Ask question

Ask question

All categories

2 years ago

15

The formula for the distance traveled over time ttt and at an average speed vvv is v\cdot tv⋅tv, dot, t. Amit ran for 404040 min

utes at a speed of about 555 kilometers per hour. What calculation will give us the estimated distance Amit ran in kilometers?

1 answer:

andrew-mc [135]2 years ago

4 0

Answer:

5 x 40/60

Step-by-step explanation:

khan explanation

You might be interested in

A company is in the business of finding addresses of long-lost friends. The company claims to have a 70% success rate. Suppose t

Galina-37 [17]

Answer:

a) Figure attached

b) $E(X) =\mu= np = 9*0.7=6.3$

$Sd(X) =\sigma= \sqrt{np(1-p)}= \sqrt{9*0.7*(1-0.7)}=1.375$

c) For the case n= 6

$P(X \leq 2) = 0.9891$

For the case n= 5

$P(X \leq 2) = 0.9692$

So then we need at least n=5 or n=6 to satisfy the condition required.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=9, p=0.7)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

For this case we can use the following R code:

> x <- seq(0,9,by = 1)

> y <- dbinom(x,9,0.7)

> plot(x,y,main="Histogram",type = "h")

And we can see on the figure attached the solution.

We see that the higher probabilities are from 4 to 9

Part b

The expected value is given by:

$E(X) =\mu= np = 9*0.7=6.3$

The variance is given by:

$Var (X) =\sigma^2= np(1-p) = 9*0.7*(1-0.7)= 1.89$

And the standard deviation is:

$Sd(X) =\sigma= \sqrt{np(1-p)}= \sqrt{9*0.7*(1-0.7)}=1.375$

Part c

First we can find the probability that at least two addresses will be found in the list of 9 that we have like this:

$P(X \geq 2)$

We can use the complement rule and we have:

$P(X \geq 2) = 1-P(X$

We find the indicidual probabilities:

$P(X=0)=(9C0)(0.7)^0 (1-0.7)^{9-0}=0.00001968$

$P(X=1)=(9C1)(0.7)^1 (1-0.7)^{9-1}=0.000413$

$P(X \geq 2) = 1-[0.00001968+0.000413]=0.9996$

If we use the case of n=8 and we find $P(X\leq 2)$ , we got:

$P(X \leq 2) = 0.9987$

For the case n= 7

$P(X \leq 2) = 0.9962$

For the case n= 6

$P(X \leq 2) = 0.9891$

For the case n= 5

$P(X \leq 2) = 0.9692$

So then we need at least n=5 or n=6 to satisfy the condition required.

5 0

2 years ago

Fast Pax Annual Salaries (Thousands of Dollars) Analysts determined that the $255,000 salary is an outlier. The box-and-whisker

SIZIF [17.4K]

The data appears slightly skewed, so the median is probably the most appropriate measure.
You have a good chance of making between $16,000 and $23,000 because that is the range for the middle 50% of employees.

7 0

2 years ago

Read 2 more answers

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the

larisa [96]

Answer:

No, at the 0.05 significance level, the number of units produced on the night shift is not larger.

Step-by-step explanation:

We are given that the mean number of units produced by a sample of 54 day-shift workers was 345. The mean number of units produced by a sample of 60 night-shift workers was 351.

Assume the population standard deviation of the number of units produced on the day shift is 21 and 28 on the night shift.

Let $\mu_1$ = population mean number of units produced on the day shift

$\mu_2$ = population mean number of units produced on the night shift

So, <u>Null Hypothesis</u>, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1\geq\mu_2$ {means that the mean number of units produced on the night shift is same or lesser on the day shift}

<u>Alternate Hypothesis,</u> $H_A$ : $\mu_1-\mu_2$ or $\mu_1$ {means that the mean number of units produced on the night shift is larger}

The test statistics that will be used here is <u>Two-sample z test statistics</u> as we know about population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }$ ~ N(0,1)

where, $\bar X_1$ = sample mean number of units produced by a sample of 54 day-shift workers = 345

$\bar X_2$ = sample mean number of units produced by a sample of 60 night-shift workers = 351

$\sigma_1$ = population standard deviation of the number of units produced on the day shift = 21

$\sigma_2$ = population standard deviation of the number of units produced on the day shift = 28

$n_1$ = sample of day-shift workers = 54

$n_2$ = sample of night-shift workers = 60

So, <em><u>test statistics</u></em> = $\frac{(345-351)-(0)}{\sqrt{\frac{21^{2} }{54}+\frac{28^{2} }{60} } }$

= -1.302

Now at 0.05 significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is more than the critical value of z as -1.302 > -1.6449 so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean number of units produced on the night shift is same or lesser than those produced on the day shift.

4 0

2 years ago

A sociologist wishes to estimate the average number of automobile thefts in a large city per day within 2 automobiles. He wishes

anzhelika [568]

Answer:

$n=(\frac{2.58(4.2)}{2})^2 =29.35 \approx 30$

So the answer for this case would be n=30 rounded up to the nearest integer

Step-by-step explanation:

Information given

$ME = 2$ the margin of error desired

$\sigma =4.2$ standard deviation from previous studies

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. The significance level is $\alpha=0.01$ and the critical value would be $z_{\alpha/2}=2.58$ , replacing into formula (b) we got:

$n=(\frac{2.58(4.2)}{2})^2 =29.35 \approx 30$

So the answer for this case would be n=30 rounded up to the nearest integer

5 0

2 years ago

A sewing club is making a quilt consisting of 25 squares with each side of the square measuring 30 centimeters. If the quilt has

Sedbober [7]

First, you must find an equation. You would first find the length of 1 square, and then multiply it by how many rows. Then on the side, you would find the width on one square, and then multiply it by how many columns. Once you find you total length and width, find the perimeter of the whole quilt! Let me know if there is any questions! <span />

5 0

2 years ago