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3 years ago

It takes21 minutes for 5 people to paint 7walls. How many minutes does it take 3people to paint 5 walls?

1 answer:

8 0

so 5 people work each one for 21 minutes, that means 21+21+21+21+21 minutes altogether, namely 105 minutes, it took that long for 7 walls, hmmmm how long for just one wall then? well, 105 ÷ 7, namely 15 minutes then, for just 1 wall.

now, if it takes 15 minutes of work to do one wall, what about 5 walls? well, that'd be 15*5 or 75 minutes.

so if we have 3 folks working, how much would each one work? well, 75 ÷ 3, namely 25 minutes, so each of them will work 25 minutes, namely 25+25+25 minutes, so in 25 minutes, they'll be done with 5 walls.

You might be interested in

Last year, Deshaun had $20,000 to invest. He invested some of it in an account that paid 9% simple interest per year, and he inv

dezoksy [38]

Answer:

the amount Deshaun invested in an account that paid 9% interest = $9,000

the amount Deshaun invested in an account that paid 5% interest = $11,000

Step-by-step explanation:

Let x = the amount Deshaun invested in an account that paid 9% interest

Let y = the amount Deshaun invested in an account that paid 5% interest.

Thus;

x + y = 20,000 - - - - (eq 1)

Now, After one year, he received a total of $1440 in interest. Thus;

0.09x + 0.05y = 1440

Multiply each term by 100 to give;

9x + 5y = 144000 ---- (eq 2)

Making x the subject in equation 1,we have;

x = 20,000 - y

Putting 20,000 - y for x in eq 2,we have;

9(20,000 - y) + 5y = 1440

180000 - 9y + 5y = 144000

180000 - 4y = 144000

180000 - 144000 = 4y

36000 = 4y

Divide both sides by 4 to give;

y = 36000/4

y = $9,000

x = 20,000 - 9000

x = $11,000

7 0

3 years ago

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. An SRS of 8 subjects was randomly selected

chubhunter [2.5K]

Answer:

e. No, the p-value is greater than 0.01, so there is insufficient evidence at the 0.01 level to conclude that pain levels are lower after hypnotism.

Step-by-step explanation:

Hello!

The study's objective is to test if hypnotism reduces pain.

A sample of n=8 subjects was taken and the pain level was recorded in each subject before and after being hypnotized.

The variable of interest was determined by calculating the difference of pain level after - before hypnosis. This is a paired sample test and the variable can be determined as:

Xd: Difference between pain level felt after hypnosis and pain level felt before hypnosis of a subject.

The sample average and standard deviation obtained were:

Xd= -3

Sd= 3

And the variable is presumed to be approximately normal.

An approximately normal distribution is enough to conduct a paired sample t-test.

If the claim is that hypnosis reduces pain, then the average pain level after hypnosis should be less than the average pain level before hypnosis, then the average difference is expected to be negative, symbolically: μd < 0

The test will be one-tailed and so will be the p-value.

Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

So first step is to calculate the value of the statistic under the null hypothesis and then you can calculate the p-value.

H₀: μd ≥ 0

H₁: μd < 0

$t_{H_0}= \frac{X_d-Mu_{d}}{\frac{S_d}{n} }= \frac{-0-0}{\frac{3}{\sqrt{8} } } = -2.828= -2.83$

The DF of the t-test are n-1= 7

Then you can calculate the p-value as:

P(t₇≤-2.83)= 0.0127

The level of the test is α: 0.01

The decision rule is:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The p-value > α the decision is to not reject the null hypothesis.

Correct option: e. No, the p-value is greater than 0.01, so there is insufficient evidence at the 0.01 level to conclude that pain levels are lower after hypnotism.

I hope this helps!

5 0

3 years ago

How many sides does each regular polygon have if the measure of each interior angle is 160

Leviafan [203]

Answer:

4 sides i think

4 0

3 years ago

Irene has 1 gallon of milk. She uses

Umnica [9.8K]

Answer:

Step-by-step explanation:

In 1 gallon, there are 128 ounces.

Divide 128 by 4 and get 32.

5 0

3 years ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

3 years ago