Let's look at an example.
We'll add the fractions 1/6 and 1/8
Before we can add, the denominators must be the same.
To get the denominators to be the same, we can...
- multiply top and bottom of 1/6 by 8 to get 8/48
- multiply top and bottom of 1/8 by 6 to get 6/48
At this point, both fractions involve the denominator 48. We can add the fractions like so
8/48 + 6/48 = (8+6)/48 = 14/48
Add the numerators while keeping the denominator the same the entire time.
The last step is to reduce if possible. In this case, we can reduce. This is because 14 and 48 have the factor 2 in common. Divide each part by 2.
The fraction 14/48 reduces to 7/24
Overall, 1/6 + 1/8 = 7/24
Answer:
480 milligrams.
Step-by-step explanation:
Divide 320 by 92, you get 3.47826087, multiply that by 138 to get your answer (480).
Answer:
48
Step-by-step explanation:
You can use a graphing calculator like desmos or draw your own graph to figure this out. Start by graphing each of those points. Then count the width and height. Multiply them to get the area. The width and height of the rectangle were 6 and 8, so the area is 48
Answer:
32m²
Step-by-step explanation:
thanks me later ,correct me if I'm wrong
Answer: 1) 0.6561 2) 0.0037
Step-by-step explanation:
We use Binomial distribution here , where the probability of getting x success in n trials is given by :-

, where p =Probability of getting success in each trial.
As per given , we have
The probability that any satellite dish owners subscribe to at least one premium movie channel. : p=0.10
Sample size : n= 4
Let x denotes the number of dish owners in the sample subscribes to at least one premium movie channel.
1) The probability that none of the dish owners in the sample subscribes to at least one premium movie channel = 

∴ The probability that none of the dish owners in the sample subscribes to at least one premium movie channel is 0.6561.
2) The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.
= ![P(X>2)=1-P(X\leq2)\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\\\\= 1-[0.6561+^4C_1(0.10)^1(0.90)^{3}+^4C_2(0.10)^2(0.90)^{2}]\\\\=1-[0.6561+(4)(0.0729)+\dfrac{4!}{2!2!}(0.0081)]\\\\=1-[0.6561+0.2916+0.0486]\\\\=1-0.9963=0.0037](https://tex.z-dn.net/?f=P%28X%3E2%29%3D1-P%28X%5Cleq2%29%5C%5C%5C%5C%3D1-%5BP%28X%3D0%29%2BP%28X%3D1%29%2BP%28X%3D2%29%5D%5C%5C%5C%5C%3D%201-%5B0.6561%2B%5E4C_1%280.10%29%5E1%280.90%29%5E%7B3%7D%2B%5E4C_2%280.10%29%5E2%280.90%29%5E%7B2%7D%5D%5C%5C%5C%5C%3D1-%5B0.6561%2B%284%29%280.0729%29%2B%5Cdfrac%7B4%21%7D%7B2%212%21%7D%280.0081%29%5D%5C%5C%5C%5C%3D1-%5B0.6561%2B0.2916%2B0.0486%5D%5C%5C%5C%5C%3D1-0.9963%3D0.0037)
∴ The probability that more than two dish owners in the sample subscribe to at least one premium movie channel is 0.0037.