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3 years ago

One gallon of paint usually covers 350 square feet.How many square feet will 2.5 gallons cover?

Mathematics

2 answers:

Mamont248 [21]3 years ago

5 0

1 gallon covers 350 sqft
350sqft * 2.5 gallons= 875 sqft

875sqft will be covered by 2.5 gallons.

Margaret [11]3 years ago

4 0

Answer:

The answer is 875

Step-by-step explanation:

Hope this helps

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Solve each equation for y (get into slope-intercept/y=mx + b form):

Gwar [14]

Answer:

Step-by-step explanation:

In order to do this we need to isolate y by performing the inverse operations on the other values like so...

a) 10x + 5y = 20 ... subtract 10x on both sides

5y = 20 - 10x ... divide both sides by 5

y = 4 - 2x ... we can move the 2x to the right to make it into y = mx + b

y = -2x + 4

b) 3x - 2y = 10 + 4x ... subtract 3x on both sides

-2y = 10 + x ... divide both sides by -2

y = -5 - 0.5x ... move -0.5 to the left so it matches y = mx + b

y = -0.5x - 5

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3 years ago

An antique collector wants to know the age of a particular chair in a shop

Nataly_w [17]

Well you gotta ask the chair politely, "how old are you chair?"

4 0

3 years ago

Read 2 more answers

Where do I put the other two

Ilya [14]

You can put h on -1 and sum on 0.4. Also, you put 1.4 in the wrong place. It should be three more places to the right.

4 0

3 years ago

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

5 0

3 years ago

A clinical psychologist wants to test whether experiencing childhood trauma affects one's self-efficacy in adulthood. He randoml

Verdich [7]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 152.5

For the alternative hypothesis,

µ ≠ 152.5

This is a two tailed test.

Since no population standard deviation is given, the distribution is a student's t.

Since n = 231

Degrees of freedom, df = n - 1 = 231 - 1 = 230

t = (x - µ)/(s/√n)

Where

x = sample mean = 148.9

µ = population mean = 152.5

s = samples standard deviation = 27.4

t = (148.9 - 152.5)/(27.4/√231) = - 2

We would determine the p value using the t test calculator. It becomes

p = 0.047

Since alpha, 0.05 > thanthere sufficient evidence to conclude that the self-efficacy of adults who have experienced childhood trauma differs from that in the general population of individuals the p value, 0.047, then we would reject the null hypothesis. Therefore, At a 5% level of significance, there is sufficient evidence to conclude that the self-efficacy of adults who have experienced childhood trauma differs from that in the general population of individuals

6 0

3 years ago