A teacher weighed 145 lbs in 1986 and weighs 190 lbs in 2007. What is the rate of change in weight? HELP

In the first 42 Super bowls, 0.16 of the MVPs were runner backs.

zhenek [66]

The answer for question (1) 16%

3 0

3 years ago

Single Factor ANOVA is a method we use when we want to compare a quantitative variable among more than two categories.

Goryan [66]

Answer:

The given statement is "True".

Step-by-step explanation:

A statistical analytical technique that divides the perceived aggregate unpredictability present throughout an information source even further into 2 components.

Systemic components
Random components

If relevant information yet another category independent variable, as well as another quantifiable variable of interest, has been obtained, are using a single-way ANOVA.

Thus the above is the true answer.

3 0

3 years ago

Will mark brainliest for whoever answers

Ostrovityanka [42]

Answer:

A

Step-by-step explanation:

A

5 0

3 years ago

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According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100:114 (46.7%

mojhsa [17]

Answer:

$z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154$

$p_v =2*P(z$

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 girls born is not significantly different from 0.467

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=63 represent the number of girls born

$\hat p=\frac{63}{150}=0.42$ estimated proportion of girls born

$p_o=0.467$ 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 if girls is 0.467.:

Null hypothesis: $p=0.467$

Alternative hypothesis: $p \neq 0.467$

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:

$z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154$

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 bilateral test the p value would be:

$p_v =2*P(z$

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 girls born is not significantly different from 0.467

3 0

3 years ago

Find the area of the following figure:

Sidana [21]

Answer: 15cm^2

Step-by-step explanation:

0.5*10*3=15

4 0

3 years ago

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