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zimovet [89]
2 years ago
7

Which quadrant will the line y=Negative one-fourthx – 15 never pass through?

Mathematics
1 answer:
Nata [24]2 years ago
6 0

Answer:

quadrant 2

Step-by-step explanation:

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5 0
2 years ago
Help me please..................
Rudiy27
253. First, I squared 8, to get 64, then multiplied that by 12. I subtracted 9 from this because 3x3 is 9 to get 759. Then, divide this by 3 to get 253. Hope this helped!
8 0
3 years ago
Read 2 more answers
Help me to answer this pls​
nlexa [21]

Problem 1

x = measure of angle N

2x = measure of angle M, twice as large as N

3(2x) = 6x = measure of angle O, three times as large as M

The three angles add to 180 which is true of any triangle.

M+N+O = 180

x+2x+6x = 180

9x = 180

x = 180/9

x = 20 is the measure of angle N

Use this x value to find that 2x = 2*20 = 40 and 6x = 6*20 = 120 to represent the measures of angles M and O in that order.

<h3>Answers:</h3>
  • Angle M = 40 degrees
  • Angle N = 20 degrees
  • Angle O =  120 degrees

====================================================

Problem 2

n = number of sides

S = sum of the interior angles of a polygon with n sides

S = 180(n-2)

2700 = 180(n-2)

n-2 = 2700/180

n-2 = 15

n = 15+2

n = 17

<h3>Answer: 17 sides</h3>

====================================================

Problem 3

x = smaller acute angle

3x = larger acute angle, three times as large

For any right triangle, the two acute angles always add to 90.

x+3x = 90

4x = 90

x = 90/4

x = 22.5

This leads to 3x = 3*22.5 = 67.5

<h3>Answers:</h3>
  • Smaller acute angle = 22.5 degrees
  • Larger acute angle = 67.5 degrees
4 0
2 years ago
M% of n is what percent of p?<br> FIND FOR PERCENT
makvit [3.9K]

Answer:

m*p/n = percent

Step-by-step explanation:

m% of n is what percent of p?

We can use ratios

m             percent

------- =  ---------------

n                    p

Using cross products

m*p = n * percent

Divide each side by n

m*p/n = percent

5 0
3 years ago
Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o
timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 1 the average number of aircraft arrival is:

\lambda t=8\times 1=8

The probability distribution of a Poisson distribution is:

P(X=x)=\frac{e^{-8}(8)^{x}}{x!}

Compute the value of P (X = 6) as follows:

P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

P(X\geq 6)=1-P(X

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

P(X\geq 10)=1-P(X

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 1.5=12

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

SD=\sqrt{\lambda t}=\sqrt{12}=3.464

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 2.5=20

Compute the value of P (X ≥ 20) as follows:

P(X\geq 20)=1-P(X

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0
3 years ago
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