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Soloha48 [4]
2 years ago
12

Solve the inequality

10(c-3)" alt="2(5c-7)\geq 10(c-3)" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
alina1380 [7]2 years ago
5 0

1 Remove parentheses.

2(5c−7)≥c−3

2 Expand.
10c−14 ≥ c −3


3 Add 14 to both sides.
10c-14+14≥ c − 3+14


4 Simplify c−3+14 to c+11
10c≥c+11

5 Subtract c from both sides.

10c−c≥11

6 Simplify 10c−c to 9c
9c≥11

7 Divide both sides by 9.
c ≥11/9
​
​​
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PilotLPTM [1.2K]

Answer:

The answer is 3

Step-by-step explanation:

7 0
3 years ago
Suppose y varies directly with x when x is 4, y is 32. What is y when x is -2
omeli [17]

Step-by-step explanation:

Since y varies directly with x,

y = kx, where k is a real constant.

When x = 4, y = 32.

=> (32) = (4)k, k = 8.

Therefore when x = -2,

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3 0
3 years ago
A certain medical test is known to detect 73% of the people who are afflicted with the disease Y. If 10 people with the disease
Setler [38]

The probability of an event is the measurement of the chance of that event's occurrence. The probabilities of considered events are:

  • P(At least 8 have the disease) ≈ 0.4378
  • P(At most 4 have the disease)  ≈ 0.0342
<h3 /><h3>How to find that a given condition can be modeled by binomial distribution?</h3>

Binomial distributions consist of n independent Bernoulli trials. Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining to a binomial distribution with parameters n and p, then it is written as

X = B(n,p)

The probability that out of n trials, there'd be x successes is given by

P(X=x)  = ^nC_xp^x(1-p)^{n-x}

Since 10 people can be either diseased or not and they be so independent of each other (assuming them to be selected randomly) , thus, we can take them being diseased or not as outputs of 10 independent Bernoulli trials.

Let we say

Success= Probability of a diseased person tagged as diseased by the clinic

Failure = Probability of a diseased person tagged as not diseased by the clinic.

Then,

P(Success) = p = 72% = 0.72 (of a single person)

P(Failure) = q = 1-p = 0.28

Let X be the number of people diagnosed diseased by the clinic out of 10 diseased people. Then we have: X ≈ B(n+10,P=0.73)

Calculating the needed probabilities, we get:

a) P(At leased 8 have disease) = P(X≥8) =P(X=8) + P(X=9) + P(X=10)

P(X≥8) = ^{10}C_8(0.73)^8(0.28)^2+^{10}C_9(0.73)^9(0.27)^1+^{10}C_{10}(0.73)^{10}(0.27)^0

P(X≥8) ≈ 0.2548 + 0.1456 + 0.0374 ≈ 0.4378

b)  P(At most 4 have the disease) = P(X≤4) = P(X=0) + P(X=1)+P(X=2)+P(X=3)+P(X=4)

P (X ≤ 4) =

^{10}C_0(0.73)^0(0.27)^{10}+^{10}{C_1(0.73)^1(0.27)^9+^{10}{C_2(0.73)^2(0.27)^8+^{10}C_3(0.73)&^3(0.27)^7 \\

+^{10}C_4(0.73)^4(0.27)^6

P (X ≤ 4) = 0.000003 + 0.000076+0.00088+0.00604+0.02719

P (X ≤ 4) =  0.0342

Thus,

The probabilities of considered events are:

  • P(At leased 8 have disease) = 0.4378 approx
  • P(At most 4 have the disease)  = 0.0342 approx

Learn more about binomial distribution here:

brainly.com/question/13609688

#SPJ1

4 0
2 years ago
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Tems11 [23]

Answer:

18 students is the amount that passed the test

so the fraction would be 18/24, which can be simplified to 3/4, which is 75% of the class.

75% of the class passed

0.75% of the class passed

3 0
3 years ago
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Answer:

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