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jekas [21]
3 years ago
13

A rectangle is removed from a right triangle to create the shaded region shown below. Find the area of the shaded region.

Mathematics
1 answer:
irina [24]3 years ago
8 0

Answer:

I can't tell what the triangle is

Step-by-step explanation:

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Plzzzz help I am stuck on number 4
Nat2105 [25]

Answer:

lol nah <3

Step-by-step explanation:

i would try and help, but not with that trump pfp my dude

8 0
3 years ago
In the figure below, two lines intersect, forming four angles. Write your answer in the space provided.
Maurinko [17]

Answer:

Step-by-step explanation:

Angle A and 50 will combined have an angle of 180

Angle A = 180-50

Angle C is opposite of angle A and both will be equal. Same with Angle B and it's opposing Angle.

You got this!

7 0
3 years ago
A presidential candidate's aide estimates that, among all college students, the proportion who intend to vote in the upcoming el
inessss [21]

Answer:

z=\frac{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24  

p_v =P(z  

If we compare  the p value and the significance level given \alpha=0.01 we see that 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 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6

Step-by-step explanation:

Assuming the following question: A presidential candidate's aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at least 60% . If 127 out of a random sample of 240 college students expressed an intent to vote, can we reject the aide's estimate at the 0.1 level of significance?

Data given and notation

n=240 represent the random sample taken

X=127 represent the college students expressed an intent to vote

\hat p=\frac{127}{240}=0.529 estimated proportion of college students expressed an intent to vote

p_o=0.6 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

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 at least 60% of students are intented to vote .:  

Null hypothesis:p \geq 0.6  

Alternative hypothesis:p < 0.6  

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{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24  

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=0.01. The next step would be calculate the p value for this test.  

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

p_v =P(z  

If we compare  the p value and the significance level given \alpha=0.01 we see that 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 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6

7 0
3 years ago
The proportion of junior executives leaving large manufacturing companies within three years is to be estimated within 3 percent
Leona [35]

Answer: 709

Step-by-step explanation:

The formulas we use to find the required sample size :-

1. n=(\dfrac{z^*\cdot \sigma}{E})^2

, where \sigma = population standard deviation,

E = Margin of error .

z* = Critical value

2. n=p(1-p)(\dfrac{z^*}{E})^2 , where p= prior estimate of population proportion.

3. If prior estimate of population proportion is unavailable , then we take p= 0.5 and the formula becomes

n=0.25(\dfrac{z^*}{E})^2

Given : Margin of error : E= 3% =0.03

Critical value for 95% confidence interval = z*= 1.96

A study conducted several years ago revealed that the percent of junior executives leaving within three years was 21%.

i.e. p=0.21

Then by formula 2., the required sample size will be :

n=0.21(1-0.21)(\dfrac{1.96}{0.03})^2

n=0.21(0.79)(65.3333)^2

n=(0.1659)(4268.44008889)\\\\ n=708.134933333\approx709 [Round to the next integer.]

Hence, the required sample size of junior executives should be studied = 709

4 0
4 years ago
If WZ is the perpendicular bisector of VY , what conclusion can you make?
BaLLatris [955]
<span>I think it's right. A. ZY=VZ</span>
8 0
3 years ago
Read 2 more answers
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