<h3>Given:</h3>
- P= $12500
- R= 10%
- T= 3 years
<h3>Note that:</h3>
- P= Principal amount
- R= Rate of interest
- T= Time period
<h3>To find:</h3>
- The simple interest
- The total amount paid
<h3>Solution:</h3>
First, we'll have to multiply, principal amount (12500), rate (10) and time period (3).
Now, we'll have to divide the amount (375000) by 100.
<em>I=$3750</em>
Now, we can find the total amount paid.
Let's substitute according to the formula.
<em>A=$16250</em>
<u>Therefore</u><u>,</u><u> </u><u>simple</u><u> </u><u>interest</u><u> </u><u>is</u><u> </u><u>$</u><u>3</u><u>7</u><u>5</u><u>0</u><u> </u><u>and</u><u> </u><u>$</u><u>1</u><u>6</u><u>2</u><u>5</u><u>0</u><u> </u><u>was</u><u> </u><u>paid</u><u> </u><u>in</u><u> </u><u>total</u><u>.</u>
Answer:
Step-by-step explanation:
Area of sector = angle/360 * pi * r^2
7.1 = 40/360 * 22/7 * r^2
7.1 = 1/9 * 22/7 * r^2
63.9 = 22r^2/7
r^2 = (7 * 63.9)/22
r^2 = 20.331818181818
r = square root of 20.331818181818
r = 4.51
d = r * 2 = 4.51 * 2 = 9.02
Answer:
Step-by-step explanation:
given that the U.S. Department of Housing and Urban Development (HUD) uses the median to report the average price of a home in the United States.
We know that mean, median and mode are measures of central tendency.
Mean is the average of all the prices while median is the middle entry when arranged in ascending order.
Mean has the disadvantage of showing undue figure if extreme entries are there. i.e. outlier affect mean.
Suppose a price goes extremely high, then mean will fluctuate more than median.
So median using gives a reliable estimate since median gives the middle price and equally spread to other sides.
Answer:
a) P=0.2503
b) P=0.2759
c) P=0.3874
d) P=0.2051
Step-by-step explanation:
We have this information:
25% of American households have only dogs (one or more dogs)
15% of American households have only cats (one or more cats)
10% of American households have dogs and cats (one or more of each)
50% of American households do not have any dogs or cats.
The sample is n=10
a) Probability that exactly 3 have only dogs (p=0.25)
b) Probability that exactly 2 has only cats (p=0.15)
c) Probability that exactly 1 has cats and dogs (p=0.1)
d) Probability that exactly 4 has neither cats or dogs (p=0.5)