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3 years ago

Express polynomial as a product of two factors.

Mathematics

1 answer:

Marat540 [252]3 years ago

4 0

Answer:

x=36/7-3

Step-by-step explanation:

see this is for nerds so i can't really show you cause i don't know how to but put this on your test and u will get it right

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What is the measure of angle b

tigry1 [53]

Answer:

36°

Step-by-step explanation:

By remote interior angle theorem:

b + 81° = 117°

b = 117° - 81°

b = 36°

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3 years ago

Find the critical point for f and then use the second derivative test to decide whether the critical point is a relative maximum

anygoal [31]

Answer:the answer is 9

Step-by-step explanation:

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3 years ago

Find the slope of the line passing through the point (-6,8) and (-9,1)

Ipatiy [6.2K]

Hello!

Y2-y1/x2-x1

1-9/8-6
-8/2

The slope is -4

Hope this helps!

-bambi

5 0

2 years ago

A salesman gets a commission of 2.4% on sales up to sh. 100,000. He gets an additional commission of 1.5% on sales above this. C

emmainna [20.7K]

Answer:

Sh. 10,920

Step-by-step explanation:

Since the sales is above sh. 100,000,his commission is (2.4+1.5) 3.9%.

3.9% of 280,000= 10920.

Therefore, his commission is sh. 10,920

6 0

3 years ago

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

Express polynomial as a product of two factors.​

Express polynomial as a product of two factors.