It is given here that there is 1/3 probability of professional baseball player will get a hit. Hence if at least three hits are gained out of 5 attempts, the calculation goes: 5C3* (1/3)^3*(2/3)^2 + 5C4* (1/3)^4*(2/3)^1 +5C5 *<span>(1/3)^5*(2/3)^0 equal to 0.21. </span>
<h2>
Answer with explanation:</h2>
We are given a semi-ellipse gate whose dimensions are as follows:
Height of 20 feet and a width of 15 feet.
Now, if a truck is loaded then:
Height of truck is: 12 feet and a width of truck is: 16 feet
The truck won't pass through the gate since the width of truck is more than that of the gate.
When the truck is not loaded then:
Height of truck is: 12 feet and a width of truck is: 10 feet
The truck would easily pass through the gate since, the dimensions of truck are less than that of the gate.
Answer:
The correct option is (A) $304.47.
Step-by-step explanation:
The formula to compute the future value (<em>FV</em>) of an amount (A), compounded daily at an interest rate of <em>r</em>%, for a period of <em>n</em> years is:
![FV=A\times [1+\frac{r\%}{365}]^{n\times 365}](https://tex.z-dn.net/?f=FV%3DA%5Ctimes%20%5B1%2B%5Cfrac%7Br%5C%25%7D%7B365%7D%5D%5E%7Bn%5Ctimes%20365%7D)
The information provided is:
A = $300
r% = 1.48%
n = 1 year
Compute the future value as follows:
![FV=A\times [1+\frac{r\%}{365}]^{n\times 365}](https://tex.z-dn.net/?f=FV%3DA%5Ctimes%20%5B1%2B%5Cfrac%7Br%5C%25%7D%7B365%7D%5D%5E%7Bn%5Ctimes%20365%7D)
![=300\times [1+\frac{0.0148}{365}]^{365}\\\\=300\times (1.00004055)^{365}\\\\=300\times 1.014911\\\\=304.4733\\\\\approx \$304.47](https://tex.z-dn.net/?f=%3D300%5Ctimes%20%5B1%2B%5Cfrac%7B0.0148%7D%7B365%7D%5D%5E%7B365%7D%5C%5C%5C%5C%3D300%5Ctimes%20%281.00004055%29%5E%7B365%7D%5C%5C%5C%5C%3D300%5Ctimes%201.014911%5C%5C%5C%5C%3D304.4733%5C%5C%5C%5C%5Capprox%20%5C%24304.47)
Thus, the balance after 1 year is $304.47.
The correct option is (A).
Answer:
a. 12
b. 7.200 and 2.683
Step-by-step explanation:
The computation is shown below:
Given that
P = 0.40 and n = 30
a)
The expected value of received e-mails is
= n × p
= 30 × 0.4
= 12
b)
The variance of emails received is
= n × p × (1 - p)
= 30 × 0.4 × 0.6
= 7.200
Now
The standard deviation of emails received is
= sqrt(variance)
= 2.683
We simply applied the above formula