We know that the price of all the fabrics is the same. However, the length of the fabrics is different. We have to determine which fabric is the longest.
Option A:
= 8.6
Option B:
= 8.25
Option C:
= 8.6667
Option D:
= 7.33....
Hence, the longest fabric is the one given in option C, and its length is 8.6667
Answer:
3y - 2x
Step-by-step explanation:
Your thoughts are correct; Ned loses 14 points overall.
The reason the correctly answered questions do not factor into the answer is because that is not what the question is asking. Correct answers are not even mentioned :)
So for this problem, we will be using the exponential equation format, which is y = ab^x. The a variable is the initial value, and the b variable is the growth/decay.
Since our touchscreen starts off at a value of 1200, that will be our a variable.
Since the touchscreen is decaying in value by 25%, subtract 0.25 (25% in decimal form) from 1 to get 0.75. 0.75 is going to be your b variable.
In this case, time is our independent variable. Since we want to know the value 3 years from now, 3 is the x variable.
Using our info above, we can solve for y, which is the cost after x years.

In context, after 3 years the touchscreen will only be worth $506.
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.