To solve this problem, we are going to set up an equation. Let the number that we are trying to find be represented by the variable x. If we plug in the numbers that we know, we get the following equation:
3x/4 = 24
To simplify this equation, we need to multiply both sides by 4, to begin getting the x alone on the left side of the equation.
3x = 96
Finally, we need to divide both sides by 3, to get rid of the coefficient that is being multiplied to x.
x = 32
Therefore, the number that you are trying to find is 32.
Answer:
10.4900
Step-by-step explanation:
pls make a brainliest
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.