Help for a few extra points!!

Sally can paint a room in 9 hours while it takes Steve 3 hours to paint the same room. How long would it take them to paint the

Archy [21]

You can solve this one by adding their rates, but you have to invert the numbers they give you to find the rate. Rate is measured in something per unit of time, and "per" indicates division. So the rates are: 1/3 of an hour (Sally) and 1/6 of an hour (Steve). So add the rates: 1/3 + 1/6 = 1/2. (1/3 is 2/6, and 2/6 + 1/6 = 3/6). Since their combined rate is 1/2 of a room in one hour, it takes them two hours to paint the room.

This answer makes sense: It takes Sally three hours by herself; with Steve's help, she's faster, but not THAT much faster, because he's pretty slow. You can often spot-check word problem answers like this by giving it a common sense once-over. If we came up with four hours, that would clearly be wrong: it should take her LONGER if she's getting help.

7 0

3 years ago

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Please help me steps please

Taya2010 [7]

To make it easier for you, kuta software has a y0ut0be channel and she posts all of the steps there! search up the name of the worksheet and it’ll be there

7 0

3 years ago

A cylinder has a diameter of 22 cm and a height of 9 cm. Identify the volume of the cylinder to the nearest tenth. HELP PLEASE!!

OlgaM077 [116]

Answer: 3421.19

Step-by-step explanation:

look up cylinder formula and r= radius which is just the diameter cut in half

7 0

3 years ago

A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec

Ket [755]

Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (n ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

$Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18$

The sample size is, n = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

$\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191$

(a)

The null hypothesis is:

H₀: The usual load is 13 credits, i.e. μ = 13.

Assume that the significance level of the test is, α = 0.05.

Construct a (1 - α) % confidence interval for population mean to check the claim.

The (1 - α) % confidence interval for population mean is given by:

$CI=\bar x\pm z_{\alpha/2}\times SE$

For 5% level of significance the two tailed critical value of z is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Construct the 95% confidence interval as follows:

$CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)$

As the null value, μ = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

$P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z$

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

3 0

3 years ago

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How to find the smallest integer k such that 60k is a perfect square

suter [353]

9514 1404 393

Answer:

k = 15

Step-by-step explanation:

Look for the missing factors that make all of the factors of 60k be squares.

60 = 2² × 3 × 5

The factors 3 and 5 each have an odd exponent (1), so those two factors must be part of k.

60k = 2² × 3 × 5 × k

is a perfect square when ...

3 × 5 = k = 15

For k = 15, 60k = 900 = 30²

6 0

3 years ago