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abruzzese [7]
4 years ago
10

Tammy deposited $520 in the bank account that earns simple interest every year after 5 years she had earned $156 and interest if

no money was deposited into or withdrawn from the account what was the annual interest rate
Mathematics
1 answer:
aleksley [76]4 years ago
6 0

Answer:

6%.

Step-by-step explanation:

We have been given that Tammy deposited $520 in the bank account that earns simple interest every year after 5 years she had earned $156.

To find the interest rate we will use simple interest formula.

I=P*r*T

I= Interest.

P= Principal amount.

r=Annual interest rate (in decimal form).

T= Time in years.

We have been given that I=156, T=5, P=520

Upon substituting our values in above formula we will get,

156=520*r*5

156=2600r

r=\frac{156}{2600}

r=0.06

Let us multiply 0.06 by 100 to convert annual interest rate in percentage.

0.06*100=6 \text{ percent}

Therefore, the annual interest rate was 6%.

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What is the solution to the inequality (20+5 &gt; 1? 0 -3&gt;n&gt; - 2 02-2 On&lt; 2 or n &gt; 3​
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Answer:

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Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Consider the following ordered data. 6 9 9 10 11 11 12 13 14 (a) Find the low, Q1, median, Q3, and high. low Q1 median Q3 high (
IrinaVladis [17]

Answer:

Low             Q1                Median              Q3                 High

6                  9                     11                      12.5                14

The interquartile range = 3.5

Step-by-step explanation:

Given that:

Consider the following ordered data. 6 9 9 10 11 11 12 13 14

From the above dataset, the highest value = 14  and the lowest value = 6

The median is the middle number = 11

For Q1, i.e the median  of the lower half

we have the ordered data = 6, 9, 9, 10

here , we have to values as the middle number , n order to determine the median, the mean will be the mean average of the two middle numbers.

i.e

median = \dfrac{9+9}{2}

median = \dfrac{18}{2}

median = 9

Q3, i.e median of the upper half

we have the ordered data = 11 12 13 14

The same use case is applicable here.

Median = \dfrac{12+13}{2}

Median = \dfrac{25}{2}

Median = 12.5

Low             Q1                Median              Q3                 High

6                  9                     11                      12.5                14

The interquartile range = Q3 - Q1

The interquartile range =  12.5 - 9

The interquartile range = 3.5

7 0
3 years ago
Hello,
Damm [24]

Answer:

  -1

Step-by-step explanation:

The axes on your graph are not labeled, so we have to assume they follow the usual convention. That is, the vertical axis is the y-axis, and the horizontal axis is the x-axis.

The x-coordinate tells how far to the left or right of the y-axis the point is. Here, point L is 1 grid square to the left of the vertical line that is the y-axis. If you follow the vertical line through L down to where it crosses the x-axis, you will see an unlabeled open circle there. (We don't know the purpose of that circle, but we call it to your attention so you know you're looking in the right place.)

Looking 4 more grid squares to the left of that point, you see the marking "-5". This tells you each grid square corresponds to one unit. Then the first one to the left of the y-axis (where the open circle is) has a value of -1. That is the value of the x-coordinate of point L.

The x-coordinate of point L is -1.

4 0
3 years ago
Julie is 3 years older than her sister. Julie's age is 11 years old.<br> How old is her sister?
Ne4ueva [31]

Answer:

14 :)

Step-by-step explanation:

4 0
3 years ago
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