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3 years ago

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Which of the following data is used to determine credit scores? A. who stays with you at your current residence B. the location

of your current residence C. the length of stay at your current residence

1 answer:

dusya [7]3 years ago

8 0

Answer:

The correct option is the length of stay at your current residence.

Step-by-step explanation:

We have been asked that which data is used to determine credit scores.

The correct option is the length of stay at your current residence.

This question is used to determine credit scores. This helps the lender to look up information about your history of payments. When a lender asks this question they are able to determine how long you typically stay in one place, and look up bills at that residence to make sure they are paid on time before giving out a loan. They can also use this information to determine a credit score based on what is being paid on time, late or not at all...

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A body moves s metres in a time t seconds so that s = t3 – 3t2 + 8. Find:

Lady bird [3.3K]

Using derivatives, it is found that:

i) $v(t) = 3t^2 - 6t$

ii) 9 m/s.

iii) $a(t) = 6t - 6$

iv) 6 m/s².

v) 1 second.

<h3>What is the role of derivatives in the relation between acceleration, velocity and position?</h3>

The velocity is the derivative of the position.

The acceleration is the derivative of the velocity.

In this problem, the position is:

$s(t) = t^3 - 3t^2 + 8$

item i:

Velocity is the <u>derivative of the position</u>, hence:

$v(t) = 3t^2 - 6t$

Item ii:

$v(3) = 3(3)^2 - 6(3) = 27 - 18 = 9$

The speed is of 9 m/s.

Item iii:

Derivative of the velocity, hence:

$a(t) = 6t - 6$

Item iv:

$a(2) = 6(2) - 6 = 6$

The acceleration is of 6 m/s².

Item v:

t for which a(t) = 0, hence:

$6t - 6 = 0$

$6t = 6$

$t = \frac{6}{6}$

$t = 1$

Hence 1 second.

You can learn more about derivatives at brainly.com/question/14800626

7 0

2 years ago

HELP! THIS IS DUE AT 10:40!!!!!!

katrin2010 [14]

Answer:

Yes

Step-by-step explanation:

Both quadrilaterals have two triangles. The triangles that coordinate with the triangles in the other quadrilateral are proven congruent by AAS and AA. That makes the quadrilaterals congruent.

3 0

3 years ago

HELP DOES ANYONE KNOW “?”

lakkis [162]

Answer:

hope it helps...any queries comment me!!!

4 0

2 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

A river discharges 4 × 106 cubic feet of water into the Pacific Ocean each second. There are 86,400 seconds in one day. To find

erik [133]

Answer:

3.456E + 14= $3.456X10^{14$

Step-by-step explanation:

The river discharges $4X10^6$ cubic feet of water into the Pacific Ocean each second.

There are 86,400 seconds in one day.

Amount of water discharges each day

$=4X10^6X86400\\=345600000000\\=3.456X10^{11$

Since this differs from what you have, I will explain the scientific notation.

3.456E + 14 simply means $3.456X10^{14$

The options are not well laid out but you can pick the appropriate option.

8 0

3 years ago