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Korolek [52]
3 years ago
12

Find the slope and y intercept of the graph of the graph of each equation.

Mathematics
2 answers:
AleksandrR [38]3 years ago
7 0
Solve each equation for y and then the slope is m and y-intercept is b as in:
y = mx+b

So, if you do that, you'll get (remember, this is <em>after</em> solving for y):
1. m = -6, b = -2
2. m = 5, b = -1
3. m = -5/3, b = 5
4. m = 0, b = -1/4
5. m = -1, b = -3
6. m = 3, b = -4
7. m = 1/2, b = -5/2
8. m = 7/2, b = 1

Hope this helps.
djyliett [7]3 years ago
5 0
1: y= -6x -2
2: y=5x-1
3: y= -5/3x +5
4: y= -1/4
5: y= -x+3
6: y=3x -4
7: y= 1/2x - 5/2
8: y= 7/2x + 1
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Probability = \frac{1}{24}

Step-by-step explanation:

Given

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Determine the probability of 1 orange, 1 apple and 1 banana

Since, order is not important:

Probability = P(Orange) * P(Apple) * P(Banana)

Probability = \frac{3}{10} * \frac{5}{9} * \frac{2}{8}

<em>The difference in the numerator is as a result of picking the fruit without replacement</em>

Probability = \frac{30}{720}

Probability = \frac{1}{24}

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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

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