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Anuta_ua [19.1K]
2 years ago
9

The last time Larissa checked her credit score, it was 760, and her only credit event since then has been applying for a store c

redit card. Which of these is most likely to be her credit score now? A. 770 B. 780 C. 750 D. 760
Mathematics
1 answer:
Pie2 years ago
8 0

Answer:

750

Step-by-step explanation:

Applying for credit from any place pulls your credit report.  When your credit report is pulled, your score decreases some.  

However, this will not have a large effect on your score, so 750 would most likely be the new score.

Getting a store credit card affects the credit score a little bit, so the answer would likely be D.

please mark brainliest :)

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The mean of a set of credit scores is Mu = 690 and sigma = 14. Which credit score is within a z-score of 3.3? 634 640 720 750
xz_007 [3.2K]

Answer:

720

Step-by-step explanation:

To solve this question, we would be using the formula for z- score

The formula for calculating a z-score is z = (x-μ)/σ

where x = raw score

μ = population mean

σ = sigma = population standard deviation.

From the question, we are given

The mean of a set of credit scores Mu = 690

sigma = population standard deviation 14

z - score = 3.3

This means we are to find x

z = (x-μ)/σ

3.3 = x - 690/ 14

Cross Multiply

3.3 × 14 = x - 690

46.2 = x - 690

x = 690 + 46.2

x = 736.2

Since we are told that it is a range, the closest out of the options given to our calculated x which is 736.2 is 720

8 0
2 years ago
Read 2 more answers
Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

3 0
1 year ago
A trampoline salesman makes $25,000 annually plus 6% commission on his total sales. If he sold $40,000 worth of trampolines this
Katarina [22]

Multiply sales by 6%

40,000 x 0.06 = 2,400

Add that to his base pay:

2400 + 25000 = $27,400 total pay.

5 0
1 year ago
Malachy, Gavin and Gavyn share £18 in a ratio 3:1:2. How much money does each person get?
Drupady [299]

Answer:

6

Step-by-step explanation:

18×(3/6) = 918×(1/6) = 318×(2/6) = 6

7 0
1 year ago
Question 1(Multiple Choice Worth 1 points)
ra1l [238]
It’s 120 minutes because it would be two hours but there not putting it into hours there putting it into minutes and 1 hour is 60 minutes so two hours is 120 minutes
5 0
2 years ago
Read 2 more answers
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