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3 years ago

Need this answer please help

Mathematics

1 answer:

HACTEHA [7]3 years ago

4 0

Answer:

Step-by-step explanation:

non linear because linear has always to the first power

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Zene deposited $50 in a savings account earning 5% interest, compounded annually. To thenearest cent, how much will he have in 3

JulsSmile [24]

Answer:

$\$57.88$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=3\ years\\ P=\$50\\ r=5\%=5/100=0.05\\n=1$

substitute in the formula above

$A=50(1+\frac{0.05}{1})^{1*3}$

$A=50(1.05)^{3}$

$A=\$57.88$

6 0

3 years ago

If a true-false test with 10 questions is given what is the probability of scoring?

yan [13]

Answer:50%

Step-by-step explanation: It doesn't matter how many questions there are because for each question, you have a 50% chance of getting it right. So, the probability is 50%.

7 0

1 year ago

20<br> X<br> 40°<br> solve for x

kirza4 [7]

Could you please formulate you answer better? So we can get a bette understanding

8 0

2 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.