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2 years ago

If F(x) = x - 5 and G(x) = x2, what is G(F(x))?

Mathematics

1 answer:

vlada-n [284]2 years ago

3 0

Answer:

here is the answer

Step-by-step explanation:

hope it helps

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Find the product. <br><br> a 4 (3 a 2 - 2 a + 1)

TEA [102]

Is the a the question or are you adding it

3 0

3 years ago

What is the difference between -17 and -7

kari74 [83]

Answer: -10

Step-by-step explanation:

Difference between -17 & -7 should be

-17 - (-7)

-10

6 0

2 years ago

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Lian needs to solve the quadratic equation: x^2 - 4x - 2 = 0. Which statement about how Lian should solve this equation is true?

Ulleksa [173]

Answer:

x = 4.45 or x = - 4.45

Step-by-step explanation:

Here are the steps:

Substitute the values,

$4+\sqrt{-4^2-4(1*-2)}$ ÷ 2 × 1

x = 2 ± √6

x = 4.45 or x = - 4.45

<em>good luck, i hope this helps :)</em>

5 0

3 years ago

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At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

3 years ago

Whats the lowest common multiple of 120 and 19600

Levart [38]

Answer:

Multiples of 120 are 120, 240, 360, 480, 600, 720, 840 etc; Multiples of 150 are 150, 300, 450, 600, 750, 900 etc; Therefore, the least common multiple of 120 and 150 is 600.

Least common multiple (LCM) of 19600 and 19619 is 384532400.

4 0

2 years ago

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