Find the mean of this set of numbers. 5, –11, 7, 12, –1, 0

2 answers:

8 0

Add all the numbers together, then divide by how much numbers were in the set.

5 + -11 + 7 + 12 + -1 = 12

There are SIX numbers in this set

12/6 = 2

Answer: 2 is the mean

RSB [31]2 years ago

7 0

If you add all the numbers, you get 12. divide this number by the number of numbers there are in total and you get 12 divided by 6. So the mean is 2.

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What is 1/2 X (6 X 4)+3 + 2 Please show your work

exis [7]

Answer:

Step-by-step explanation:

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

$\dfrac{1}{2}\times\underbrace{(6\times4)}_{1}+3+2\\\\=\underbrace{\dfrac{1}{2}\times24}_{2}+3+2\\\\=12+3+2=17$

4 0

2 years ago

each of the 20 balls is tossed independently and at random into one of the 5 bins. let p be the probability that some bin ends u

amm1812

if p is the probability that some bin ends up with 3 balls and q is the probability that every bin ends up with 4 balls. pq is 16.

First, let us label the bins with 1,2,3,4,5.

Applying multinomial distribution with parameters n=20 and p1=p2=p3=p4=p5=15 we find that probability that bin1 ends up with 3, bin2 with 5 and bin3, bin4 and bin5 with 4 balls equals:

5−2020!3!5!4!4!4!

But of course, there are more possibilities for the same division (3,5,4,4,4) and to get the probability that one of the bins contains 3, another 5, et cetera we must multiply with the number of quintuples that has one 3, one 5, and three 4's. This leads to the following:

p=20×5−2020!3!5!4!4!4!

In a similar way we find:

q=1×5−2020!4!4!4!4!4!

So:

pq=20×4!4!4!4!4!3!5!4!4!4!=20×45=16

thus, pq = 16.

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7 0

1 year ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Kitty [74]

Answer: (-2, 5) and (2, -3)

<u>Step-by-step explanation:</u>

Graph the line y = -2x + 1 (which is in y = mx + b format) by plotting the y-intercept (b = 1) on the y-axis and then using the slope (m = -2) to plot the second point by going down 2 and right 1 unit from the first point:

y - intercept = (0, 1) 2nd point = ( -1, 1).

Graph the parabola y = x² - 2x - 3 by first plotting the vertex and then plotting the y-intercept (or some other point):

$y = x^2-2x-3\quad \rightarrow \quad a=1,\ b=-2,\ c=-3\\\\\text{axis of symmetry:}\ x = \dfrac{-b}{2a}\ \longrightarrow \ x=\dfrac{-(-2)}{2(1)}=\dfrac{2}{2}=1\\\\\text{y-value of vertex:}\ f(1) = (1)^2-2(1)-3\quad \longrightarrow \quad y = 1 - 2 - 3=-4\\\\\text{y-intercept:}\ f(0)= (0)^2-2(0)-3\ \longrightarrow \ y=0 - 0 - 3 = -3 \\$