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QveST [7]
3 years ago
14

There is a line that includes the point (6,9) and has a slope of 3. What is its equation in

Mathematics
1 answer:
Nostrana [21]3 years ago
6 0

Answer:

y=3x-9

Step-by-step explanation:

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4 years ago
Find the slope of the line<br> (2,3) (13,8)
alexandr1967 [171]

Answer:

\boxed{m=\frac{5}{11}}

Step-by-step explanation:

\left(2,\:3\right)\:\left(13,\:8\right)

\mathrm{Slope}=\cfrac{y_2-y_1}{x_2-x_1}

\left(x_1,\:y_1\right)=\left(2,\:3\right)

\left(x_2,\:y_2\right)=\left(13,\:8\right)

Slope(m)=\frac{8-3}{13-2}

m=\frac{5}{11}

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2 years ago
Please help me!!
klemol [59]
Answer: g(x)=(3/4)x, points at (-4,-3), (0,0), and (4,3)
Step by step:
We know what the equation is, so we just have to multiply it by 3/4.
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8 0
4 years ago
.4^3Grafting, the uniting of the stem of one plant with the stem or root of another, is widely used commercially to grow the ste
11111nata11111 [884]

Answer:

0.784 = 78.4% probability that there will be at least one failed graft in the next three done

Step-by-step explanation:

To solve this question, we need to understand the binomial probability distribution and conditional probability.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: No failed grafts in the first seven

Event B: At least one fail in the next three.

Intersection of events A and B:

Since the probability of a graft failling is independent of other grafts, we have that:

P(A \cap B) = P(A)*P(B)

So

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)*P(B)}{P(A)} = P(B)

So we just have to find the probability of one fail in three trials.

Three trials means that n = 3.

The probability is

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{3,0}.(0.4)^{0}.(0.6)^{3} = 0.216

Then

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.216 = 0.784

0.784 = 78.4% probability that there will be at least one failed graft in the next three done

3 0
3 years ago
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