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tekilochka [14]

3 years ago

12

There are 3 students to 1 adult on a field trip.You need to find the number of adults needed for 24 students. What would the sec

ond ratio of the proportion look like if the first proportion is as follows

1 answer:

AysviL [449]3 years ago

4 0

Answer: 24 : 8 <em>or </em>24 to 8

Step-by-step explanation:

Let the number of adults be x

3/1 = 24/x . We have to do cross multiplication,

24 = 3x

x = 8

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How many tacos could you buy with your money? On Tuesday you can buy two tacos and you get one free.

tester [92]

Answer:

due to not much information given and the way you ask the question I would definitely say no

8 0

2 years ago

Read 2 more answers

Suppose you have a bag of colored plastic disks and you choose one without looking. The probability the disk you choose is yello

Ostrovityanka [42]

Answer: 0.2

Step-by-step explanation:

Since the probability that the disk chosen is 5/25, thus can be written in decimal as 0.2.

This means that in a bag of 15 colored plastics discs with different colors, the probability of picking a yellow colored disk is 0.2.

8 0

3 years ago

An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 180180 engin

tiny-mole [99]

Answer:

$z=\frac{4.6-4.8}{\frac{0.9}{\sqrt{180}}}=-2.98$

$p_v =P(Z$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=4.6$ represent the sample mean

$\sigma=0.9$ represent the population standard deviation

$n=180$ sample size

$\mu_o =4.8$ represent the value that we want to test

$\alpha=0.1$ 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 thetrue mean is below the specifications, the system of hypothesis would be:

Null hypothesis: $\mu \geq 4.8$

Alternative hypothesis: $\mu < 4.8$

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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:

$z=\frac{4.6-4.8}{\frac{0.9}{\sqrt{180}}}=-2.98$

P-value

Since is a one sided test the p value would be:

$p_v =P(Z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.

4 0

3 years ago

___ L = 240 kL ? <br><br> A. 24,000<br> B. 240,000<br> C. 0.240<br> D. 2.4

koban [17]

It's b 240000
I'm sorry if wrong I'm not too good at these things

6 0

2 years ago

Read 2 more answers

Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is

nadya68 [22]

I'll leave the computation via R to you. The $W_i$ are distributed uniformly on the intervals $[0,10i]$ , so that

$f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}$

each with mean/expectation

$E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i$

and variance

$\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2$

We have

$E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3$

so that

$\mathrm{Var}[W_i]=\dfrac{25i^2}3$

Now,

$E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30$

and

$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

$E[(W_1+W_2+W_3)^2]$

$=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])$

because $W_i$ and $W_j$ are independent when $i\neq j$ , and so

$E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3$

giving a variance of

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3$

and so the standard deviation is $\sqrt{\dfrac{350}3}\approx\boxed{116.67}$

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

8 0

3 years ago