IM1 Test Exponential Functions

IM1 Test Exponential Functions

9th - 10th Grade

40 Qs

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IM1 Test Exponential Functions

IM1 Test Exponential Functions

Assessment

Quiz

Mathematics

9th - 10th Grade

Hard

Created by

Elvira Avila

Used 4+ times

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40 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Rhonda deposited $3000 in an account in the Merrick National Bank, earning 4.2% interest, compounded annually. She made no deposits or withdrawals. Write an equation that can be used to find B, her account balance after t years.

B(t) = 3000(1 - 4.2)t

B(t) = 3000(1 + 4.2)t

B(t) = 3000(1 - 0.042)t

B(t) = 3000(1 + 0.042)t

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Marilyn collects old dolls. She purchases a doll for $450. Research shows this doll's value will increase by 2.5% each year. Write an equation that determines the value, V, of the doll t years after purchase.

V(t) = (1 + 0.025)t

V(t) = (1 - 0.025)t

V(t) = (1 + 2.5)t

V(t) = (1 - 2.5)t

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

A student invests $500 for x years in a savings account that earns 4% interest per year. No further deposits or withdrawals are made during this time. Write an function f(x) to represent the amount of money earned after x years.

f(x) = 500 (.60)x

f(x) = 500 (.96)x

f(x) = 500 (1.04)x

f(x) = 500 (1.4)x

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

A car was purchased for $25,000. Research shows that the car has an average yearly depreciation rate of 18.5%. Create a function that will determine the value, V(t), of the car t years after purchase.

V(t) = 25000(1 - 0.185)t

V(t) = 25000(1 + 0.185)t

V(t) = 25000(1 - 18.5)t

V(t) = 25000(1 + 18.5)t

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

John invested $900 in a savings account at a 2.3% annual interest rate. He made no deposits or withdrawals on the account. Write a function J(t) to represent the amount of money in the account after t years.

J(t) = 900(1 - 0.23)t

J(t) = 900(1 + 0.23)t

J(t) = 900(1 + 0.023)t

J(t) = 900(1 - 0.023)t

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