Evaluate ( r - s ) 3 t r 2 r =-3 s=-4

What is the area of this figure. 128 in 136 in 153 in 258 in

aliya0001 [1]

So, to find the solution to this problem, we will we using pretty much the same method we used in your previous question. First, let's find the area of the rectangle. The area of a rectangle is length x width. The length in this problem is 16 and the width is 3, and after multiplying these together, we have found 48 in^2 to be the area of the square. Next, we can find the area of the trapezoid. The area of a trapezoid is ((a+b)/2)h where a is the first base, b is the second base, and h is the height. In this problem, a=16, b=5, and h=10. So, all we have to do is plug these values into the area formula. ((16+5)/2)10 = (21/2)10 = 105. So, the area of the trapezoid is 105 in^2. Now after adding the two areas together (48in^2 and 105in^2), we have found the solution to be 153in^2. I hope this helped! :)

6 0

3 years ago

Construct a triangle STU in which LT=60o Lu=70° and TU = 7.5 cm 60°

disa [49]

Ok first you have to multiply and get scammed

6 0

3 years ago

Use this information to answer the questions. University personnel are concerned about the sleeping habits of students and the n

Oksanka [162]

Answer:

$z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097$

$p_v =P(Z>2.097)=0.018$

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 students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Step-by-step explanation:

1) Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

$\hat p=\frac{209}{377}=0.554$ estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

$p_o=0.5$ 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)

2) 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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097$

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 right tailed test the p value would be:

$p_v =P(Z>2.097)=0.018$

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 students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

6 0

3 years ago

Find how X and y compare when log X = 6.6 and log y = 3.6

asambeis [7]

Answer

x = 3981071.7055

y = 3981.0717

x > y

Step by step explanation

First let's find the value of x and y by taking anti - log.

log X = 6.6

X = log^ -1 (6.6)

X = 3981071.7055

Similarly,

log Y = 3.6

Y = log ^(-1) 3.6

Y = 3981.0717

Once we found the values of X and Y , we can easily compare them.

Here x > y

Hope you will understand the concept.

Thank you. :)

8 0

3 years ago

A survey asked 816 college freshmen whether they had been to a movie or eaten in a restaurant during the past week. The followin

Nikitich [7]

Answer:

the answer is 90

Step-by-step explanation:

296-96=90

3 0

3 years ago