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3 years ago

It takes 60 minutes for 7 people to paint 5 walls. How many minutes does it take for 10 people to paint 10 walls?

Mathematics

1 answer:

sweet-ann [11.9K]3 years ago

7 0

60 = 5k/7

420 = 5k

84 = k

m = 10(84)/10
m = 84

Hope this helps!

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Simplify. [6.6 ÷ (–5 + 3)] • (–1)

eimsori [14]

Hi there, we use PEMDAS to solve this problem, PEMDAS is parenthesis, Exponents, Multiply, Divide, Add, and Subtract. In this case, we have [6.6÷(-5+3)]*(-1), 6.6/-5+3(-1)=6.6/-2(-1)=(-3.3)(-1)=3.3. So, your answer is 3.3

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3 years ago

How can i do this please help me

Aleksandr-060686 [28]

The tangent of an angle is the ratio of the opposite side to the adjacent side.

... tan(45°) = (height above eye level)/(20 m)

... height above eye level = (20 m)·tan(45°) = (20 m)·1 = 20 m

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Eye level is 1.7 m above ground, so the height of the flagpole above ground is

... height of pole = height above eye level + height of eye level

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3 years ago

Which triangle has hypotenuse BF?

Sav [38]

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Step-by-step explanation:

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3 years ago

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Find the equation of the line that has a slope of 5 and passes through the point (-9,6)

Aleks [24]

Answer:

y-6=5(x-(-9))

Step-by-step explanation:

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3 years ago

A group of 54 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic

Ket [755]

Answer:

$t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601$

P-value

First we find the degrees of freedom given by:

$df = n-1= 54-1=53$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{53}>0.9601)=0.3414$

Step-by-step explanation:

Data given and notation

$\bar X=5.25$ represent the sample mean

$s=3.98$ represent the sample standard deviation

$n=54$ sample size

$\mu_o =4.73$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean differs from 4.73, the system of hypothesis would be:

Null hypothesis: $\mu =4.73$

Alternative hypothesis: $\mu \neq 4.73$

Since we know don't the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601$

P-value

First we find the degrees of freedom given by:

$df = n-1= 54-1=53$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{53}>0.9601)=0.3414$

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3 years ago