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Irina-Kira [14]
3 years ago
14

the table compares the time at which jayden got on the bus to school (in minutes after 8 am) and the duration of the ride (in mi

nutes) for several different days. Can the duration of the ride be represented as a function of the time jayden got on the bus
Mathematics
2 answers:
DiKsa [7]3 years ago
7 0

Answer:

No

Step-by-step explanation:

seropon [69]3 years ago
4 0

Answer: The answer is no, the duration of the ride can’t be represented as a function of the time Jayden got on the bus.

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Answer:

y = 1/3x + 2

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If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f
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Part A) Annual \$66,480.95  

Part B) Semiannual \$66,862.38  

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Step-by-step explanation:

we know that

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

so

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in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=1  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{1})^{1*5}  

A=\$47,400(1.07)^{5}  

A=\$66,480.95  

Part B) Semiannual

in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=2  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{2})^{2*5}  

A=\$47,400(1.035)^{10}  

A=\$66,862.38  

Part C) Monthly

in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=12  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{12})^{12*5}  

A=\$47,400(1.0058)^{60}  

A=\$67,195.44  

Part D) Daily

in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=365  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{365})^{365*5}  

A=\$47,400(1.0002)^{1,825}  

A=\$67,261.54  

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Answer:

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3 years ago
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