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allsm [11]
3 years ago
13

John and Annie have two distinguishable ponds. Initially, each of the ponds contains four ducks and five geese. John first picks

a bird uniformly at random from the left pond and moves it to the right pond. Then, Annie picks a bird uniformly at random from the right pond. What is the probability that Annie picks a duck?
Mathematics
1 answer:
Sladkaya [172]3 years ago
4 0

Answer: 4/9

Step-by-step explanation:

The probability that annie picks a duck Pt= the probability of picking a duck without accounting for the added one (Po)+ probability of picking the added bird and it's a duck(Pi)

Pt = Po+Pi

Since the total number of birds in the right pond is 10 after the addition of one by john

Po= 4/10

Pi= the of john adding a duck × the probability of annie picking the added bird

Pi= 4/9 × 1/10

Pi = 4/90

Pt= 4/10 + 4/90

Pt = (36+4)/90

Pt=40/90

Pt= 4/9

(This implies that the probability of picking a duck remain the same even after the addition of one bird from the left ponds because they both have equal proportions of duck and geese i.e the initial number of duck and geese in both right and left ponds are 4 and 5 respectively)

Thanks.

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Answer:

The value of the bond when Tyler's mom purchased it was $150

Step-by-step explanation:

we know that

In this problem we have a exponential function of the form

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

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b is the base

r is the rate

b=(1+r)

In this problem

r=4%=4/100=0.04

b=1+0.04=1.04

substitute

f(x)=a(1.04)^{x}

where

x is the number of years since the savings bond was purchased

f(x) is the value of the savings bond

For x=1

f(x)=$156

substitute

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Solve for a

156=a(1.04)

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therefore

The value of the bond when Tyler's mom purchased it was $150

8 0
4 years ago
A six-sided die is rolled two times. What is the theoretical fractional probability that the first number is greater than the se
Alekssandra [29.7K]

Given:

A six-sided die is rolled two times.

To find:

The theoretical fractional probability that the first number is greater than the second number.

Solution:

If a six-sided die is rolled two times, then the total possible outcomes are 36.

S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

The outcomes where the first number is greater than the second number are

A = {(2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), (6,5)}

So, the favorable outcomes = 15.

Now, the theoretical fractional probability that the first number is greater than the second number is

Probability=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}

Probability=\dfrac{15}{36}

Probability=\dfrac{5}{12}

Therefore, the required probability is \dfrac{5}{12}.

5 0
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What is the slope of the line shown below?​
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Hi there!

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Use slope formula to solve for the slope of the line:

slope = \frac{y_{2}-y_{1}  }{x_{2} -x_{1} }

Plug in points into the equation. We can use the points (0, -10) and (25, -5):

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Answer: B&E

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We know it is between 8.5 and 9

8.8^2 = 77.44

This is a good approximation

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