Answer: 
Step-by-step explanation:
<h3>
The exercise is: " Evaluate 
when 
and 
</h3>
Given the following expression:
You can follow these steps in order to evaluate it:
1. Substitute and into the expression provided in the exercise:
2. Solve the multiplications. Remember that:
Then:
3. Reduce the fraction. Notice that the numerator 6 and the denomiantor 4 can be both divided by 2. Then:
4. Solve the addition:
Since the number 21 has a denominator 1, the Least Common Denominator is:
Then, the sum is:
Answer:
378.5 or just 378
Step-by-step explanation:
This is a linear model with x representing the number of generations that's gone by, y is the number of butterflies after x number of generations has gone by, and the 350 represents the number of butterflies initially (before any time has gone by. When x = 0, y = 350 so that's the y-intercept of our equation.)
The form for a linear equation is y = mx + b, where m is the rate of change and b is the y-intercept, the initial amount when x = 0.
Our rate of change is 1.5 and the initial amount of butterflies is 350, so filling in the equation we get a model of y = 1.5x + 350.
If we want y when x = 19, plug 19 in for x and solve for y:
y = 1.5(19) + 350
y = 378.5
Since we can't have .5 of a butterfly we will round down to 378
Answer:
1) The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Step-by-step explanation:
Given : Assume there are 365 days in a year.
To find : 1) What is the probability that ten students in a class have different birthdays?
2) What is the probability that among ten students in a class, at least two of them share a birthday?
Solution :
Total outcome = 365
1) Probability that ten students in a class have different birthdays is
The first student can have the birthday on any of the 365 days, the second one only 364/365 and so on...
The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday
P(2 born on same day) = 1- P( 2 not born on same day)
![\text{P(2 born on same day) }=1-[\frac{365}{365}\times \frac{364}{365}]](https://tex.z-dn.net/?f=%5Ctext%7BP%282%20born%20on%20same%20day%29%20%7D%3D1-%5B%5Cfrac%7B365%7D%7B365%7D%5Ctimes%20%5Cfrac%7B364%7D%7B365%7D%5D)
![\text{P(2 born on same day) }=1-[\frac{364}{365}]](https://tex.z-dn.net/?f=%5Ctext%7BP%282%20born%20on%20same%20day%29%20%7D%3D1-%5B%5Cfrac%7B364%7D%7B365%7D%5D)
The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Answer: x=4,y=5
Step-by-step explanation:
