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Andrej [43]
3 years ago
12

$1,250 was deposited into an account for 2 years with a 2.5% interest rate. What is the in

Mathematics
1 answer:
raketka [301]3 years ago
4 0
I will assume it is 2.5% interest rate over 2 years and not per year. If this isn’t the case then let me know.
-
-
Well $1250 divided by 100 = $12.50
This is 1%-
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-
To make 2.5% we do-> $12.50 x 2.5= $31.25
This makes our interest rate $31.25
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-
-
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To get our total amount.....
$1250 + $31.25 = $1281.25
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Answer:

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Herman and Jackie are saving money to pay for college. Herman currently has $15,000 and is working hard to save $1000 per month.
Mashutka [201]

Answer:

Part 1) 10\ months

Part 2) \$25,000

Step-by-step explanation:

Let

y ----> the total amount of savings

x ----> the number of months

we know that

The linear equation in slope intercept form is equal to

y=mx+b

where

m is the slope or unit rate

b is the y-intercept or initial value

In this problem we have

<em>Herman</em>

The slope is equal to m=\$1,000\ per\ month

The y-intercept is b=\$15,000

substitute

y=1,000x+15,000 ----> equation A

<em>Jackie</em>

The slope is equal to m=\$1,300\ per\ month

The y-intercept is b=\$12,000

substitute

y=1,300x+12,000 ----> equation B

Part 1) In how many months will they have the same amount of savings?

equate equation A and equation B

1,300x+12,000=1,000x+15,000

solve for x

1,300x-1,000x=15,000-12,000

300x=3,000

x=10\ months

Part 2) How much will each of them have saved?

substitute the value of x=10 months in any of the equations

equation A

y=1,000(10)+15,000=\$25,000

equation B

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The probability distribution of the amount of memory X (GB) in a purchased flash drive is given below. x 1 2 4 8 16 p(x) .05 .10
Rama09 [41]

Answer:

a) E(X) = 6.45

b) E(X^{2} )= 57.25

c) V(X) = 15.648

d) E(3X + 2) = 21.35

e) E(3X^{2} +2) = 173.75

f) V(3X+2) = 140.832

g) E(X+1) = 7.45

h) V(X+1) = 15.648

Step-by-step explanation:

a) E(X) = \sum xP(x)

E(X) = (1*0.05) + (2*0.10) + (4*0.35) + (8*0.40) + (16*0.10)\\E(X) = 6.45

b)

E(X^{2} ) = (1^{2} *0.05) + (2^{2} *0.10) + (4^{2} *0.35) + (8^{2} *0.40) + (16^{2} *0.10)\\  E(X^{2} )= 57.25

c)

V(X) = E(X^{2} ) - (E(X))^{2} \\V(X) = 57.25 - 6.45^{2} \\V(X) = 15.648

d)

E(3X+2) = 3E(X) + 2\\E(3X+2) = (3*6.45) + 2 \\E(3X+2) = 21.35

e)

E(3X^{2} +2) = 3E(X^{2} ) + 2\\E(3X^{2} +2) = (3*57.25) + 2 \\E(3X^{2} +2) = 173.75

f)

V(3X+2) = 3^{2} V(X)\\V(3X+2) = 9*15.648\\V(3X+2) = 140.832

g)

E(X+1) = E(X) + 1\\E(X+1) = 6.45 + 1\\E(X+1) =7.45

h)

V(X+1) = 1^{2} V(X)\\V(X+1) = 15.648

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