Given:
Mean, μ = 22
Standard deviation, σ = 7
Let's answer the following questions.
a. Given:
Sample size, n = 25
Let's find the probability that the sample mean is between 21.5 and 22.5.
We have:
![\begin{gathered} P(21.5Thus, we have:[tex]\begin{gathered} P(\frac{21.5-22}{\frac{7}{\sqrt[]{25}}}Using the standard normal table (NORMSDIST), we have:[tex]\begin{gathered} P(0.3571)=0.6395 \\ P(-0.3571)\text{ = }-0.3605 \\ \\ P(1.7857)-P(-0.3571)=0.6395-0.3605=0.279 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20P%2821.5Thus%2C%20we%20have%3A%5Btex%5D%5Cbegin%7Bgathered%7D%20P%28%5Cfrac%7B21.5-22%7D%7B%5Cfrac%7B7%7D%7B%5Csqrt%5B%5D%7B25%7D%7D%7DUsing%20the%20standard%20normal%20table%20%28NORMSDIST%29%2C%20we%20have%3A%5Btex%5D%5Cbegin%7Bgathered%7D%20P%280.3571%29%3D0.6395%20%5C%5C%20P%28-0.3571%29%5Ctext%7B%20%3D%20%7D-0.3605%20%5C%5C%20%20%5C%5C%20P%281.7857%29-P%28-0.3571%29%3D0.6395-0.3605%3D0.279%20%5Cend%7Bgathered%7D)
Therefore, the probability that sample mean is between 21.5 and 22.5 is 0.279
b. Given:
n = 25
Let's find the probability that the sample mean is between 21 and 22 minutes.
We have:
[tex]\begin{gathered} P(21Using the standard normal table, we have:[tex]\begin{gathered} P(-0.714286
Therefore, the probability that sample mean is between 21 and 22 is 0.2625
c. Given:
n = 144
Let's find the probability the sample mean is between 21.5 and 22.5
[tex]\begin{gathered} P(21.5
Therefore, the probability that sample mean is between 21.5 and 22.5 given a sample of 144 is 0.6086
d. Given:
Sample size in a = 25
Sample size in c = 144
The sample size in c is greater than the sample size in a so the standard error of the mean in (c) should be less than the standard error in (a).
As the standard error values become more concentrated
Answer:
The answer to the question is
The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is (-∞, 4)
Step-by-step explanation:
To apply look for the interval, we divide the ordinary differential equation by (t-4) to
y'' + 
y' + 
y = 
Using theorem 3.2.1 we have p(t) = , q(t) = , g(t) =
Which are undefined at 4. Therefore the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution, that is where p, q and g are continuous and defined is (-∞, 4) whereby theorem 3.2.1 guarantees unique solution satisfying the initial value problem in this interval.
Answer: 6928
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Explanation:
We have two areas we need to find: The area of the trapezoid and the area of the rectangle. Let's call these areas A1 and A2.
Area of Trapezoid = (height)*(base1+base2)/2
A1 = h*(b1+b2)/2
A1 = 80*(150+100)/2
A1 = 80*250/2
A1 = 20000
A1 = 10000
Area of Rectangle = (length)*(width)
A2 = L*W
A2 = 48*64
A2 = 3072
Subtract the two areas (A1-A2) to get the difference D
D = A1 - A2
D = 10000 - 3072
D = 6928
This difference D is exactly equal to the shaded area.
Answer:
a
Step-by-step explanation:
hope this helps yout just have to put the negative letters with the positive!!!
