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Bogdan [553]
3 years ago
10

What is the answer to f(x)=400(1/2)^4x

Mathematics
1 answer:
Vlad1618 [11]3 years ago
4 0

Answer:

25x

Step-by-step explanation:

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The rabbit population in a small zoo starts with 150 rabbits. The rabbit population is decreasing at a rate of 15% each week. Wh
Mariulka [41]

Answer:

Option D F(x)=150(0.85)^{x}

Step-by-step explanation:

In this problem we have a exponential function of the form

F(x)=a(b)^{x}

where

F(x) -----> the rabbit population

x ----> the number of weeks

a is the initial value

b is the base

In this problem we have

a=150 rabbits

b=1-r

r=15%=15/100=0.15

so

b=1-0.15=0.85

substitute the values

F(x)=150(0.85)^{x}

8 0
3 years ago
Read 2 more answers
If the shaded sector is 75m what would the area be of the circle? A. 30 B. 36 C.144 D. 108
fredd [130]
The correct answer is letter B
8 0
3 years ago
Drone INC. owns four 3D printers (D1, D2, D3, D4) that print all their Drone parts. Sometimes errors in printing occur. We know
USPshnik [31]

Answer:

Step-by-step explanation:

Hello!

There are 4 3D printers available to print drone parts, then be "Di" the event that the printer i printed the drone part (∀ i= 1,2,3,4), and the probability of a randomly selected par being print by one of them is:

D1 ⇒ P(D1)= 0.15

D2 ⇒ P(D2)= 0.25

D3 ⇒ P(D3)= 0.40

D4 ⇒ P(D4)= 0.20

Additionally, there is a chance that these printers will print defective parts. Be "Ei" represent the error rate of each print (∀ i= 1,2,3,4) then:

P(E1)= 0.01

P(E2)= 0.02

P(E3)= 0.01

P(E4)= 0.02

Ei is then the event that "the piece was printed by Di" and "the piece is defective".

You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)

Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:

P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06

This is a conditional probability and you can calculate it as:

P(A/B)= \frac{P(AnB)}{P(B)}

To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.

P(D1∩E)= P(E1)= 0.01

P(D2∩E)= P(E2)= 0.02

P(D3∩E)= P(E3)= 0.01

P(D4∩E)= P(E4)= 0.02

Now you can calculate the probability of the piece bein printed by each printer given that it is defective:

P(D1/E)= \frac{P(E1)}{P(E)} = \frac{0.01}{0.06}= 0.17

P(D2/E)= \frac{P(E2)}{P(E)} = \frac{0.02}{0.06}= 0.33

P(D3/E)= \frac{P(E3)}{P(E)} = \frac{0.01}{0.06}= 0.17

P(D4/E)= \frac{P(E4)}{P(E)} = \frac{0.02}{0.06}= 0.33

P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.

The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)

I hope this helps!

8 0
3 years ago
Which line is represented by y = 3/2x - 3?
const2013 [10]

Answer:

The answer is the third picture

Step-by-step explanation:

6 0
3 years ago
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120/210 *60 (give the answer in hour and minutes)
Katyanochek1 [597]
120/210 = 0.57
0.57 x 60 = 34.2

34 hours and 12 minutes
8 0
3 years ago
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