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3 years ago

HELP ILL GIVE BRAINLIEST QUICK!!

Mathematics

1 answer:

Semenov [28]3 years ago

3 0

Answer:

Dan is correct

Step-by-step explanation:

First, I calculated the amount of hours that the graph shows in total, coming up with a total of 3,085 hours. I do this by multiplying the number of adults surveyed by the number of hours they spent. So for the first column, (80+70)*1, second column, (90+80)*2, third column, (135+130)*3, fourth column, (140+135)*4, and fifth column, (65+75)*5 if you get what I mean. Then I add up all values. Knowing that there were two sets of 500 American adults interviewed already, I came to the conclusion that 1,000 total adults were surveyed. Finally, I divided the total number of hours(3,085) by the total number of adults interviewed (1,000) to get a mean of 3.085 hours, rounding it to 3 hours and proving that Dan is correct.

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30 tens - 3 tens 3 tenths

lubasha [3.4K]

300-30.3=269.7 because you barrow from 3 in 300 because you can't round into 0 and that equalled 7 and than you do 9-0=9 and there's 009.7 than you do 9-3=6 and that's 069.7 and you do 2-0=2 and there's 269.7

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3 years ago

PLS HELP WILL GIVE BRAINLY!!!!!

yKpoI14uk [10]

Isn’t it choice B …….…………

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3 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:

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

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3 years ago

The values of x and y are directly and one pair of values are given.

Anestetic [448]

Answer:

The relation is $y = -1.5x$