There are 6 squares and 9 circles. What is the simplest ratio of squares to total shapes?

A group of retired admirals, generals, and other senior military leaders, recently published a report, "Too Fat to Fight". The r

weqwewe [10]

Answer:

$z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735$

$p_v =P(z$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .

Step-by-step explanation:

1) Data given and notation

n=180 represent the random sample taken

X=125 represent the number of americans between 17 to 24 that not qualify for the military

$\hat p=\frac{125}{180}=0.694$ estimated proportion of americans between 17 to 24 that not qualify for the military

$p_o=0.75$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :

Null hypothesis: $p\geq 0.75$

Alternative hypothesis: $p < 0.75$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735$

4) 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 left tailed test the p value would be:

$p_v =P(z$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .

6 0

3 years ago

85 is greater than the difference of x and63

miss Akunina [59]

85 > x-63

This is how you would arrange the problem,

isolate the X by adding 63 to 85 to get 148,
Flip the inequality to <

resulting in X<148.

8 0

4 years ago

Find the slope between these two points: (5,0) and (-1,12)

Ede4ka [16]

The slope of this function is -2

7 0

3 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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7 0

4 years ago

Each type of cable has a prescribed bend radius , which is the radius of the maximum arc into which you can loop a cable without

Solnce55 [7]

The answer is true i believe.

6 0

4 years ago