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Ratling [72]
2 years ago
11

Draw a line tangent to the circle at the point (0, 4). Write an equation for this tangent line. Explain your work.PLEASE HELP

Mathematics
1 answer:
kirill [66]2 years ago
5 0

Answer:

y = 1/4x + 4

Step-by-step explanation:

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If two boxes of cereal and a jug of milk cost $8.50, and three boxes of cereal and two jugs of milk cost $14.00, how much does a
Citrus2011 [14]

Answer:

$3

Step-by-step explanation:

Let c represent the cost of a box of cereal and let m represent the cost of a jug of milk.

Create a system of equations:

2c + m = 8.5

3c + 2m = 14

Solve by elimination by multiplying the top equation by -2:

-4c - 2m = -17

3c + 2m = 14

Add them together and solve for c:

-c = -3

c = 3

So, a box of cereal costs $3

3 0
2 years ago
Is 49 and 770 relatively prime
natita [175]
No nether are a prime number because they have a remainder and prime numbers <span> can only be divided without a remainder by itself and one
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6 0
3 years ago
sylvia has $100. She spent 4/10 of her money on a jacket and 20/100 of her money on jeans. What fraction of her money did sylvia
8_murik_8 [283]
60/100 that's what I got .
4 0
3 years ago
Read 2 more answers
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
2 3/4 divided by 4 1/8 in fraction form
dmitriy555 [2]

Answer:

5/4

Step-by-step explanation:

5 0
2 years ago
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