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Marrrta [24]
2 years ago
10

James had $ 6500 in a bank account that paid 4% interest at the end of each year. How much money did Mr. James have in his accou

nt at the end of 1 year ?
$6700

$ 6760

$ 6750

$6775​
Mathematics
2 answers:
Kazeer [188]2 years ago
8 0
He would have $6,760 in his account after year one.
Do 6,500*0.04=260
6,500+260=6,760
Elena L [17]2 years ago
3 0

Answer:

6760

Step-by-step explanation:

4% of 6500

= (6500 x 4) ÷ 100

= 26000 ÷ 100

= 260 $

6500 + 260 = 6760$

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Step-by-step explanation:

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4 0
2 years ago
What is the mean of this data set? 12,17,16,10,20,13,14,14,12,12,19,18
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Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
The mean sat verbal score is 486, with a standard deviation of 95. use the empirical rule to determine what percent of the score
Pavel [41]
Find the z-scores for the two scores in the given interval.

z=\frac{x-\mu}{\sigma}

For the score x =391, z=\frac{391-486}{95}=\frac{-95}{95}=-1.

For the score x = 486, z=\frac{486-486}{95}=0

Now you want the area (proportion of data) under the normal distribution from z = -1 to z = 0. The Empirical Rule says that 68% of the data falls between z = -1 to z = 1. But the curve is symmetrical around the vertical axis at z = 0, so the answer you want is HALF of 68%.

8 0
3 years ago
X + 2y = 33<br> x-y = 11
Tasya [4]

Answer:

x=\frac{55}{3},\:y=\frac{22}{3}

Step-by-step explanation:

Solve by Substitution ;

\begin{bmatrix}x+2y=33\\ x-y=11\end{bmatrix}\\\\\mathrm{Isolate}\:x\:\mathrm{for}\:x+2y=33:\quad x=33-2y\\\\\mathrm{Subsititute\:}x=33-2y\\\\\begin{bmatrix}33-2y-y=11\end{bmatrix}\\\\Simplify\\\begin{bmatrix}33-3y=11\end{bmatrix}\\\\\mathrm{Isolate}\:y\:\mathrm{for}\:33-3y=11:\quad y=\frac{22}{3}\\\mathrm{For\:}x=33-2y\\\\\mathrm{Subsititute\:}y=\frac{22}{3}\\x=33-2\times\frac{22}{3}\\\\x=\frac{55}{3}\\\\x=\frac{55}{3},\:y=\frac{22}{3}

7 0
3 years ago
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