To see which ordered pairs work in the inequality, substitute the ordered pairs in for the variables. If you get a true statement, this means the pair is a solution to the inequality.

A. 3, 18

-3x + y > 7

-3(3) + 18 > 7

-9 + 18 > 7

9 > 7 Correct :)

3, 18 is a solution to this inequality.

B. -2, -3

-3(-2) + -3 > 7

6 + -3 > 7

3 > 7 Incorrect :(

-2, -3 is not a solution to this inequality.

C. -1, 8

-3(-1) + 8 > 7

3 + 8 > 7

11 > 7 Correct :)

-1, 8 is a solution to this inequality

D. 1, 9

-3(1) + 9 > 7

-3 + 9 > 7

6 > 7 Incorrect :(

1, 9 is not a solution to this inequality.

E. 0, 7

-3(0) + 7 > 7

0 + 7 > 7

7 > 7 Correct :)

0, 7 is a solution to this inequality.

PLEASE MARK BRAINLIEST

4 0

3 years ago

I did half the answer but what is e= ?

OLga [1]

Answer:

See below.

Step-by-step explanation:

e=12c+6

e=12(1)+6

e=12+6

e=18

e=12(2)+6

e=24+6

e=30

e=12(3)+6

e=36+6

e=42

e=12(4)+6

e=48+6

e=54

-hope it helps

5 0

2 years ago

8.4x-5=20.2 solve for x

tresset_1 [31]

Answer:

x=3

Step-by-step explanation:

6 0

3 years ago

A quiz consists of 10 true or false questions. To pass the quiz a student must answer at least eight questions correctly.

lys-0071 [83]

Answer:

The probability of the student will pass the quiz = .0546

Step-by-step explanation:

Given -

Total no of question = 10

If the student guesses on each question there are two outcomes true of false

the probability of guesses question correctly = $\frac{1}{2}$

the probability of success is (p) = $\frac{1}{2}$

the probability of guesses question incorrectly = $\frac{1}{2}$

the probability of failure is (q) = 1- p = $\frac{1}{2}$

If the student guesses on each question he must answered at least 8 question correctly

the probability of the student will pass the quiz = $P(X\geq8 )$

= P(X = 8 ) + P(X = 9) + P(X = 10 )

= $\binom{10}{8}(p)^{8}(q)^{10 - 8} + \binom{10}{9}(p)^{9}(q)^{10 - 9} + \binom{10}{10}(p)^{10}(q)^{10 - 10}$

= $\frac{10!}{(2!)(8!)}(\frac{1}{2})^{8}(\frac{1}{2})^{10 - 8} +\frac{10!}{(1!)(9!)} (\frac{1}{2})^{9}(\frac{1}{2})^{10 - 9} + \frac{10!}{(0!)(10!)}(\frac{1}{2})^{10}(\frac{1}{2})^{10 - 10}$

= $45\times\frac{1}{2^{10}} + 10\times\frac{1}{2^{10}} + 1\times\frac{1}{2^{10}}$

= $\frac{56}{2^{10}}$

= .0546

5 0

3 years ago