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Lana71 [14]
3 years ago
9

When h has the value 5 calculate 6h+9

Mathematics
2 answers:
jonny [76]3 years ago
6 0
The answer is 39 because 6 times 5 is 30. 30 plus 9 is 39. :)
tankabanditka [31]3 years ago
4 0
39. If H equals 5 you would multiply 6 by 5 and add it by 9 which equals 39.
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The absolute value of a number is
makvit [3.9K]

Answer: the magnitude of a real number without regard to its sign.

Step-by-step explanation:

Meaning whether it is 5 or -5 the absolute value is 5 for both because that is their distance from zero.

7 0
3 years ago
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Katrina drinks 0.5 gallons of water per day. Which expression shows how to find the number of cups of water she drinks in a
Hunter-Best [27]

Answer:

Well the options are not copy pasted properly so im just gonna answer the question

Step-by-step explanation:

1 Gallon contains 16 cups,

so 16 cups for 1 day

so for 7 days,

    No. of cups = 7 x 16 = 112

so the expression would be,

16 cups x 7 days

7 0
3 years ago
F(x) = x5 + 5x4 - 5x3 - 25x2 + 4x + 20
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Answer:

-4(5-x)

Step-by-step explanation:

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

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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
Could you please explain <br><br> find an equation for i, -i, -4, 1 = x as its solution
Virty [35]

We can do this easily using 0s.

(x - i) (x + i) (x + 4) (x - 1) = 0

If you plug in any of the numbers, you'll get 0, making the equation true.

5 0
3 years ago
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