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defon
2 years ago
15

E. Lindsey sold stationery to her family and her mother's friends. She deposited the $125

Mathematics
1 answer:
IRISSAK [1]2 years ago
3 0

Answer:

$134.71

Step-by-step explanation:

This is a simple interest question

A = P(1 + rt)

A = Amount after t months or years

P = Principal or amount saved

r = interest rate = 5.18% = 0.0518

t = time.in years = 18 months

= 1 year and 6 month

= 1.5 year

A = $125( 1 + 0.0518 × 1.5)

A = $125 ( 1 + 0.0777)

A = $125(1.0777)

A = $134.71

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The yearbook club is handing out T-shirts to its members. There are 5 blue, 7 green, 9 red, and 4 yellow T-shirts in all. If Jac
damaskus [11]

Answer:

n(R)=9

n(Y)=4

n(B)=5

n(G)=7

n(s)25

probability of getting red =n(R)/n(S)=9/25

6 0
2 years ago
Mr North spent $144,00 to build a fence around the perimeter of bis vegetable garden.He paid $6.oo per yard for fencing.
NikAS [45]
First, you could see the amount of fence he could buy, or 144/6, which would be 24, so Mr. North can buy 24 yards of fencing.
So now to find the possible plans, we know that there are four sides, but the width and the length occur twice since it's a rectangle.
So since we know that, we can just split 24 in half to find the possibilities for one of the width sides and one of the length sides, if that makes any sense. 24/2 = 12.
So now, you could say some possibilities are length = 6 and width = 6, or length = 4 and width = 8.
And now, to consider which plan would be the best, it would probably be a 6x6 design, because it gives the biggest area to the vegetable garden, and is easy to move around.
width = 6
length = 6
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5 0
3 years ago
Vince weighs 160 pounds, and his friend Nadir weighs 140 pounds. Nadir calculated that his weight on another planet would be abo
Oksanka [162]

Answer:76

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4 0
2 years ago
A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t
Licemer1 [7]
If A(t) is the amount of salt in the tank at time t, then the rate at which this amount changes over time is given by the ODE

A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}

A'(t)+\dfrac1{100}A(t)=15

We're told that the tank initially starts with no salt in the water, so A(0)=0.

Multiply both sides by an integrating factor, e^{t/100}:

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}
\left(e^{t/100}A(t)\right)'=15e^{t/100}
e^{t/100}A(t)=1500e^{t/100}+C
A(t)=1500+Ce^{-t/100}

Since A(0)=0, we have

0=1500+C\implies C=-1500

so that the amount of salt in the tank over time is given by

A(t)=1500(1-e^{-t/100})

After 10 minutes, the amount of salt in the tank is

A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}
8 0
3 years ago
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