Given:
μ = 70, the mean
σ = 10, the standard deviation
The random variable is x = 65.
Calculate the z-score.
z = (x - μ)/σ = (65 -70)/10 = - 0.5
From standard ables, obtain
P(x<5) = 0.3085 ≈ 30%
The score of 65 is in the 30th percentile.
Answer: 30th
<u>ANSWER:
</u>
Rate per annum at which CI will amount from RS 2000 to RS 2315.35 in 3 years is 5%
<u>SOLUTION:
</u>
Given,
P = RS 2000
C.I = RS 2315.35
T = 3 years
We need to find the rate per annum. i.e. R = ?
We know that,
When interest is compound Annually:
![Amount $=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{n}$](https://tex.z-dn.net/?f=Amount%20%24%3D%5Cmathrm%7BP%7D%5Cleft%281%2B%5Cfrac%7B%5Cmathrm%7BR%7D%7D%7B100%7D%5Cright%29%5E%7Bn%7D%24)
Where p = principal amount
r = rate of interest
n = number of years
![$2315.35=2000 \times\left(1+\frac{R}{100}\right)^{3}$](https://tex.z-dn.net/?f=%242315.35%3D2000%20%5Ctimes%5Cleft%281%2B%5Cfrac%7BR%7D%7B100%7D%5Cright%29%5E%7B3%7D%24)
![$\left(1+\frac{R}{100}\right)^{3}=\frac{2315.35}{2000}$](https://tex.z-dn.net/?f=%24%5Cleft%281%2B%5Cfrac%7BR%7D%7B100%7D%5Cright%29%5E%7B3%7D%3D%5Cfrac%7B2315.35%7D%7B2000%7D%24)
![$\left(1+\frac{R}{100}\right)^{3}=1.157$](https://tex.z-dn.net/?f=%24%5Cleft%281%2B%5Cfrac%7BR%7D%7B100%7D%5Cright%29%5E%7B3%7D%3D1.157%24)
![$1+\frac{R}{100}=\sqrt[3]{1.157}$](https://tex.z-dn.net/?f=%241%2B%5Cfrac%7BR%7D%7B100%7D%3D%5Csqrt%5B3%5D%7B1.157%7D%24)
![$1+\frac{R}{100}=1.0500$](https://tex.z-dn.net/?f=%241%2B%5Cfrac%7BR%7D%7B100%7D%3D1.0500%24)
![$\frac{R}{100}=1.05-1$](https://tex.z-dn.net/?f=%24%5Cfrac%7BR%7D%7B100%7D%3D1.05-1%24)
![$\frac{R}{100}=0.05$](https://tex.z-dn.net/?f=%24%5Cfrac%7BR%7D%7B100%7D%3D0.05%24)
R = 5%
Hence, rate per annum is 5 percent.
Answer: 77 pounds each month I think
Step-by-step explanation:
Answer: Center: (5,0)
Radius:5
Step-by-step explanation: Happy to help :)
Answer:
i
Step-by-step explanation:
dontkow