Assume the random variable X is normally distributed with mean μ = 50 and standard deviation σ= 7. Compute the probability.
Complete question is;
Assume the random variable X is normally distributed with mean μ = 50 and standard deviation σ = 7. What is the probability P (X > 42)?
Answer:
P(X > 42) = 0.1271
Step-by-step explanation:
We are given;
mean; μ = 50
Standard deviation; σ = 7
Formula for the z-score is;
z = (x - μ)/σ
Thus;
z = (42 - 50)/7
z = -1.14
Since we are looking for P (X > 42), then let's look up this z-value from the z-distribution table attached.
We have;
P(X > 42) = 0.1271
Which equation shows the point-slope form of the line that passes through (3, 2) and has a slope of 1/3
O y + 2 =1/3(x + 3)
O y-2=1/3(x-3)
O y + 3 = 1/3(x+ 2)
O y-3= 1/3(x-2)
Answer:
y - 2 = 1/3(x - 3)
Step-by-step explanation:
Point slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.
Plug in the slope:
y - y1 = m(x - x1)
y - y1 = 1/3(x - x1)
Plug in the given point:
y - y1 = 1/3(x - x1)
y - 2 = 1/3(x - 3)
So, the correct answer is y - 2 = 1/3(x - 3)
simplify the expression (2r^4y4xy^2) completely
Step-by-step explanation:
we can simplify the stuff inside the parentheses to
[tex] \frac{ {x}^{3} }{2y} [/tex]
now we need to multiply it with itself, giving us
[tex] \frac{ {x}^{6} }{4 {y}^{2} } [/tex]
so yeah, D is the correct answer
3. Two dice are rolled. What’s the conditional probability that both dice are 5’s if it’s known that the sum of points is divisible by 5?
Answer:
[tex]Pr =\frac{1}{3}[/tex]
Step-by-step explanation:
Given
[tex]S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)[/tex]
[tex](3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)[/tex]
[tex](5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)\}[/tex] --- sample space
First, list out all outcomes whose sum is divisible by 5
[tex]A = \{(4,6), (5,5),(6,4)\}[/tex]
So, we have:
[tex]n(A) = 3[/tex]
Next, list out all outcomes that has an outcome of 5 in both rolls
[tex]B = \{(5,5)\}[/tex]
[tex]n(B) =1[/tex]
The required conditional probability is:
[tex]Pr =\frac{n(B)}{n(A)}[/tex]
[tex]Pr =\frac{1}{3}[/tex]
