Let P be the final value
Let r be the rate
Let t be the time in years
Let C be the initial value
Let n be the number of times interest is given per year
P = C (1 + r/n) ^ (nt)
r = 0.075 (7.5%)
t = 3/4 years
C = $1500
r = 4 (4 quarters in one year)
P = C (1+r/n)^ (t/n)
= 1500 * (1 + 0.075/4) ^ (3/4 * 4)
= 1585.97
Have an awesome day!
If you are wondering, a quarter means 1/4 of a year.
Answer:
b= (2a+11)/5
Step-by-step explanation:
5x + 2y = 9
2y=9-5x
y= 9/2 - 5x/2
slope, m1= -5/2
for perpendicularity, m1 × m2 = -1
slope for QR= m2=
![= - 1 \div \frac{ - 5}{2} = - 1 \times \frac{2}{ - 5} = \frac{ - 2}{ - 5} = \frac{2}{5}](https://tex.z-dn.net/?f=%20%3D%20%20-%201%20%20%5Cdiv%20%20%5Cfrac%7B%20-%205%7D%7B2%7D%20%20%3D%20%20-%201%20%5Ctimes%20%20%5Cfrac%7B2%7D%7B%20-%205%7D%20%20%3D%20%20%5Cfrac%7B%20-%202%7D%7B%20-%205%7D%20%20%3D%20%20%5Cfrac%7B2%7D%7B5%7D%20)
m2 is 2/5
![qr = \frac{b - 3}{a - 2}](https://tex.z-dn.net/?f=qr%20%3D%20%20%5Cfrac%7Bb%20-%203%7D%7Ba%20-%202%7D%20)
![\frac{2}{5} = \frac{b - 3}{a - 2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B5%7D%20%3D%20%20%5Cfrac%7Bb%20-%203%7D%7Ba%20-%202%7D%20)
5(b-3)=2(a-2)
5b-15=2a-4
5b=2a-4+15
5b= 2a+11
![b = \frac{2a + 11}{5}](https://tex.z-dn.net/?f=b%20%3D%20%20%5Cfrac%7B2a%20%2B%2011%7D%7B5%7D%20)
Answer:
nsjzjzk
Step-by-step explanation:
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Answer:
Option C) 0.25
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 22 minutes
Standard Deviation, σ = 11.9 minutes
We are given that the distribution of response time is a bell shaped distribution that is a normal distribution.
Formula:
P(response times is over 30 minutes)
P(x > 30)
Calculation the value from standard normal z table, we have,
![P(x > 30) = 1 - 0.749= 0.251 = 25.1\%](https://tex.z-dn.net/?f=P%28x%20%3E%2030%29%20%3D%201%20-%200.749%3D%200.251%20%3D%2025.1%5C%25)
Thus, the correct answer is
Option C) 0.25