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mr_godi [17]
3 years ago
10

What is (-1,2) and (1,-4) in slope-intercept form?

Mathematics
1 answer:
omeli [17]3 years ago
7 0
You will need the point slope formula if you don't know it already which is x2-x1 over y2-y1. so in this case, it would be 1-(-1) over -4 -(2) which gives you 1 over 3 simplified and that is your slope. you are then going to substitute one of the point in the slope formula, y=mx+b. so let's say you go with the first point and substitute it, that will give you, 2=1/3(-1)+b and you would solve for b. after you figure out your equation, you multiply 1/3 by -1 to give you -1/3+b=2. Then you would add -1/3 to both sides to get b by itself. You would have b= 2+(-1/3). you with then calculate the difference to get b= 2 1/3 or b= 2.3.
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Based on data from the U.S Center for Health Statistic, the death rate for males in the 15-24 age bracket is 114.4 per 100,000 p
antiseptic1488 [7]

Answer:

a) 0.998856

b) 0.997713

c) 0.999120

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Male death rate

114.4 per 100,000 population

So the survival rate is:

(100000 - 114.4)/(100,000) = 0.998856

Female death rate

44.0 per 100,000 population

So the survival rate is

(100000 - 44)/(100,000) = 0.99956

a) If a male in that bracket are randomly selected, what is the probability that he will survive? ( Express the answer with six decimal places.)

The survival rate for males is of 0.998856. So this is the probability that he will survive.

b) If two males in that ages bracket are randomly selected, what is the probability that they both survive?

Each one has a 0.998856 probability of surviving. So

P = (0.998856)^{2} = 0.997713

c) If two females in that age bracket are randomly selected, what is the probability that they both survive?

Each female has a 0.99956 probability of surviving. So

P = (0.99956)^{2} = 0.999120

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B3x%20-%206%7D%20%20%5C%20%5Cdiv%20%5Cfrac%7Bx%20%2B%209%7D%7Bx%20-%202%7D%
mojhsa [17]
Simplify the following:
x/((x - 2) (x + 9) (3 x - 6))

Factor 3 out of 3 x - 6:
x/(3 (x - 2) (x - 2) (x + 9))

Combine powers. x/((x - 2) (x + 9)×3 (x - 2)) = (x (x - 2)^(-1 - 1))/((x + 9)×3):
(x (x - 2)^(-1 - 1))/(3 (x + 9))

-1 - 1 = -2:
Answer: (x (x - 2)^(-2))/(3 (x + 9))
7 0
3 years ago
Pls help me out ill give brainlyist​
Anna35 [415]

Answer:

B) y=3x+4.5

Step-by-step explanation:

7 0
4 years ago
Solve the equation: p2 - 4p + 4 = 0.​
iragen [17]

Perfect square trinomial

\boxed{ a^2 \pm 2ab+b^2  = (a \pm b)^2 }

Using the perfect square trinomial

p^2 - 4p + 4 = 0

p^2 - 2(2p) + (2)^2 = 0

(p-2)^2 = 0

(p-2) = 0

p = 2

7 0
3 years ago
Read 2 more answers
For each function f(n) and time t in the table, determine the largest size n of a problem that can be solved in time t, assuming
egoroff_w [7]
First recall that a microsecond is 10^−6 seconds. Hence, one second = 10^6 microseconds, one hour = 3600000000 = 3.6 • 10^9 microseconds, one month (assume a month has 30 days) = 2592000000000 = 2.592 • 10^12 microseconds, and one century = 3110400000000000 = 3.1104 • 10^15 microseconds.

Row 1: f(n) = log n In this case, we need to determine the largest n such that log n ≤ 1000000. To solve this inequality, we need to rewrite the inequality as 2^logn ≤ 2^1000000 or n ≤ 2^1000000. Recall from lecture that 2^10 ≈ 10^3, thus we have that 2^1000000 = 2^10•100000 = (2^10)^100000 ≈ (10^3)^100000 = 10^300000. This is the result given in the textbook. For one hour, we have that n ≤ 2^3600•1000000 and thus n ≈ 10^1080000000.

Row 8: f(n) = n! For us to see that the largest sized input is 12 that can be processed within an hour when f(n) = n! one can simply, compute 12! And verify that it is less than the number of microseconds in one hour, but that 13! is greater than the number of microseconds in an hour.

Row 4: f(n) = n log n In this case, use Maple to solve equations like n log n−1000000 = 0. The Maple command for solving this equation is fsolve(n*log[2](n) - 1000000 = 0)
8 0
4 years ago
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