All categories

2 years ago

Pls help me ill give brainliest Do all 3 to get brainliest

Mathematics

1 answer:

vlada-n [284]2 years ago

8 0

Answer:

7. B 8. C 9. C

Step-by-step explanation:

7. y is greater than because the area above the line is shaded and since the line is dashed there it's not greater than or equal to.

8. C provides the correct y values for all of the x values.

9. The point (6, 3) produces -3 ≥ -8 which is the only one that is correct out of all the points provided.

You might be interested in

If a is half b , and b is twice c , then a:c=

faust18 [17]

Answer: a/c = 1

a/b = 1/2 => a = 1/2 .b

b/c = 2 => c = 1/2 .b

=> a/c = (1/2.b)/(1/2.b) = 1

Step-by-step explanation:

4 0

3 years ago

Please help me!!

Ratling [72]

Answer:

E=1/2mv2, v/10./17

Step-by-step explanation:

<h2>i need brainlest <3</h2>

4 0

3 years ago

The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

5 0

3 years ago

Chocolate chip cookies have a distribution that is approximately normal with a mean of 24.1 chocolate chips per cookie and a sta

garik1379 [7]

P5

From Z tables and at P5 = 5% = 0.05, Z = -1.645
Therefore,
True value = mean +Z*SD = 24.1+(-1.645*2.1) = 20.6455 Chips per cookie

P95

From Z table and at P95=95%=0.95, Z= 1.645
Therefore,
True value = 24.1 +(1.645*2.1) = 27.5545 Chips per cookie

These values shows the percentages of amount of chips in the cookies. Thus, the more the percentage considered, the more the amount of chips in the cookies. This can be used to control the number of chips in cookies during production.

7 0

3 years ago

Read 2 more answers

PLZ HELP ME NOW.....

Likurg_2 [28]

A- solve bracket first so

3/4 + 2/9 +5/12 then multiply by 3.6 = 5

do the same with the rest and u get

A=5

B=1

C=15

D=3

please mark brainliest

5 0

3 years ago