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2 years ago

What best determines whether a borrower's interest rate on an adjustable rate loan goes up or down?

Mathematics

2 answers:

antiseptic1488 [7]2 years ago

7 0

Answer:

-a market's condition is the correct answer

Step-by-step explanation:

edge 2021 :)

RoseWind [281]2 years ago

3 0

Answer:

C on edg

Step-by-step explanation:

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PLZ HELP I WILL GIVE THE BRAINLEST AND 22 PTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! An expression is shown below: 6(m + 2 +

n200080 [17]

Answer:

part A: expression 1: 6(8m+2)

expression 2: 6m+12+42m

Part B: 6(m+2+7m)= 6(8m+2)

combine like terms is 6(m+2+7m), so it is 6(2+8m) or 6(8m+2)

6(8m+2)= 6(8m+2)

Part C: 6m+12+42m=6(m+2+7m)

m=0

6(0+2+7(0))=6(0)+12+42(0)

6(2)= 12

12=12

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3 years ago

Find the sample mean on the outputs. Does the sample mean provide some evidence that the mean of the population of all possible

soldi70 [24.7K]

I know it and it is 7

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2 years ago

Suki is making lemonade to bring to the beach. She has two containers. One holds one gallon and the other holds 7 quarts. If she

Alex73 [517]

Answer:

40 cups

Step-by-step explanation:

7 0

2 years ago

If you roll 3 dice (one white, one black, and one red), find the probability that

zheka24 [161]

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3 0

3 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.

To find just the percentage above the mean, we divide this value by 2

So:

$P = {0.9545 - 0.6827}{2} = 0.1359$

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0

3 years ago