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Gnoma [55]
2 years ago
15

Help me with this please I will give you extra points

Mathematics
1 answer:
Ierofanga [76]2 years ago
6 0

\qquad\qquad\huge\underline{{\sf Answer}}♨

Total outcomes in which Billy can get 5 or more : 2 <u>outcomes</u>

Total outcomes = 6

So, the probability of getting 5 or higher is ~

\qquad \sf  \dashrightarrow \: \dfrac{favorable \: outcomes}{total \: outcomes}

\qquad \sf  \dashrightarrow \: \dfrac{2}{6}

\qquad \sf  \dashrightarrow \: \dfrac{1}{3}

Therefore, the correct choice is C

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

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What is 6,051 x 10^5 written in standard form?
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Answer:

605100

Step-by-step explanation:

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At the beginning of each of her four years in college, Miranda took out a new Stafford loan. Each loan had a principal of $5,500
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Answer:

D. $31,337.27

Step-by-step explanation:

We have that the initial amount of the loan is $5500.

Miranda took the loan for 4 years. So, the total present value is $5500×4 = $22,000.

The rate of interest on the loan is 7.5% i.e. 0.075 and it was for the duration of 10 years.

Also, it is given that the loan was compounded annually.

We have the formula as,

P=\frac{\frac{r}{n}\times PV}{1-(1+\frac{r}{n})^{-t\times n}}

i.e. PV=\frac{P\times [1-(1+\frac{r}{n})^{-t\times n}]}{\frac{r}{n}}

Substituting the values, we get,

i.e. PV=\frac{P\times [1-(1+\frac{0.075}{12})^{-10\times 12}]}{\frac{0.075}{12}}

i.e. 22000=\frac{P\times [1-(1+0.00625)^{-120}]}{0.00625}

i.e. 22000=\frac{P\times [1-(1.00625)^{-120}]}{0.00625}

i.e. 22000=\frac{P\times [1-0.4735]}{0.00625}

i.e. 22000=\frac{P\times 0.5265}{0.00625}

i.e. P=\frac{22000\times 0.00625}{0.5265}

i.e. P=\frac{137.5}{0.5265}

i.e. P=261.16

Thus, the total lifetime cost to pay of the loans compounded annually  = 261.16 × 120 = $31,339.2

Hence, the total cost close to the answer is $31,337.27

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