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3 years ago

- 9x = 81 plsssss help me will cash app put it below the answer !

Mathematics

1 answer:

Firdavs [7]3 years ago

5 0

Divide both sides of the equation by the same factor. -9/-9 = 81/-9. simplify x= -9 and the answer is x= -9

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How do you solve this and i have other questions like this one i really need to know how to do it

Natalija [7]

Answer:

Step-by-step explanation:

it's because 5 is squared so 5x5 is 25 and x squared

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2 years ago

What is greater 64kg or 64000g

Aleksandr [31]

Answer: neither

Step-by-step explanation: because 64 kg and 640000g are equal masses.

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2 years ago

Read 2 more answers

The weight, in pounds, of a newborn baby tt months after birth can be modeled by the function W(t)=1.25t+6.W(t)=1.25t+6. What is

ollegr [7]

Answer:

$6$

Step-by-step explanation:

GIVEN: The weight, in pounds, of a newborn baby $t$ months after birth can be modeled by the function $W(t)=1.25t+6$ .

TO FIND: What is the y-intercept of the function and what is its interpretation in the context of the problem.

SOLUTION:

in $W(t)=1.25t+6$

if $t=0$

$W(0)=6$

Hence y-intercept is $6$

the y-intercept in this represents the weight of baby just after he is born.

8 0

2 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

PLEASE ANSWER NUMBER 15 ‼️‼️‼️‼️

elena-s [515]

Answer:

Step-by-step explanation:

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2 years ago