Rewrite the equation by completing the square <br> X^2+6x+9=0<br> (X+?)^2=?

For each container, there are only two possible outcomes. Either it contains a golden sheet of paper, or it does not. The probability of a container containing a golden sheet of paper is independent of other containers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper containers.

This means that $p = 0.3$

14 of these containers of paper.

This means that $n = 14$

What is the probability that John finds less than 7 golden sheets of paper?

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{14,0}.(0.3)^{0}.(0.7)^{14} = 0.0068$

$P(X = 1) = C_{14,1}.(0.3)^{1}.(0.7)^{13} = 0.0407$

$P(X = 2) = C_{14,2}.(0.3)^{2}.(0.7)^{12} = 0.1134$

$P(X = 3) = C_{14,3}.(0.3)^{3}.(0.7)^{11} = 0.1943$

$P(X = 4) = C_{14,4}.(0.3)^{4}.(0.7)^{10} = 0.2290$

$P(X = 5) = C_{14,5}.(0.3)^{5}.(0.7)^{9} = 0.1963$

$P(X = 6) = C_{14,6}.(0.3)^{6}.(0.7)^{8} = 0.1262$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0068 + 0.0407 + 0.1134 + 0.1943 + 0.2290 + 0.1963 + 0.1262 = 0.9067$

90.67% probability that John finds less than 7 golden sheets of paper

7 0

2 years ago

Which set of points on the graph represents equivalent ratios?

Elenna [48]

Answer:c,d,f I think

Step-by-step explanation:

8 0

2 years ago

You want to enlarge a 4 x 5 inch photo to fit into a 1 inch wide frame

skelet666 [1.2K]

The 4•x by 5•x inner dimensions of the 53 inches outer perimeter picture frame, the 1 inch wide frame and the 4 × 5 inch picture to be enlarged, gives;

a. The outer perimeter equation is presented as follows;

18•x + 8 = 53 inches

b. The scale factor required for the enlargement of the picture is 2.5

<h3>Which method can be used to write the equation and find the scale factor?</h3>

The dimensions of the photo = 4 × 5 inch

Width of the frame = 1 inch

Outer perimeter of the frame = 53 inches

The width of the inside of the picture frame = 4•x

The height of the inside of the picture frame = 5•x

a. The equation for the outer perimeter of the picture frame is therefore;

2 × (4•x + 1 + 1) + 2 × (5•x + 1 + 1) = 53

Which gives;

2 × (4•x + 2) + 2 × (5•x + 2) = 53

8•x + 4 + 10•x + 4 = 53

A simplified equation for the outer perimeter of the picture frame is therefore;

18•x + 8 = 53

b. Solving the above equation, we have;

18•x + 8 = 53

18•x = 53 - 8 = 45

18•x = 45

Therefore;

x = 45 ÷ 18 = 2.5

x = 2.5

The width and height of the picture frame are therefore;

Width = 4•x = 4 × 2.5 = 10

Height = 5•x = 5 × 2.5 = 12.5

Which gives;

Width = 10 inches

Height = 12.5 inches

The enlargement factor, sf, is given by the ratio of a side of the picture frame to a side of the picture as follows;

sf = 12.5/5 = 2.5

The factor by which to enlarge the picture frame is 2.5

Learn more about linear scale factors here:

brainly.com/question/19381630

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3 0

1 year ago

Rewrite the equation by completing the square X^2+6x+9=0 (X+?)^2=?