Answer:
The probability that can afford to spend between $800 and $900
P(800≤X≤900) = 0.6826
The percentage of that can afford to spend between $800 and $900
P(800≤X≤900) = 68 percentage
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean of the Normal distribution = $850
Given that the standard deviation of the Normal distribution = $50
Let 'X' be a random variable in a normal distribution
Let x₁ = 800
![z_{1} = \frac{x_{1}-mean }{S.D} = \frac{800-850}{50} = -1](https://tex.z-dn.net/?f=z_%7B1%7D%20%3D%20%5Cfrac%7Bx_%7B1%7D-mean%20%7D%7BS.D%7D%20%3D%20%5Cfrac%7B800-850%7D%7B50%7D%20%3D%20-1)
Let x₂ =850
![z_{2} = \frac{x_{2}-mean }{S.D} = \frac{900-850}{50} = 1](https://tex.z-dn.net/?f=z_%7B2%7D%20%3D%20%5Cfrac%7Bx_%7B2%7D-mean%20%7D%7BS.D%7D%20%3D%20%5Cfrac%7B900-850%7D%7B50%7D%20%3D%201)
<u><em>Step(ii):-</em></u>
The probability that can afford to spend between $800 and $900
P(800≤X≤900) = P(-1≤Z≤1)
= P(Z≤1) - P(Z≤-1)
= 0.5 + A(1) - (0.5 - A(-1))
= A(1) +A(-1)
= 2× A(1) (∵ A(-1) =A(1)
= 2 × 0.3413
= 0.6826
The percentage of that can afford to spend between $800 and $900
P(800≤X≤900) = 68 percentage
Answer:
29
Step-by-step explanation:
First, replace v with 6 and k with 12
12/6 + 5 - 4(6) + 12
12/6= 2
2 + 5 - 4(6) + 12
4(6) = 24
2 + 5 - 24 + 12
2+ 5 = 7
24 + 12 = 36
7 - 36 = 29
hope this helps!
I believe it would be sixteen people. Hope it helps! :)
Answer:
n = 3.8
Step-by-step explanation:
First, we would distribute 0.3 with n and 1.5.
0.3(n - 5) = 0.4 - 0.2n → 0.3n - 1.5 = 0.4 - 0.2n
Next, we can move the number with the variable to one side.
0.2n + 0.3n - 1.5 = 0.4 - 0.2n + 0.2n → 0.2n + 0.3n - 1.5 = 0.4
(Because we added 0.2n to the negative 0.2n on the right side, it cancels it out)
We can also move 1.5 to the right side of the equation as well.
0.2n + 0.3n - 1.5 + 1.5 = 0.4 + 1.5 → 0.2n + 0.3n = 0.4 + 1.5
Then, we combine like terms
0.2n + 0.3n = 0.5n & 0.4 + 1.5 = 1.9 → 0.5n = 1.9
Finally, we divide both sides by 0.5.
n = 3.8