<span>The amount P as a function of t (in years) is given by
P(t) = P0 (1 + r/n)^(t n)
So if n = 4, and r = 0.02, and P0 = 1000, then
P(t) = 1000 (1 + 0.02/4)^(4 t) = 1000 (1 + 0.005)^(4 t)
At the end of the first quarter, t = 1/4, so
P(1/4) = $1000 (1.005)^(1) = $1005
At the end of the second quarter, t = 1/2 , therefore
P(1/2) = $1000 (1.005)^(2) = $1000 (1.010025) = $1010.03
At the end of the third quarter , t = 3/4, therefore
P(3/4) = $1000 (1.005)^(3) = $1000 (1.015075125) = $1015.08
At the end of the year, t = 4, therefore
P(1) = $1000 (1.005)^4 = $1000 (1.020150500625) = $1020.15
As for the second question, after the first period (quarter),
the formula becomes
P = P0 (1.005)^1 = 1.005 P0
which is choice A. </span>
Answer:
2
Step-by-step explanation:
You would use PEMDAS( parentheses exponents multiply adding subtracting) (1/2)4.
^
o.5×4
^
2
Answer:
10
Step-by-step explanation:
P(black bean soup) = cans of black bean / total = 6/12 =1/2
out of the next 20
20 *P(black bean)
20 * 1/2 = 10
A) it is left skewed
B) the median is 5
C) the mean is 5.15
D) the mean would be more affected (a change of 1.05 versus a change of 0.5).
The majority of the data is to the right of the graph; this means it is left skewed.
To find the median, write all of the data values out:
2, 3, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7
The middle value is 5.
We find the sum of this set of values and divide by 13, the number of data points, to find the mean:
2+3+4+4+5+5+5+6+6+6+7+7+7 = 67/13 = 5.15
If we added an additional data value at 20, the new median would be 5.5. The new mean would be (67+20)/14 = 6.2. The mean changes more than the median.
Answer:
18√2
Step-by-step explanation:
2√18 + 3√2 + √162
= 2√(9 * 2) + 3√2 + √(81 * 2)
= (2 * 3)√2 + 3√2 + 9√2
= 6√2 + 3√2 + 9√2
= (6 + 3 + 9)√2
= 18√2
