Answer:
Step-by-step explanation:
$27.63 / 2 = $13.82
$13.82 * 0.20 = $2.76
$13.82 + $2.76 = $16.58
Answer:
25
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 175cm
Standard deviation = 6 cm
Percentage of students below 163 cm
163 = 175 - 2*6
So 163 is two standard deviations below the mean.
By the Empirical rule, 95% of the heights are within 2 standard deviations of the mean. The other 100-95 = 5% are more than 2 standard deviations of the mean. Since the normal distribution is symmetric, 2.5% of them are more than 2 standard deviations below the mean(so below 163cm) and 2.5% are more than two standard deviations above the mean.
2.5% of the students have heights less than 163cm.
Out of 1000
0.025*1000 = 25
25 is the answer
Answer:
Step-by-step explanation:
given is the Differential equation in I order linear as
Take Laplace on both sides
![L(y') +4L(y) = 48L(t)\\sY(s)-y(0) +4Y(s) = 48 *\frac{1}{s^2} \\Y(s) [s+4]=\frac{48}{s^2}+9\\Y(s) = \frac{1}{s^2(s+4)}+\frac{9}{s+4}](https://tex.z-dn.net/?f=L%28y%27%29%20%2B4L%28y%29%20%3D%2048L%28t%29%5C%5CsY%28s%29-y%280%29%20%2B4Y%28s%29%20%3D%2048%20%2A%5Cfrac%7B1%7D%7Bs%5E2%7D%20%5C%5CY%28s%29%20%5Bs%2B4%5D%3D%5Cfrac%7B48%7D%7Bs%5E2%7D%2B9%5C%5CY%28s%29%20%3D%20%5Cfrac%7B1%7D%7Bs%5E2%28s%2B4%29%7D%2B%5Cfrac%7B9%7D%7Bs%2B4%7D)
Now if we take inverse we get y(t) the solution
Thus the algebraic equation would be
We have been given angle A as 75 degrees and sides a = 2 and b = 3.
Using Sine rule, we can set up:
Upon substituting the given values of angle A, and sides a and b, we get:
Upon solving this equation for B, we get:
Since we know that value of Sine cannot be more than 1. Hence there are no values possible for B.
Hence, the triangle is not possible. Therefore, first choice is correct.
