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2 years ago

12

ASAP PEEPS!!! Mr. Smith received a bonus of $596 from his employer. He has to pay 34% of his bonus to taxes. About how much will

Mr. Smith pay in taxes. Select the best answer.

2 answers:

aleksandrvk [35]2 years ago

7 0

$202.64 34% of 596 is $202.64

irinina [24]2 years ago

7 0

The answer is 202.64

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Which of the following is a root of the polynomial shown below f(x)=x^3+2x^2-x-2

Aleks [24]

F(x) = x^3 + 2x^2 - x - 2
<span>f(x) = x^2(x + 2) - (x + 2) </span>
<span>f(x) = (x^2 - 1)(x + 2) </span>
<span>f(x) = (x + 1)(x - 1)(x + 2) </span>
<span>Roots: -1, 1, -2 </span>

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3 years ago

Read 2 more answers

there are 105 teachers in a school 19 more are femal than male how what is the proprtion of female teachers

IRINA_888 [86]

Answer: 62

Explanation: 105 - 19 = 86 (Find out how much the college would have without the more)
86 ÷ 2 = 43 (Find out male and female teachers in a seperate way without more)
43 + 19 = 62 (Female teachers)

8 0

2 years ago

Please help. I’ll mark you as brainliest if correct.

OlgaM077 [116]

Answer:

4600

Step-by-step explanation:

it is simply the sum of home credit which is

=3000 + 1600

=4600

5 0

3 years ago

I’m suppose to round 9955 to the nearest hundred but I don’t know how help please

Kobotan [32]

Answer:

The answer would be 10,000.

Hope this helps!

4 0

3 years ago

Suppose a poll is taken that shows that 765 out of 1500 randomly selected, independent people believe the rich should pay more

Zanzabum

Answer:

$z=\frac{0.51 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1500}}}=0.775$

$p_v =P(z>0.775)=0.219$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly higher than 0.5

Step-by-step explanation:

Data given and notation

n=1500 represent the random sample taken

X=765 represent the successes

$\hat p=\frac{765}{1500}=0.51$ estimated proportion of successes

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.51 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1500}}}=0.775$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>0.775)=0.219$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly higher than 0.5

8 0

3 years ago