5TH GRADE MATH! HELP! What is written in standard form? A. 0.8 B. 0.871 C. 871 D. 8,710

Think of all of the different games that have been presented in this lesson: rolling a die, using a

jarptica [38.1K]

Hahahhahshhshshshsjsmsnbdbdmdmmsorry

3 0

2 years ago

-6 - (-7) The answer is not -13 Or -1 Or 13 I’m not sure how to do it :(

murzikaleks [220]

its 1, if its not, your school's thing sucks and needs to be reprogramed

3 0

3 years ago

Read 2 more answers

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

It is generally claimed that the average body temperature for healthy human adults is 98.6°F. To test this claim a simple random

Rzqust [24]

Answer:

Explained below.

Step-by-step explanation:

The information provided is as follows:

$\mu=98.6^{o}F\\\bar x=98.1^{o}F\\s=0.9^{o}F\\n=9$

(1)

A single mean test is to be performed in this case.

As the population standard deviation is not provided, a one-sample t-test will be used.

The correct option is b.

(2)

The null hypothesis is:

H₀: The average temperature in the population is 98.6°F, i.e. μ = 98.6°F.

The correct option is b.

(3)

The alternative hypothesis is:

Hₐ: The average temperature in the population is less than 98.6°F, i.e. μ < 98.6°F.

The correct option is c.

(4)

The standard deviation of the sample mean is as follows:

$SD_{\bar x}=\frac{s}{\sqrt{n}}=\frac{0.9}{\sqrt{9}}=0.3^{o}F$

Thus, the value of SD is 0.3°F.

(5)

Compute the value of test statistic as follows:

$t=\frac{\bar x-\mu}{SD_{\bar x}}$

$=\frac{98.1-98.6}{0.3}\\\\=-1.666666667\\\\\approx -1.67$

Thus, the value of test statistic is -1.67.

(6)

The degrees of freedom of the test are:

df = n - 1

= 9 - 1

= 8

Thus, the degrees of freedom of the test is 8.

8 0

3 years ago

IF CORRECT ILL MARK BRAINLIEST! USE THE NET TO FIND THE SURFACE AREA OF THE REGULAR PYRAMID

Phoenix [80]

Answer:

275.63 mm^2

Step-by-step explanation:

The measurements of the pyramids are

Edge length: 10
Egde width: 8.66 (because 10x8.66/2 = 43.3)
Height: 9

To find the surface area, use this formula, and you will get 275.63:

3 0

2 years ago