Answer:
X × $75000 = $75000
Step-by-step explanation:
Brainliest would be appreciated
The amount A resulting from a principal amount P being invested at rate r compounded continuously for time t is given by
... A = P·e^(rt)
FIll in your given values and solve for P.
... 25000 = P·e^(0.0525·12) = P·e^0.63
... P = 25000/e^0.63 ≈ 13314.80 . . . . . divide by the coefficient of P
The amount that must be invested is $13,314.80.
An initial investment (P) compounded continuously with a rate of interest (r) in time (t) will grow to amount (Q) is given by:
Q = P * e^(rt)
Q=25000, r=0.0525, t=12
25000 = P * e^(0.0525*12)
1.8776P = 25000
P = 13314.8
The new price of the stock which increases 3/4 per share is $ 7 1/2.
Given data:
A stock is selling on a stock exchange for 6 3/4 dollars per share.
If the price of the stock increases by 3/4 dollars per share, you can add 3/4 to the current price of 6 3/4 dollars per share:
6 3/4 + 3/4
To add the whole numbers and fractions separately:
6 + 0 + 3/4 + 3/4
Adding the whole numbers and fractions:
6 + 1 1/2
A = 7 1/2
Hence, the price of the new stock is $ 7 1/2
To learn more about fractions, refer:
#SPJ3
Step-by-step explanation:
If the price increases 3/4 per share, the new price of the stock =
6¾ + ¾ =
27 /4 + ¾ =
30 /4 =
15/2or7½
Answer: The answer is everytime you mutiply in by 0.6
Step-by-step explanation:
Answer:
The 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).
Step-by-step explanation:
The net change in 7 students' scores on the exam after completing the course are:
S = {37 ,12 ,12 ,17 ,13 ,32 ,23}
Compute the sample mean and sample standard deviation as follows:
As the population standard deviation is not known, a t-interval will be formed.
Compute the critical value of t for 80% confidence interval and 6 degrees of freedom as follows:
*Use a t-table.
Compute the 80% confidence interval for the average net change in a student's score after completing the course as follows:
Thus, the 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).