True because<span> the common factor to both numbers would be an even number.
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Answer:
24/35, about 69%
Step-by-step explanation:
The data given can be put into a 2-way table (attached). It shows that 0.48 of all calls were answered and resulted in a mortgage application. Altogether, 0.70 of all calls resulted in a mortgage application. Thus the conditional probability of interest is ...
p(spoke to attendant | applied for a mortgage) = p(spoke & applied)/p(applied)
= 0.48/0.70 = 24/35 ≈ 69%
Answer:
7) A school group is assigned to survey vast museum collections of all macroscopic (not microscopic) animal species collected from various regions around the world. What observation are the students most likely to make? A) The majority of the animals are wormlike. B) The majority of the animals possess a …
Step-by-step explanation:
Actually, yes, it is possible for two different numbers to give the same result when squared.
In my last answer, I wrote that it wasn't, but I realize now where my mistake was made.
When a number like positive 4 is squared, the answer is 16. When a number like negative 4 is squared, the answer is also 16. I think that the only time when two different squared numbers have the same result is when they are the same number but have a different positive/negative sign.
I hope this helps.
Answer:
The mean is also increased by the constant k.
Step-by-step explanation:
Suppose that we have the set of N elements
{x₁, x₂, x₃, ..., xₙ}
The mean of this set is:
M = (x₁ + x₂ + x₃ + ... + xₙ)/N
Now if we increase each element of our set by a constant K, then our new set is:
{ (x₁ + k), (x₂ + k), ..., (xₙ + k)}
The mean of this set is:
M' = ( (x₁ + k) + (x₂ + k) + ... + (xₙ + k))/N
M' = (x₁ + x₂ + ... + xₙ + N*k)/N
We can rewrite this as:
M' = (x₁ + x₂ + ... + xₙ)/N + (k*N)/N
and (x₁ + x₂ + ... + xₙ)/N was the original mean, then:
M' = M + (k*N)/N
M' = M + k
Then if we increase all the elements by a constant k, the mean is also increased by the same constant k.
