20 Here is a list of five numbers.

The probability that a customer of a network operator has a problem about you needing technical staff's help in a month is 0.01.

snow_tiger [21]

Answer:

(a) average calls = 5

(b) probability that there is exactly one call in 6 consecutive monts = 0.038

Step-by-step explanation:

Let event of a customer requiring help in a particular month = H

and event of a customer not requiring help in a particular month = ~H

Given

p= 0.01, therefore

Number of households, n = 500.

Binomial distribution:

x = number of households requiring help in a particular month

P(x,n,p) = C(x,n)*p^x*(1-p)^(n-x)

where

C(x,n) = n!/(x!(n-x)!) is the the number of combinations of x objects out of n

(a) Average number of households requiring help = np = 500*0.01 = 5

(b)

Probability that there are no calls requiring help in a particular month

P(0), q= C(0,n)*p^0(1-p)^(n-0)

= 1*1*0.99^500

= 0.006570483

Applying binomial probability over six months,

q = 0.006570483

n = 6

x = 1

P(x,n,q)

= C(x,n)*q^x*(1-q)^(n-x)

= 6!/(1!*5!) * 0.006570483^1 * (1-0.006570483)^5

= 0.038145

Therefore the probability that in 6 consecutive months there is exactly one month that no customer has called for help = 0.038

3 0

2 years ago

A gym offers two different membership options. In the first option, new members pay a one-time fee of $50 and a monthly

Mumz [18]

Answer:

needs points sorry hehe

7 0

2 years ago

What is the equation of the following graph?

dangina [55]

There are a couple of different ways you could do this, but I'll show the simpler way. We will use the formula

$y=a(x-h)^2+k$

along with the fact that the vertex has h and k coordinates of 1 and 4 respectively, and that a point on the graph is (3, 5). We could have used any point on the graph where there is a definite integer coordinate pair. We will fill in accordingly and solve for a.

$5=a(3-1)^2+4$ and

5 = 4a + 4. If we subtract 4 from both sides we get that $a=\frac{1}{4}$ . Now we will fill in the formula and expand as needed:

$y=\frac{1}{4}(x-1)^2+4$ and

$y=\frac{1}{4}(x^2-2x+1)+4$ . If we distribute the 1/4 in and then add the constants the final equation for that graph will be

$y=\frac{1}{4}x^2-\frac{1}{2}x+\frac{17}{4}$

7 0

3 years ago

Find which one is f f' f"

Olegator [25]

f - green (A)

f' - red (B)

f'' - blue (C)

The graph of f attains local extrema at the same points that f' = 0. There's evidence of this at x = 0 and between x = 3 and x = 3.5. (At both points, curve A attains a maximum, while curve B crosses the x-axis.)

Similarly, the graph of f' attains local extrema at the same points that f'' = 0. This is seen between x = 1.5 and x = 2. (B has a max, C crosses axis)

Also, the graph of A is continuous at x = 1.5 but exhibits a sharp turn, so at this point f is not differentiable and the graph of f' (and subsequently f'') is not continuous.

7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B2%7D%7B3%7D%28x-4%29%28x%2B5%29%3D1%5Cimplies%20%5Ccfrac%7B2%28x-4%29%28x%2

adell [148]

Answer:

x = - 6 3/10

We need to show at least two of these in workings including one in simplified format.

x = - 63/10 (minus in front of whole fraction)

mixed number form x = -6 3/10 (is simplified answer)

decimal form x = -6.3 (is simplified answer)

Step-by-step explanation:

Just in case you ever need to write these in ans search answers already answered in same format.

You can;

Go to upper right corner and click the three dots : another menu opens.

Click the summation sign (4th from left) : you are now in math mode until you deselect it (which sometimes happens automatically). Type a^{2} and it will look like a2 .

Note : the brackets { and } aren’t necessary when your exponent is a natural number between 0 and 9, both included.

However, when you want to type e.g. a15 , you need to place them, because otherwise you will get a15 .

it will look in same way as we are taught with a pen, its the stupid copy paste thing hit and miss if it works on messages here, Obviously it didnt work when i copied just now.

Adding a Range of Cells Using the Summation Button

The Summation button will copy a formula that calculates the sum of the selected numeric cells to the clipboard.

For alphanumeric cells, clicking the Summation button will concatenate the text of the selected cells.

Select the cells by dragging the mouse over the selected area or dragging while holding Alt.

On the Document tab, click Summation.

Position the insertion point in the total cell.

On the Home tab, click Paste.

The Equation bar displays the addition of all cells in the selection.

Note: Only Numeric cells can be summed. Alphanumeric cells will have their text concatenated.

6 0

2 years ago