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3 years ago

9

if the interest rate is 3% and a total of $4,370.91 will be paid to you at the end of 3 years, what is the present value of that

sum?

1 answer:

Sonja [21]3 years ago

7 0

The original value of the sum is $131.1273...
Rounded up that is $131.13

PLEASE RATE AS THE BRAINLIEST ANSWER! THANK YOU! :)

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PLEASE HELP! Find an equation of the line whose intercepts are twice those of the graph of 5y+2x=10.

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5y+2x=10y=−25x+2246810−2−4−6−8−10246810−2−4−6−8−10Let's solve for x.5y+2x=10Step 1: Add -5y to both sides.2x+5y+−5y=10+−5y2x=−5y+10Step 2: Divide both sides by 2.2x2=−5y+102x=−52y+5Answer:x=−52y+5

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What is the value of 2 over 3 to the power of 0 to the power of -3

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Answer:

$((\frac{2}{3})^0)^{-3}=1$

Step-by-step explanation:

We need to find the value of $((\frac{2}{3})^0)^{-3}$

Solving:

We know, $(a^b)^n = a^{b*n}$

$((\frac{2}{3})^{0*-3})$

$(\frac{2}{3})^0$

a^0 = 1

so,

$(\frac{2}{3})^0=1$

So, the value of $((\frac{2}{3})^0)^{-3}=1$

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The heights of a certain type of tree are approximately normally distributed with a mean height p = 5 ft and a standard

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Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean

Step-by-step explanation:

⇒ $1^s^t$ statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)

an X value is found Z standard deviations from the mean mu if:

$\frac{X-\mu}{\sigma} = Z$

In this case we have: $\mu=5\ ft$ $\sigma=0.4\ ft$

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1$

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒ $2^n^d$ statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1$

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒ $3^r^d$ statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean (FALSE)

For X =5.8 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2$

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒ $4^t^h$ statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

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$Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3$

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

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