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nikdorinn [45]
4 years ago
8

For question 2 not 3 :)

Mathematics
1 answer:
BabaBlast [244]4 years ago
6 0

Answer:

2.5

Step-by-step explanation:

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What is 10x5thousands
ella [17]
It will be 50,000 =10*5,000
8 0
3 years ago
Set up a system of linear equations to represent the scenario. Solve the system by using Gaussian elimination or Gauss-
lapo4ka [179]

She invested $11,250 in the stock, $3,750 in the CD and $12,000 in the bond fund.

<h3><u>Distributions</u></h3>

Given that Sylvia invested a total of $27,000, and she invested part of the money in a certificate of deposit (CD) that earns 3% simple interest per year, she invested in a stock that returns the equivalent of 7% simple interest, and she invested in a bond fund that returns 2%, and she invested three times as much in the stock as she did in the CD, and earned a total of $1140 at the end of 1 yr, to determine how much principal did she put in each investment, the following calculation must be made:

  • 9000 x 0.07 + 3000 x 0.03 + 15000 x 0.02 = 630 + 90 + 300 = 1020
  • 9900 x 0.07 + 3300 x 0.03 + 13800 x 0.02 = 693 + 99 + 276 = 1068
  • 12,000 x 0.07 + 4,000 x 0.03 + 11,000 x 0.02 = 840 + 120 + 220 = 1,180
  • 11400 x 0.07 + 3800 x 0.03 + 11800 x 0.02 = 798 + 114 + 236 = 1148
  • 10800 x 0.07 + 3600 x 0.03 + 12600 x 0.02 = 756 + 108 + 252 = 1116
  • 11160 x 0.07 + 3720 x 0.03 + 12120 x 0.02 = 781.2 + 111.6 + 242.4 = 1135.2
  • 11190 x 0.07 + 3730 x 0.03 + 12080 x 0.02 = 783.3 + 111.9 + 241.6 = 1136.8
  • 11250 x 0.07 + 3750 x 0.03 + 12000 x 0.02 = 787.5 + 112.5 + 240 = 1140

Therefore, she invested $11,250 in the stock, $3,750 in the CD and $12,000 in the bond fund.

Learn more about distribution in brainly.com/question/10250387

7 0
2 years ago
Determine the range of the graph
VashaNatasha [74]

Answer:

is 20

Step-by-step explanation:

hope this helps

8 0
3 years ago
Find the radius of convergence, r, of the series. ∞ xn 2n − 1 n = 1 r = 1 find the interval, i, of convergence of the series. (e
Bingel [31]
Assuming the series is

\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}

The series will converge if

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|

We have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|

So the series will certainly converge if -1, but we also need to check the endpoints of the interval.

If x=1, then the series is a scaled harmonic series, which we know diverges.

On the other hand, if x=-1, by the alternating series test we can show that the series converges, since

\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0

and is strictly decreasing.

So, the interval of convergence for the series is -1\le x.
6 0
4 years ago
Pleas help me, this is confusing.
Dmitry_Shevchenko [17]
Ask your teacher or a family member
7 0
3 years ago
Read 2 more answers
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