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2 years ago

F (x) = -x + 7 identifiy 0

Mathematics

1 answer:

jekas [21]2 years ago

7 0

Answer:

Step-by-step explanation:

$f (x) = -x + 7 \\ \\ plug \: f (x) = 0 \\ \\ 0 = - x + 7 \\ \\ x = 7 \\ \\ zero \: is \: 7$

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The answer show work please

Romashka [77]

Well you need to do 10 plus 13 which equals 23 you don't count the inside number

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3 years ago

A taxi cab in myrtle beach charges $2 per mile and $1 for every person. If a taxi cab ride for two people costs $12, how far did

Oxana [17]

Answer:

5 Miles

Step-by-step explanation:

5 Miles = 1 Miles = $2 X 5 = $10

1 Person = $1

Therefore 2 People = $1 X 2

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3 years ago

A person's metabolic rate is the rate at which the body consumes energy. Metabolic rate is important in studies of weight gain,

cupoosta [38]

Let m be mean

Mean= sum/ n

Mean= (1720+1687+1367+1614+1460+1867+1436) / 7

m= 11151 / 7

M= 1593

Mean= 1593

Standard deviation

|x-m|^2

For 1st: |1720-1593|^2=8836

For 2nd: |1687-1593|^2=10201

For 3rd: |1367-1593|^2=51076

For 4th: |1614-1593|^2=441

For 5th: |1460-1593|^2=1689

For 6th: |1867-1593|^2=75076

For 7th: |1436-1593|^2=24649

Summation of |x-m|^2 = 171968

Standard deviation sample formula is:

S.D = sqrt((summation of |x-m|^2) / n-1)

S.D=sqrt(171968/6)

S.D=sqrt(28661.33)

S.D=169.30

Standard deviation is 169.30

3 0

3 years ago

22.700 nearest hundredth/6413202/1c463fd9?utm_source=registration

Vesnalui [34]

You already have the answer in the question

4 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

F (x) = -x + 7 identifiy 0​

F (x) = -x + 7 identifiy 0