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Mathematics

1 answer:

iren [92.7K]2 years ago

4 0

I’m pretty sure it’s C and E

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Find the simple interest rate needed in order for an investment of $2000 to grow to an account of $5000 in 3 years

Nutka1998 [239]

$\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$5000\\ P=\textit{original amount deposited}\dotfill&\$2000\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &3 \end{cases} \\\\\\ 5000=2000(1+r3)\implies \cfrac{5000}{2000}=1+3r\implies \cfrac{5}{2}=1+3r \\\\\\ 5=2+6r\implies 3=6r\implies \cfrac{3}{6}=r\implies 0.5=r\implies \stackrel{\textit{converting to percent}}{0.5\cdot 100\implies 50\%}$

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3 years ago

The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0 (Think of these as

bija089 [108]

Answer:

0.477 is the probability that the average score of the 36 golfers was between 70 and 71.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 70

Standard Deviation, σ = 3

Sample size, n = 36

Let the average score of all pro golfers follow a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(score of the 36 golfers was between 70 and 71)

$\text{Standard error of sampling} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{1}{2}$

$P(70 \leq x \leq 71) = P(\displaystyle\frac{70 - 70}{\frac{3}{\sqrt{36}}} \leq z \leq \displaystyle\frac{71-70}{\frac{3}{\sqrt{36}}}) = P(0 \leq z \leq 2)\\\\= P(z \leq 2) - P(z \leq 0)\\= 0.977 - 0.500 = 0.477= 47.7\%$