Please help!!!!! will mark brainliest

Lori is running in a marathon, which is 28.4 miles she ran 0.1 of it already. How far has Lori run?

hoa [83]

So she ran the 0.1 of the 28.4
That is she ran 1% of 28.4 miles
0.1 * 28.4 is 2.84

Hope this helps :)

7 0

2 years ago

Find each function value 1. f(7) if (x)=5x 2. f(4) if f(x)=3x-1

Inessa [10]

ANSWER

$1. \: \: f(7) = 35$

$2. \: f(4) = 11$

EXPLANATION

To find the value of a function, f(x) at a given value x=a, we plug in x=a into f(x) to obtain f(a) in simplified form.

The first function is

$f(x) = 5x$

To find f(7), we substitute x=7, into the function.

$\implies f(7) =5(7) = 5 \times 7 = 35$

The second function is

$f(x) = 3x - 1$

To find f(4), we replace x with 4 to obtain,

$f(4) = 3(4) - 1$

$f(4) = 12 - 1 = 11$

6 0

2 years ago

A nationwide study of American homeowners revealed that 65% have one or more lawn mowers. A lawnequipment manufacturer, located

GenaCL600 [577]

Answer:

$z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589$

$p_v =P(z>1.589)=0.056$

If we compare the p value obtained with the significance level assumed $\alpha=0.05$ 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 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

Step-by-step explanation:

Data given and notation

n=497 represent the random sample taken

X=340 represent the homes in Omaha with one or more lawn mowers

$\hat p=\frac{340}{497}=0.684$ estimated proportion of homes in Omaha with one or more lawn mowers

$p_o=0.65$ is the value that we want to test

$\alpha$ represent the significance level

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 true proportion of homes in Omaha with one or more lawn mowers is higher than 0.65.:

Null hypothesis: $p\leq 0.65$

Alternative hypothesis: $p > 0.65$

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:

$z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589$

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

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

$p_v =P(z>1.589)=0.056$

If we compare the p value obtained with the significance level assumed $\alpha=0.05$ 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 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

7 0

2 years ago

In a sample of 388 people in a certain, 22% had red hair. Find a 99% confidence interval for the proportion of people in the cou

Anit [1.1K]

Answer: (0.1658, 0.2742)

Step-by-step explanation:

Formula to find the confidence interval for population proportion is given by :-

$\hat{p}\pm z^*\sqrrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$