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3 years ago

How do you get 8(c-9)=6(2c-12)-4c

1 answer:

6 0

So first you factor it out:

8c - 72 = 12c - 72 - 4c

Then you isolate the c's:

-72 +72 = 12c - 4c -8c

Then you add both sides:

0 = 0c

C is equal to 0

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Value<br>5. What per cent of 84 is 14?

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14/84 x100 =. 1/6 x 100 = 16.66%

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2 years ago

Graph the ellipse with equation x squared divided by 9 plus y squared divided by 25 = 1.

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3 years ago

Read 2 more answers

A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o

balu736 [363]

Answer:

$P(X \ge 1) = 0.509$

$P(X \ge 1) = 0.6807$

Step-by-step explanation:

From the question we are told that

The number of students in the class is N = 20 (This is the population )

The number of student that will cheat is k = 3

The number of students that he is focused on is n = 4

Generally the probability distribution that defines this question is the Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally probability mass function is mathematically represented as

$P(X = x) = \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}$

Here C stands for combination , hence we will be making use of the combination functionality in our calculators

Generally the that he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

Here

$P(X \le 0) = \frac{ ^{3} C_0 * ^{20 - 3} C_{4- 0}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ ^{3} C_0 * ^{17} C_{4}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ 1 * 2380}{ 4845}$

$P(X \le 0) = 0.491$

Hence

$P(X \ge 1) = 1 - 0.491$

$P(X \ge 1) = 0.509$

Generally the that he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

$P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]$

Here n = 6

$P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{20 -3}C_{6-0}}{^{20}C_6}]$

$P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]$

$P(X \ge 1) =1- [ \frac{1 * 12376}{38760}]$

$P(X \ge 1) =1- 0.3193$

$P(X \ge 1) = 0.6807$

5 0

3 years ago

Please help fast elementary school all questions

VMariaS [17]

Write the highlighted digit's place and value.

3. 567; ones place 4. 6,327; thousandth place

5. 9,325; hundredth place 6. 8,281; tenth place

Write each number in standard form.

7. 5,000 + 500 + 3; 5,503 8. 2,000 + 300 + 20 + 9; 2,329

9. 4,000 + 600 + 8; 4,608 9. 9,000 + 300 + 70 + 2; 9,372

11. 1,221

We know Sierra has the same numbers in the thousandth and ones place, so keep does numbers: 1,?2?

It also has two more hundreds and 3 fewer ones place: Take the original number; 1,024 (Make sure you keep the numbers) into a equation; (1024+200)-3. First Add as it's in the ( ), then subtract, 1,221

I hope I've helped!

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2 years ago

I went to the store and bought 4 packs of nerds for $0.98 each and a pack of Little Debbie cakes for $2.15. How

AlexFokin [52]

Answer:

Kindly refer to the attachment

<h2>Thank U Next</h2>

8 0

2 years ago