Factor to find the zeros of the function defined by the quadratic expression. −13x2 − 130x − 273

Carlotta has 12 2/3 yd of ribbon to be divided among 19 girls in the cheerleading squad.how much ribbon will each girl get if it

trasher [3.6K]

Answer:

Each girl gets 2/3 yd of ribbon.

Step-by-step explanation:

12 2/3=38/3

(38/3)/19

(38/3)(1/19)

(2/3)(1/1)=2/3

4 0

2 years ago

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

Korvikt [17]

Answer:

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

Step-by-step explanation:

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

Please I need help it’s an emergency

Fofino [41]

Answer:

the answer is 36

Step-by-step explanation:

beacuse if we use the pythagorean theorem for 9 and 12 we get 15 which means that the perimeter is P=9+15+12=36

8 0

2 years ago

(10 points for the answer)<br><br> y=

Nat2105 [25]

Answer: y=1x+4
Solving Steps:

5 0

2 years ago

Suppose that 50 identical batteries are being tested. After 8 hours of continuous use, assume that a given battery is still oper

timama [110]

<span>Binomial Problem with n = 50 and P(op) = 0.0.7

P(31<=50) = 1 - P(0<=x<=30) = 1 - binomcdf(50,0.7,30) = 1-0.0848 = 0.9152

</span>

5 0

2 years ago