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Yakvenalex [24]
3 years ago
5

What is an equation of a line,in point-slope from, that passes through (1, -7) and has a slope of -2/3

Mathematics
1 answer:
il63 [147K]3 years ago
3 0

Answer:

y + 7 = - \frac{2}{3}(x - 1)

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

Where m is the slope and (a, b) a point on the line

Here m = - \frac{2}{3} and (a, b) = (1, - 7), thus

y - (- 7) = - \frac{2}{3}(x - 1), that is

y + 7 = - \frac{2}{3}(x - 1) ← in point- slope form

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Name the x-intercept
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Answer:

There are no x-intercepts

Step-by-step explanation: This is because the parabola does not pass through the x-axis at all.

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Select the correct answer.
gulaghasi [49]
The correct answer is B.
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3 years ago
A particular concentration of a chemical found in polluted water has been found to be lethal to 40% of the crayfish that are exp
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Answer:

a) P(X=13)=(24C13)(0.4)^{13} (1-0.4)^{24-13}=0.0608

P(X=17)=(24C17)(0.4)^{17} (1-0.4)^{24-17}=0.0017

P(X=13 U X=17) = 0.0608+0.0017=0.0624

b) P(X \geq 17)=0.000427

c) P(X \geq 16)= 1-0.000427=0.9996

d) E(X)= np = 22*0.4=8.8

e) Var(X) = np(1-p)= 22*0.4*(1-0,4)=5.28

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=24, p=0.4)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

3) Part a

P(X=13)=(22C13)(0.4)^{13} (1-0.4)^{22-13}=0.0336

P(X=17)=(22C17)(0.4)^{17} (1-0.4)^{22-17}=0.00035

P(X=13 U X=17) = 0.0336+0.00035=0.0340

4) Part b

P(X \geq 17)=P(X=17)+P(X=18)+P(X=19)+P(X=20)+P(X=21)+P(X=22)

P(X=17)=(22C17)(0.4)^{17} (1-0.4)^{22-17}=0.00035

P(X=18)=(22C18)(0.4)^{18} (1-0.4)^{22-18}=0.0000651

P(X=19)=(22C19)(0.4)^{19} (1-0.4)^{22-19}=0.00000914

P(X=20)=(22C20)(0.4)^{20} (1-0.4)^{22-20}=0.000000914

P(X=21)=(22C21)(0.4)^{21} (1-0.4)^{22-21}=5.81x10^{-8}

P(X=22)=(22C22)(0.4)^{22} (1-0.4)^{22-22}=1.76x10^{-9}

P(X \geq 17)=0.000427

5) Part c

P(X \leq 16)=1-P(X>16)=1-P(X\geq 17) = 1-[P(X=17)+P(X=18)+P(X=19)+P(X=20)+P(X=21)+P(X=22)]

P(X \geq 16)= 1-0.000427=0.9996

6) Part d

The expected value is:

E(X)= np = 22*0.4=8.8

7) Part e

The variance is given by:

Var(X) = np(1-p)= 22*0.4*(1-0,4)=5.28

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James can run 10 miles per hour.

The equation is J*2=A
Let J = James
Let A = Jamil
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Answer:

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