Answer:
0, 1, 2, 3
Step-by-step explanation:
-1 < x < 4 means that x has to be an integer (whole number aka no fractions) between -1 and 4. Note that the < and > means that x cannot be -1 or 4 since it has to be greater than -1, or smaller than 4.
Answer:
27,736.5
Step-by-step explanation:
18,491 / (2/3)
When dividing by a fraction
flip it over and multiply
18,491 * 3/2
55,473/2
27,736.5
Two consecutive integers...
n and n+1 have a sum of 137 so
n+n+1=137
2n+1=137
2n=136
n=68
So the numbers are 68 and 69.
Answer:
0.0157
Step-by-step explanation:
From the information given:
The sample size = 70
The expected no. of days of year that are birthday of exactly 4 people is:![P = \bigg [ \dfrac{1}{365} \bigg]^4](https://tex.z-dn.net/?f=P%20%3D%20%5Cbigg%20%5B%20%5Cdfrac%7B1%7D%7B365%7D%20%5Cbigg%5D%5E4)
The expected number of days with 4 birthdays = 
![\sum \limits ^{365}_{i=1} E(x_i) = 365 \times \bigg[ \ ^{70}C_{4} \times ( \dfrac{1}{365})^4 ( 1 - \dfrac{1}{365})^{70-4} \bigg]](https://tex.z-dn.net/?f=%5Csum%20%5Climits%20%5E%7B365%7D_%7Bi%3D1%7D%20%20E%28x_i%29%20%3D%20365%20%5Ctimes%20%5Cbigg%5B%20%20%5C%20%5E%7B70%7DC_%7B4%7D%20%5Ctimes%20%28%20%5Cdfrac%7B1%7D%7B365%7D%29%5E4%20%28%201%20-%20%5Cdfrac%7B1%7D%7B365%7D%29%5E%7B70-4%7D%20%5Cbigg%5D)
![\sum \limits ^{365}_{i=1} E(x_i) = 365 \times \bigg[ \ \dfrac{70!}{4!(70-4)!} \times ( \dfrac{1}{365})^4 ( 1 - \dfrac{1}{365})^{66} \bigg]](https://tex.z-dn.net/?f=%5Csum%20%5Climits%20%5E%7B365%7D_%7Bi%3D1%7D%20%20E%28x_i%29%20%3D%20365%20%5Ctimes%20%5Cbigg%5B%20%20%5C%20%5Cdfrac%7B70%21%7D%7B4%21%2870-4%29%21%7D%20%5Ctimes%20%28%20%5Cdfrac%7B1%7D%7B365%7D%29%5E4%20%28%201%20-%20%5Cdfrac%7B1%7D%7B365%7D%29%5E%7B66%7D%20%5Cbigg%5D)
![\sum \limits ^{365}_{i=1} E(x_i) = 365 \times \bigg[ \ 916895 \times 5.6342 \times 10^{-11} \times 0.8343768898 \bigg]](https://tex.z-dn.net/?f=%5Csum%20%5Climits%20%5E%7B365%7D_%7Bi%3D1%7D%20%20E%28x_i%29%20%3D%20365%20%5Ctimes%20%5Cbigg%5B%20%20%5C%20916895%20%5Ctimes%205.6342%20%5Ctimes%2010%5E%7B-11%7D%20%5Ctimes%200.8343768898%20%5Cbigg%5D)
= 0.0157
Therefore, the required probability = 0.0157
For this question we are looking at arithmetic sequences. A sequence is any set of numbers in a particular order. An arithmetic sequence is a specific type of sequence where the difference between 2 consecutive terms is constant. The difference is called the common difference. In your problem...
t_n = 3n + 2
we need to find the common difference, then we can find the 1st four terms. Find the 1st term by replacing n with 1.
t_1 = 3(1) +2
t_1 = 3 +2 = 5
Find the second term by replacing n with 2
t_2 = 3(2) + 2 = 8
and so on
t_3 = 3(3) + 2 = 11
t_4 = 3(4) + 2 = 14
The 1st 4 numbers in the sequence are 5,8,11, and 14.
You can see the common difference is d=3. You could plug any number into the equation to find that number in the sequence. For example, the 100th number in the sequence would be
t_100 = 3(100) + 2 = 302
