SOMEONE PLZ HELP I HAVE TILL TMR TO GET A 49% up

Solve xy − 2 = k for x.

MA_775_DIABLO [31]

Isolate x
So xy - 2 = k
Then
xy = k + 2

x = (k+2)/y

6 0

2 years ago

Pairs that satisfy function y=2x+1

evablogger [386]

Test a pair from each table by substituting their values into the given equation and solving.

A) y = 2x + 1 .......... 2 = 2(0) + 1 .......... 2 ≠ 1

B) y = 2x + 1 .......... 1 = 2(0) + 1 .......... 1 = 1

C) y = 2x + 1 .......... -1 = 2(0) + 1 .......... -1 ≠ 1

D) y = 2x + 1 .......... -2 = 2(0) + 1 .......... -2 ≠ 1

The only pair that satisfied the equation was from answer choice B. Therefore, B is the correct answer.

8 0

3 years ago

Write the rules for determining the sign of the product when multiplying integers. Then, write the the rules for determining the

kondor19780726 [428]

For multiplication, if both integers have like signs, it is always positive. If they have unlike signs it is negative.
For division, if both integers have like signs it is positive. If they have unlike signs it is negative.

7 0

3 years ago

I need help with this

Allisa [31]

Answer:

d

Step-by-step explanation:

8 0

2 years ago

In Illinois, 9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders; that is, they have been arre

Veseljchak [2.6K]

Answer:

a) 21.58% probability that exactly 3 people are repeat offenders

b) 97.91% probability that at least one person is a repeat offender

c) 3.69

d) 1.83

Step-by-step explanation:

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders

This means that $p = 0.09$

41 people arrested for DUI in Illinois are selected at random.

This means that $n = 41$

a. What is the probability that exactly 3 people are repeat offenders?

This is P(X = 3).

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

$P(X = 3) = C_{41,3}.(0.09)^{3}.(0.91)^{38} = 0.2158$

21.58% probability that exactly 3 people are repeat offenders

b. What is the probability that at least one person is a repeat offender?

Either none are repeat offenders, or at least one is. The sum of the probabilities of these outcomes is 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ .

Then

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = 0) = C_{41,0}.(0.09)^{0}.(0.91)^{41} = 0.0209$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0209 = 0.9791$

97.91% probability that at least one person is a repeat offender

c. What is the mean number of repeat offenders?

$E(X) = np = 41*0.09 = 3.69$

d. What is the standard deviation of the number of repeat offenders?

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{41*0.09*0.91} = 1.83$

4 0

3 years ago