Is each question Statistical or Not statistical to classify each question.

Ahhhh i need help like now !!!!!!!!!!!

ozzi

Answer:

x = 2

Step-by-step explanation:

Simplify both sides of the equation.

−20 = − 4x − 6x

−20 = −4x − 6x

−20 = (−4x − 6x)(Combine Like Terms)

−20 = −10x

Flip the equation.

−10x = −20

Divide both sides by -10.

-10x/-10 = -20/-10

x = 2

7 0

3 years ago

Is it right And why

antiseptic1488 [7]

Yes it is indeed correct,
B. 7.47 is the correct answer!
Brainliest pls! thx! :)

7 0

3 years ago

A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 15 drivers li

svetlana [45]

Answer:

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

Step-by-step explanation:

For each driver surveyed, there are only two possible outcomes. Either they were involved in a car accident last year, or they were not. This means that we solve this problem using binomial probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And $\pi$ is the probability of X happening.

In this problem

15 drivers are randomly selected, so $n = 15$ .

A success consists in finding a driver that was involved in an accident. A car insurance company has determined that 8% of all drivers were involved in a car accident last year. This means that $\pi = 0.08$ .

What is the probability of getting 3 or more who were involved in a car accident last year?

This is $P(X \geq 3)$ .

Either less than 3 were involved in a car accident, or 3 or more were. Each one has it's probabilities. The sum of these probabilities is decimal 1. So:

$P(X < 3) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{15,0}.(0.08)^{0}.(0.92)^{15} = 0.2863$

$P(X = 1) = C_{15,1}.(0.08)^{1}.(0.92)^{14} = 0.3734$

$P(X = 2) = C_{15,2}.(0.08)^{2}.(0.92)^{13} = 0.2273$

So

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2863 + 0.3734 + 0.2273 = 0.887$

Finally

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.887 = 0.113$

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

5 0

3 years ago

A mathematics teacher wanted to see the correlation between test scores and

SpyIntel [72]

Answer:

The homework grade, to the nearest integer, for a student with a test grade of 68 is 69.

Step-by-step explanation:

The general form of the linear regression equation is:

$y=a+bx$

Here,

y = dependent variable = test grade

x = independent variable = homework grade

a = intercept

b = slope

Compute the value of a and b as follows:

$a=\frac{\sum Y\cdot \sum X^{2}-\sum X\cdot\sum XY}{n\cdot \sum X^{2}-(\sum X)^{2}}\\\\=\frac{(592\times 44909)-(591\times45227)}{(8\times44909)-(591)^{2}}\\\\=-14.316$ $b=\frac{n\cdot \sum XY-\sum X\cdot\sum Y}{n\cdot \sum X^{2}-(\sum X)^{2}}\\\\=\frac{(8\times 45227)-(591\592)}{(8\times44909)-(591)^{2}}\\\\=1.195$