Answer:
The area lies to the right of the z-score 0.48 means all the values greater than it. This can be calculated on a graphing calculator using the function normCdf, where
- Lower bound = 0.48
- Upper bound = 9999
- Mean = 0
- Standard deviation = 1
<u>The result would be normCdf(0.48,9999,0,1) ≈ </u><u>0.315614</u>
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The area lies to the left of the z-score 0.79 means all the values less than it. This can be calculated on a graphing calculator using the function normCdf, where
- Lower bound = -9999
- Upper bound = 0.79
- Mean = 0
- Standard deviation = 1
<u>The result would be normCdf(-9999,0.79,0,1) ≈</u><u> 0.785236</u>
Answer:
9.219544457292887
Step-by-step explanation:
Use distance formula
√(−5−2)^2+(−3−3)^2
√(−7)^2+(−6)^2
√49+36
√85
√≈9.219544457292887
Hoped this helped!
: )
Answer:
-8x^2 -4x
Step-by-step explanation:
f(x) = –4x – 2
g(x) = 2x
(f · g)(x) = (-4x-2) * (2x)
distribute
= -2x*4x -2x*2
= -8x^2 -4x
For this case we have the following functions:

We must find:
By definition we have to:

So:

Thus, we have to:

So, the value is -169
ANswer:
Option A
Answer: The equation would NOT help us solve for the length and width of the classroom is "
".
Step-by-step explanation:
Let y be the width of the rectangular classroom floor and x be the length of the rectangular classroom floor .
Given ,
The perimeter of a rectangular classroom floor is 90 feet.
The length of the floor is twice the width.
i.e. Length =2 (width)
i.e. x= 2y ..(i)
Also, perimeter = 2(length+width)
When we put value of x from (i), we get

Hence, the equation would NOT help us solve for the length and width of the classroom is "
".