Assuming this is the equation you want to operate on:



Before final operation we have to assume that x is not equal to 0. :)
Answer: 0.8238
Step-by-step explanation:
Given : Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with
and
.
Let x denotes the scores on a certain intelligence test for children between ages 13 and 15 years.
Then, the proportion of children aged 13 to 15 years old have scores on this test above 92 will be :-
![P(x>92)=1-P(x\leq92)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{92-106}{15})\\\\=1-P(z\leq })\\\\=1-P(z\leq-0.93)=1-(1-P(z\leq0.93))\ \ [\because\ P(Z\leq -z)=1-P(Z\leq z)]\\\\=P(z\leq0.93)=0.8238\ \ [\text{By using z-value table.}]](https://tex.z-dn.net/?f=P%28x%3E92%29%3D1-P%28x%5Cleq92%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B92-106%7D%7B15%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq%20%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq-0.93%29%3D1-%281-P%28z%5Cleq0.93%29%29%5C%20%5C%20%5B%5Cbecause%5C%20P%28Z%5Cleq%20-z%29%3D1-P%28Z%5Cleq%20z%29%5D%5C%5C%5C%5C%3DP%28z%5Cleq0.93%29%3D0.8238%5C%20%5C%20%5B%5Ctext%7BBy%20using%20z-value%20table.%7D%5D)
Hence, the proportion of children aged 13 to 15 years old have scores on this test above 92 = 0.8238
The answer to this question is
B. (x+7)²+(y-4)²=64
I hope this helps!!!!!!!
Answer: x = 34
Step-by-step explanation:
⁴ ⁻ˣ/₅ + ˣ + 2/₃ = 6
Now resolve into fraction and convert to a linear equation
4 - x /₅ + x + 2/₃ = 6
3( 4 -x ) + 5( x + 2 )/15 = 6
3( 4 - x ) + 5( x + 2 ) = 6 x 15
12 - 3x + 5x + 10 = 90
2x + 22 = 90
2x = 90 - 22
2x = 68
x = 34
Answer:
find y intercept: x= 0 => y = 12
y = -x + 12 (1)
y = x + 12 (2) is answer
Step-by-step explanation: