All categories

2 years ago

write the equation of the line in point slope form that passes through the given point with the given slope 6. (-4, 3); m = 2

Mathematics

1 answer:

zysi [14]2 years ago

3 0

Analyze the problem

ignoring the "6."

using point slope form,

y-3=2(x+4)

multiplication

y-3=2x+8

move the three and add

y=2x+11

You might be interested in

A national college researcher reported that 65% of students who graduated from high school in 2012 enrolled in college. Twenty n

Usimov [2.4K]

Answer:

a) The probability that exactly 17 of them enroll in college is 0.116.

b) The probability that more than 14 enroll in college is 0.995.

c) The probability that fewer than 11 enroll in college is 0.001.

d) It would be be unusual if more than 24 of them enroll in college since the probability is 0.009.

Step-by-step explanation:

We can model this with a binomial distribution, with n=29 and p=0.65.

The probability that k students from the sample who graduated from high school in 2012 enrolled in college is:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{29}{k} 0.65^{k} 0.35^{29-k}\\\\\\$

a) The probability that exactly 17 of them enroll in college is:

$P(x=17) = \dbinom{29}{17} p^{17}(1-p)^{12}=51895935*0.0007*0=0.116\\\\\\$

b) The probability that more than 14 of them enroll in college is:

$P(X>14)=\sum_{15}^{29} P(X=k_i)=1-\sum_{0}^{14} P(X=k_i)\\\\\\P(x=0)=0\\\\P(x=1)=0\\\\P(x=2)=0\\\\P(x=3)=0\\\\P(x=4)=0\\\\P(x=5)=0\\\\P(x=6)=0\\\\P(x=7)=0\\\\P(x=8)=0\\\\P(x=9)=0\\\\P(x=10)=0.001\\\\P(x=11)=0.002\\\\P(x=12)=0.005\\\\P(x=13)=0.013\\\\P(x=14)=0.027\\\\\\P(X>14)=1-0.005=0.995$

c) Using the probabilities calculated in the point b, we have:

$P(X$

d) The probabilities that more than 24 enroll in college is:

$P(X>24)=\sum_{25}^{29}P(X=k_i)\\\\\\ P(x=25) = \dbinom{29}{25} p^{25}(1-p)^{4}=23751*0*0.015=0.007\\\\\\P(x=26) = \dbinom{29}{26} p^{26}(1-p)^{3}=3654*0*0.043=0.002\\\\\\P(x=27) = \dbinom{29}{27} p^{27}(1-p)^{2}=406*0*0.123=0\\\\\\P(x=28) = \dbinom{29}{28} p^{28}(1-p)^{1}=29*0*0.35=0\\\\\\P(x=29) = \dbinom{29}{29} p^{29}(1-p)^{0}=1*0*1=0\\\\\\\\P(X>24)=0.007+0.002+0+0+0=0.009$

6 0

3 years ago

y=-3x+9 and goes through point P (-4,1). Then write the equation for the perpendicular line in point slope and slope intercept f

Sindrei [870]

This is the correct answer

4 0

3 years ago

What is .88 divided by 17

Marina86 [1]

The answer is .05176470588 or 0.052 (rounded answer).

7 0

4 years ago

Read 2 more answers

(1-sinx+cosx)^2 = 2(1+sinx)(1+cosx)

bogdanovich [222]

If you're trying to establish an identity, the given equation is not an identity. The proper identity would be as follows:

(1 - sin(x) + cos(x))² = (1 - sin(x))² + 2 (1 - sin(x)) cos(x) + cos²(x)

… = (1 - 2 sin(x) + sin²(x)) + 2 (1 - sin(x)) cos(x) + cos²(x)

… = 2 - 2 sin(x) + 2 (1 - sin(x)) cos(x)

… = 2 - 2 sin(x) + 2 cos(x) - 2 sin(x) cos(x)

… = 2 (1 - sin(x) + cos(x) - sin(x) cos(x))

… = 2 (1 - sin(x) + cos(x) (1 - sin(x)))

… = 2 (1 - sin(x)) (1 + cos(x))

But if you're trying to solve an equation:

(1 - sin(x) + cos(x))² = 2 (1 + sin(x)) (1 + cos(x))

2 (1 - sin(x)) (1 + cos(x)) = 2 (1 + sin(x)) (1 + cos(x))

(1 - sin(x)) (1 + cos(x)) - (1 + sin(x)) (1 + cos(x)) = 0

(1 + cos(x)) (1 - sin(x) - 1 - sin(x)) = 0

-2 sin(x) (1 + cos(x)) = 0

sin(x) = 0 or 1 + cos(x) = 0

sin(x) = 0 or cos(x) = -1

[x = arcsin(0) + 2nπ or x = arcsin(0) + π + 2nπ] or

… [x = arccos(-1) + 2nπ]

We have arcsin(0) = 0 and arccos(-1) = π, so the solution set reduces to

x = 2nπ or x = (2n + 1)π

(where n is any integer)

7 0

3 years ago

Have a math problem in algebra 2 where can i take a picture to get some help

Veseljchak [2.6K]

Answer:

Download photomath to picture your math problems.

8 0

4 years ago

write the equation of the line in point slope form that passes through the given point with the given slope 6. (-4, 3); m = 2​

write the equation of the line in point slope form that passes through the given point with the given slope 6. (-4, 3); m = 2