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3 years ago

Math mystery rogue runner 7th graders

Mathematics

1 answer:

Blababa [14]3 years ago

8 0

Answer:

Answer:

Step-by-step explanation:

Step-by-step explanation:

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PLS HELP ASAP! WILL MARK BRAINLIEST sin ∠A = 3/5 4/5 3/4 5/4

Dafna11 [192]

Answer:

3/5

Step-by-step explanation:

sohcahtoa

since we're finding sine, we need the opposite and hypotenuse.

opposite of <A=3, hyp of <A=5.

opp comes before hyp in "soh", so sin<A=3/5

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3 years ago

The circumference of a circle is 8pi What is the radius?

Mashcka [7]

Answer:

Step-by-step explanation:

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3 years ago

A store had 150 laptops in the month of January. Every month, 20% of the laptops were sold and 10 new laptops were stocked in th

Softa [21]

Answer-

$\boxed{\boxed{f(n)=0.8(f(n-1))+10,\ f(0)=150,\ \ n>0}}$

Solution-

Here, n represents the number of months and f(n) represents the number of laptops in the store after n months.

As the store had 150 laptops in the month of January or at the beginning.

So $f(0)=150$

Every month, 20% of the laptops were sold and 10 new laptops were stocked in the store.

As 20% of laptops were sold, so 80% were in the store.

So, after one month total number of laptops in the store,

$\Rightarrow f(1)=0.8(150)+10$

$\Rightarrow f(1)=0.8(f(0))+10$ $(\because f(0)=150)$

Again after one month total number of laptops in the store,

$\Rightarrow f(2)=0.8\times (0.8(150)+10)+10$

$\Rightarrow f(2)=0.8(f(1))+10$ $(\because f(1)=0.8(150)+10)$

Analyzing the pattern, the recursive function f(n) will be,

$f(n)=0.8(f(n-1))+10,\ f(0)=150,\ \ n>0$

3 0

3 years ago

Read 2 more answers

Add parentheses this equation true 2×2+12÷4+3=4

trasher [3.6K]

Answer:is this true or false or something?

Step-by-step explanation:

5 0

4 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

4 years ago