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babymother [125]
2 years ago
13

Clare made $160 babysitting last summer. She put the money in a savings account that pays 3% interest per year. If clare doesn't

touch the money in her account, she can find the amount she'll have the next year by multiplying her current amount 1.03 a. Write an expression for the amount of money clare would have after 30 years if she never withdraws money from her account.
Mathematics
1 answer:
AVprozaik [17]2 years ago
8 0

We have been given that Clare made $160 babysitting last summer. She put the money in a savings account that pays 3% interest per year. If Clare doesn't touch the money in her account, she can find the amount she'll have the next year by multiplying her current amount 1.03.

We are asked to write an expression for the amount of money Clare would have after 30 years if she never withdraws money from her account.

We will use exponential growth function to solve our given problem.

An exponential growth function is in form y=a(1+r)^x, where

y = Final value,

a = Initial value,

r = Growth rate in decimal form,

x = Time.

3\%=\frac{3}{100}=0.03

We can see that initial value is $160. Upon substituting our given values in above formula, we will get:

y=160(1+0.03)^x

y=160(1.03)^x

To find amount of money in Clare's account after 30 years, we need to substitute x=30 in our equation.

y=160(1.03)^{30}

Therefore, the expression 160(1.03)^{30} represents the amount of money that Clare would have after 30 years.

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For this case assuming that the random variable is X

df = n-1

And replacing n = 24 we got:

df = 24-1=23

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Step-by-step explanation:

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Solution to the problem

For this case assuming that the random variable is X

df = n-1

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df = 24-1=23

And we notate the distribution we got: X \sim t_{n-1}= t_{23}

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What is the exact value of the cos F?
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Step-by-step explanation:

The triangles are similar, so you know that ...

... cos(D) = cos(A) = 40/41.

From trig relations, you know ...

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and

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So ...

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The ratio for cos(A) tells you that you can consider AB=40, AC=41. Then, using the Pythagorean theorem, you can find BC = √(41² -40²) = √81 = 9.

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2 years ago
In high-school 135 freshmen were interviewed.
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Answer:

a) n(none) = 25

b) n(PE but not Bio) = 25

c) n(ENG but not both BIO and PE) = 55

d) n(students that did not take Eng or Bio) = 40

e) P( Students did not take exactly two subjects) = 0.65

Step-by-step explanation:

From the Venn diagram drawn:

a) Number of students that took none

n(Freshmen) = 135

n(all three) = 5

n (PE and Bio) = 10

n(PE and Eng) = 15

n(Bio and Eng) = 7

n (PE and Bio only) = 10 - 5 = 5

n(PE and Eng only) = 15 - 5 = 10

n(Bio and Eng only) = 7 - 5 = 2

n(PE only) = 35 - 5 - 5 - 10 = 15

n(Bio only) = 42 - 5 - 5 - 2 = 30

n(Eng only) = 60 - 10 - 5 -2 = 43

n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)

135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)

135 = 110 + n(none)

n(none) = 135 - 110

n(none) = 25

b)Number of students that too PE but not Bio

n(PE but not bio)= n(PE only) + n(PE and Eng only)

n(PE but not Bio) = 15 + 10

n(PE but not Bio) = 25

c) Number of students that took ENG but not both BIO and PE

n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10

n(ENG but not both BIO and PE) = 55

d) Number of students that did not take ENG or BIO

n( students that did not take Eng or Bio) = n(PE only) + n(none)

n(students that did not take Eng or Bio) = 15 + 25

n(students that did not take Eng or Bio) = 40

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n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)

n( Students that did not take exactly two subjects) = 15 + 30 + 43

n( Students that did not take exactly two subjects) = 88

P( Students did not take exactly two subjects) = 88/135

P( Students did not take exactly two subjects) = 0.65

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