx}{d\theta} = \dfrac{336\sin\theta}{2\sqrt{340 - 336\cos\theta}} = \dfrac{168\sin\theta}{\sqrt{340 - 336\cos\theta}}$

Rate of change of $\theta$ is 2

$\dfrac{\theta}{dt} = 2$

Rate of change of x is

$\dfrac{dx}{dt} = \dfrac{dx}{d\theta}\times\dfrac{d\theta}{dt}$

$\dfrac{dx}{dt} = \dfrac{168\sin\theta}{\sqrt{340 - 336\cos\theta}} \times2=\dfrac{336\sin\theta}{\sqrt{340 - 336\cos\theta}}$

At 60°,

$\dfrac{dx}{dt} = \dfrac{336\sin60}{\sqrt{340 - 336\cos60}} = 1.692 \text{ m/min}$

4 0

2 years ago

A newly hired basketball coach promised a high-paced attack that will put more points on the board than the team’s previously te

puteri [66]

Answer:

a. z = 2.00

Step-by-step explanation:

Hello!

The study variable is "Points per game of a high school team"

The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)

The hypothesis is:

H₀: μ ≤ 99

H₁: μ > 99

α: 0.01

There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.

The statistic value under the null hypothesis is:

Z= X[bar] - μ = 101 - 99 = 2

σ/√n 6/√36

I don't have σ, but since this is an approximation I can use the value of S instead.

I hope it helps!

7 0

2 years ago

Part of a book fell out. The first page of this part has number 143. The last page has a number written with the same digits. Ho

Gnom [1K]

Answer:

142 pages

Step-by-step explanation:

The parameters given are

First page of part of the book available = 143

The last is numbered with the digits 143

Since the book is said to have been split into two parts with, we have that one part of the book starts from the beginning, while the other part continue from the first part stops

Number on the pages on the first part = from 1 to number on the first page on the second part - 1

Hence, the part of the book available is the second part and the number of pages in the first part = 1 to 142 or 142 pages.

3 0

1 year ago