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3 years ago

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Mrs.Barkley has 75 feet ribbon. If each of the 18 students in her class gets an equal length of ribbon, how many inches of ribbo

n will each student get?

1 answer:

Black_prince [1.1K]3 years ago

3 0

Answer:

Each student would get 4 inches of ribbon

Step-by-step explanation:

Mrs. Barkley has 75 inches of ribbon and is splitting it between 18 students so you would divide 75 by 18. When you divide you get 4.166666... etc. So the answer would be 4

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Audrey, an astronomer is searching for extra-solar planets using the technique of relativistic lensing. Though there are believe

Stolb23 [73]

Answer:

Step-by-step explanation:

The model $N (t)$ , the number of planets found up to time $t$ , as a Poisson process. So, the $N (t)$ has distribution of Poison distribution with parameter $(\lambda t)$

a)

The mean for a month is, $\lambda = \frac{1}{3}$ per month

$E[N(t)]= \lambda t\\\\=\frac{1}{3}(24)\\\\=8$

(Here. t = 24)

For Poisson process mean and variance are same,

$Var[N (t)]= Var[N(24)]\\= E [N (24)]\\=8$

(Poisson distribution mean and variance equal)

The standard deviation of the number of planets is,

$\sigma( 24 )] =\sqrt{Var[ N(24)]}=\sqrt{8}= 2.828$

b)

For the Poisson process the intervals between events(finding a new planet) have independent exponential distribution with parameter $\lambda$ . The sum of $K$ of these independent exponential has distribution Gamma $(K, \lambda)$ .

From the given information, $k = 6$ and $\lambda =\frac{1}{3}$

Calculate the expected value.

$E(x)=\frac{\alpha}{\beta}\\\\=\frac{K}{\lambda}\\\\=\frac{6}{\frac{1}{3}}\\\\=18$

(Here, $\alpha =k$ and $\beta=\lambda$ )

C)

Calculate the probability that she will become eligible for the prize within one year.

Here, 1 year is equal to 12 months.

P(X ≤ 12) = (1/Г (k)λ^k)(x)^(k-1).(e)^(-x/λ)

$=\frac{1}{Г (6)(\frac{1}{3})^6}(12)^{6-1}e^{-36}\\\\=0.2148696\\=0.2419\\=21.49%$

Hence, the required probability is 0.2149 or 21.49%

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2 years ago

Which inequality is true?

Andreas93 [3]

Answer:

2 awnser is the right on

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2 years ago

PLEASE HELP WILL GIVE BRAINLY<br><br>(USE BAR NOTATION IF NEEDED!!!!!!<br>CONVERT TO DECIMAL

emmainna [20.7K]

Answer:

3.34 is the decimal form but the question is unclear, what is it that you need?

Step-by-step explanation:

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2 years ago

Read 2 more answers

german

Ok do the original price minus the new price. 90-55=35. Divide that by the original number and multiply it by 100 to find the price decrease. 35/90=.3888*100=38.89

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2 years ago

In 2008 the Better Business Bureau settled 75% of complaints they received (USA Today, March 2, 2009). Suppose you have been hir

Ede4ka [16]

Answer:

Explained below.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}= p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

(a)

The sample selected is of size <em>n</em> = 450 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

$\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204$

So, the sampling distribution of sample proportion is $\hat p\sim N(0.75,0.0204^{2})$ .

(b)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

$P(p-0.04$

$=P(-1.96$

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.95.

(c)

The sample selected is of size <em>n</em> = 200 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

$\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{200}}=0.0306$

So, the sampling distribution of sample proportion is $\hat p\sim N(0.75,0.0306^{2})$ .

(d)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

$P(p-0.04$

$=P(-1.31$

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.81.

(e)

The probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 450 is 0.95.

And the probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 200 is 0.81.

So, there is a gain in precision on increasing the sample size.

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2 years ago