For this we just need to keep dividing 64 until we are left with just the prime numbers:
64
=2*32
=2*2*16
=2*2*2*8
=2*2*2*2*4
=2*2*2*2*2*2
Therefore, 64 = 2^6
Jason’s pool is 7 inches wide because you have to multiply Length and Width so 4x7=28 so the answer is G.7 inches
Answer:
Probability that at most 50 seals were observed during a randomly chosen survey is 0.0516.
Step-by-step explanation:
We are given that Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during each survey.
The numbers of seals observed were normally distributed with a mean of 73 seals and a standard deviation of 14.1 seals.
Let X = <u><em>numbers of seals observed</em></u>
The z score probability distribution for normal distribution is given by;
Z =
~ N(0,1)
where,
= population mean numbers of seals = 73
= standard deviation = 14.1
Now, the probability that at most 50 seals were observed during a randomly chosen survey is given by = P(X
50 seals)
P(X
50) = P(
) = P(Z
-1.63) = 1 - P(Z < 1.63)
= 1 - 0.94845 = <u>0.0516</u>
The above probability is calculated by looking at the value of x = 1.63 in the z table which has an area of 0.94845.
Answer:
50 pages
Step-by-step explanation:
We can set up an equation for this in slope-intercept form.
The slope is 2 due to her expecting to write 2 pages an hour.
The y-intercept is 2 because she has already written 2 pages.
The full equation is y = 2x + 2.
We can plug 24 in for x to get y = 48 + 2, which simplifies to y = 50.