I need help plz thank you

What is the value of x in the equation 2.5 x + 10 = x minus 0.5?

irinina [24]

Answer:

x=7

Step-by-step explanation:

$2.5x+10=x-0.5\\2.5x-(x)+10-(10)=x-(x)-0.5-(10)\\1.5x=-10.5\\x=7$

6 0

3 years ago

Read 2 more answers

For each pair of jeans i will buy 3 sweaters. how many sweaters can i buy if i have $255

Zepler [3.9K]

You can buy 85 sweaters. Plze mark me brainiest:)

7 0

4 years ago

Help me plzz please please

Snowcat [4.5K]

Answer:

132 square feet

Step-by-step explanation:

a+b/2*h

20+24/2*6=132

4 0

3 years ago

One again i really am bad at these

Ivenika [448]

It’s 39 yards squared because the rectangle is 24+15 which is the triangle

6 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