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4 years ago

Find the nth term of 14,22,30,38

1 answer:

5 0

First find the gap which is 8.
Then you write 8N. To get 14 what do you have to do to the 8? Add 6 so the nth term is 8N+6

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Andre says I'm multiplied 4 by 5 then cubed the results select all expressions are equal Andres answer a 4×5 to the power of 3B4

Charra [1.4K]

Answer:

(4 * 5)^3 ;

8000

Step-by-step explanation:

Andre's statement :

Multiply 4 by 5 = 4 * 5 = 20

Cube the result = 20³

The equal expression from the option :

A.) (4 * 5)^3

Also, (4 * 5)^3 = 20^3 = 20 * 20 *20 = 8000

5 0

3 years ago

The number N of cars produced at a certain factory in 1 day after t hours of operation is given by Upper N (t )equals 800 t minu

Elden [556K]

Answer:

The cost C as a function of t is C(t) = 30,000 + 6,400,000 t - 40,000 t²

Step-by-step explanation:

The function N(t) = 800 t - 5t², represents the number of cars produced at a time t hours in a day, where 0 ≤ t ≤ 10

The function C(N) = 30,000 + 8,000 N, represents the cost C (in dollars) of producing N cars

We need to find The cost C as a function of the time t

That means Substitute N in C by its function by other word find the composite function (C о N)(t)

∵ C(N) = 30,000 + 8,000 N

∵ N(t) = 800 t - 5 t²

- Substitute N in C by 800 t - 5 t²

∴ C(N(t)) = 30,000 + 8000(800 t - 5 t²)

- Multiply the bracket by 8000

∴ C(N(t)) = 30,000 + 8000(800 t) - 8000(5 t²)

∴ C(N(t)) = 30,000 + 6,400,000 t - 40,000 t²

- C(N(t) = C(t)

∴ C(t) = 30,000 + 6,400,000 t - 40,000 t²

The cost C as a function of t is C(t) = 30,000 + 6,400,000 t - 40,000 t²

5 0

3 years ago

Which of the following number sentences is an example of the identity property? Select all that apply.

ArbitrLikvidat [17]

I think one is 8.7 + 0 = 8.7

7 0

3 years ago

a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade

son4ous [18]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

$P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})$