What are the solutions of the equation 9x^4 – 2x^2 – 7 = 0? Use u substitution to solve.

36 is 25% of what number?

DedPeter [7]

Answer:

36 is 25% of 144

Step-by-step explanation:

25% means 25/100 or 1/4. Thus, 36 is 1/4 of 36*4 = 144

4 0

2 years ago

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A department store buys 300 shirts for a total cost of $7,200 and sells them for $30 each. Find the percent markup.

Novosadov [1.4K]

Answer:

80%

Step-by-step explanation:

because 7200 / 300 equals 24 and 24 / 30 equals .8 so its 80% and if you want to check the answer you take 30 * .8 OR 80% and it gives you 24

3 0

3 years ago

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The time it takes Natalie to replace x cell phone screens is given by the equation t =

Alexxx [7]

Answer: 25 hours

Step-by-step explanation:

$t=1+3x$

8 screens

or

$x = 8$

Replace;

$t=1+3(8)$

$t=1+24$

$t=25h$

6 0

2 years ago

9 minutes =how many seconds

iogann1982 [59]

there is 60 seconds in 1 min so if you do 60x9 you should get 540

glad to help!

7 0

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:

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

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

3 years ago