The probability that a randomly selected adult has an IQ less than
135 is 0.97725
Step-by-step explanation:
Assume that adults have IQ scores that are normally distributed with a mean of mu equals μ = 105 and a standard deviation sigma equals σ = 15
We need to find the probability that a randomly selected adult has an IQ less than 135
For the probability that X < b;
- Convert b into a z-score using z = (X - μ)/σ, where μ is the mean and σ is the standard deviation
- Use the normal distribution table of z to find the area to the left of the z-value ⇒ P(X < b)
∵ z = (X - μ)/σ
∵ μ = 105 , σ = 15 and X = 135
∴ 
- Use z-table to find the area corresponding to z-score of 2
∵ The area to the left of z-score of 2 = 0.97725
∴ P(X < 136) = 0.97725
The probability that a randomly selected adult has an IQ less than
135 is 0.97725
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Answer:
It is d
Step-by-step explanation:
Answer:
I got x<equal to -11 or x>equal to4
Step-by-step explanation:
x^2+15x+44=0
(x+4)(x+11)=0(Factor left side of equation)
x+4=0 or x+11=0(Set factors equal to 0)
x=−4 or x=−11
Check intervals in between critical points. (Test values in the intervals to see if they work.)
x<−11(Works in original inequality)
−11<x<−4(Doesn't work in original inequality)
x>−4(Works in original inequality)
Answer:
x<−11 or x>−4
An equation has an infinite number of solutions if the left and right sides are identical. Let's simplify them both and then decide the value of ?.
Left hand side:
Right hand side:
So, both sides have a numeric part of -20. We can use the "?" value to make sure that the literal part is the same as well. So, we want
1.
Draw a circle with center C And radius CA, as shown in the attached picture.
Let the lengths of radii AO, OB, OC be R. Triangle ABC is inscribed in the circle with center O and one of its sides is a diameter, this means that the angle ACB is a right angle.
|AO|=|OC|=R, by the Pythagorean theorem |AC|=
.
these are all shown in the picture.
2.
Area of triangle ABC is 1/2 * 2R * R= R^2
3.
Let the area between arc BXA and chord AB be Y. (the yellow region).
and let G be the shaded region between arcs AB and AXB.
G=1/2(Area circle with center O)-Y
=
To find Y:
Notice that the area of the sector ACB is 1/4 of the area of circle with center C, since m(ACB) is 1/4 of 360 degrees.
So Area of sector ACB =
Y =area of sector ABC-Area(triangle ABC)=
4.
Finally,
This proves that the 2 shaded regions have equal area.
