4 is to 3 as x is to 20

What is the average acceleration of a southbound train that slows down from 15 m/s to 8.6 m/s in 1.2 s?

horrorfan [7]

Answer:

The average acceleration of train is $5.34\ m/s^2$

Step-by-step explanation:

We have,

A train that slows down from 15 m/s to 8.6 m/s in 1.2 s. It means that 15 m/s is its initial velocity and 8.6 m/s is its final velocity.

It is required to find the average acceleration of the train.

$a=\dfrac{v-u}{t}\\\\a=\dfrac{(8.6-15)\ m/s}{1.2\ s}\\\\a=-5.34\ m/s^2$

The average acceleration of train is $5.34\ m/s^2$ . Negative signs shows that the train is decelerating.

5 0

3 years ago

100 points please help

choli [55]

Answer:

She has $3.74 in quarters. I'm kind of confused but hopefully this helps??

Step-by-step explanation:

0.75 x 5= $3.75

$3.75/0.25= 15 quarters

4 0

2 years ago

Read 2 more answers

A forestry researcher wants to estimate the average height of trees in a forest near Atlanta, Georgia. She takes a random sample

marusya05 [52]

Answer:

The margin of error for her confdence interval is of 0.3757.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 18 - 1 = 17

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 2.8982

The margin of error is:

$M = T\frac{s}{\sqrt{n}}$

In which s is the standard deviation of the sample and n is the size of the sample.

Standard deviation of 0.55 meters.

This means that $s = 0.55$

What is the margin of error for her 99% confidence interval?

$M = T\frac{s}{\sqrt{n}}$

$M = 2.8982\frac{0.55}{\sqrt{18}}$

$M = 0.3757$

The margin of error for her confdence interval is of 0.3757.

7 0

3 years ago

What is the sample space for each one of the following statements? (7.5 points each)

photoshop1234 [79]

We can define the sample space for a given experiment as the set of all the possible outcomes for the experiment.

For example, tossing a coin has a space sample of 2 elements, tails and heads.

The spaces are:

a) {WW, BB, WBB, BWW, WBWW, BWBB,... }

b) { 1-2-3-4, 1-2-3-5,..., 7-6-5-4,...}

a) We can write one element in this sample space as:

WBB

This would mean that first we draw a white ball, then a black ball, and then the second black ball, thus we stop drawing.

So all the elements in this set will have the form

{WW, BB, WBB, BWW, WBWW, BWBB,... }

As we have infinite balls in the bag, we have infinite elements in this set.

b) Now we have only 10 balls numbered from 1 to 10, and we draw 4.

So each element in the sample space can be written as:

1-2-3-4

This element means:

"first we draw ball number 1, then ball number 2, then ball number 3, then ball number 4"

Then the sample space will have the form:

{ 1-2-3-4, 1-2-3-5,..., 7-6-5-4,...}

If you want to learn more, you can read:

brainly.com/question/23125812

6 0

3 years ago

A distributor of personal computers has five locations in a city. In the year's first quarter, the sales in units were: Location

Deffense [45]

Answer:

The correct option is (c) 13.277.

Step-by-step explanation:

The observed data is:

Location Observed sales (units)

Northside 70

Pleasant Township 75

Southwick 70

I-90 50

Venice Avenue 35

TOTAL 300

The test statistic for the Goodness of fit test is:

$\chi^{2}=\sum\frac{(Observed-Expected)^{2}}{Expected}$

For k independent samples this test statistic follows a Chi-square distribution with degrees of freedom (k - 1).

The sample size in this case is, k = 5.

The degrees of freedom is, (k - 1) = 4.

The level of significance is, α = 0.01.

The critical value of the test is:

$\chi^{2}_{\alpha ,k-1}=\chi^{2}_{0.01,4}=13.277$

**Use the Chi-square table.

Thus, the critical value is 13.277.

7 0

3 years ago