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3 years ago

What is equivalent to 7/8 and don't write 14/16

Mathematics

1 answer:

Margaret [11]3 years ago

6 0

7/8 is equivalent to 56/64

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Let the demand function for a product be q = 100 - 2p. the inverse demand function of this demand function is:_______

Westkost [7]

The inverse demand function of the given demand function is p = 50 - q/2.

A graph that depicts the relationship between a product's price and demand is called a demand curve. On a demand graph, the horizontal axis represents the amount desired, while the vertical axis represents the product's price.

The price is a function of the quantity required when there is an inverse demand curve. The inverse of a demand curve indicates that variations in the amount required cause changes in price levels. The formula for calculating the demand curve for a product yields the graph of an inverse demand curve.

Given demand function: q = 100 - 2p.

To find the inverse demand function, we find the inverse of the equation, by isolating p, to get:

q = 100 - 2p,

or, 2p = 100 - q,

or, p = 100/2 - q/2,

or, p = 50 - q/2.

Thus, the inverse demand function of the given demand function is p = 50 - q/2.

Learn more about inverse functions at

brainly.com/question/3831584

#SPJ4

3 0

2 years ago

6 high school seniors choose from among 20 quotes for their yearbook. What is the probability that at least 2 of them choose the

shusha [124]

Using the binomial distribution, it is found that there is a 0.0328 = 3.28% probability that at least 2 of them choose the same quote.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem, we have that:

There are 6 students, hence n = 6.

There are 20 quotes, hence the probability of each being chosen is p = 1/20 = 0.05.

The probability of one quote being chosen at least two times is given by:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

P(X < 2) = P(X = 0) + P(X = 1).

Then:

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

$P(X = 0) = C_{6,0}.(0.05)^{0}.(0.95)^{6} = 0.7351$

$P(X = 1) = C_{6,1}.(0.05)^{1}.(0.95)^{5} = 0.2321$

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.7351 + 0.2321 = 0.9672.

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9672 = 0.0328$

0.0328 = 3.28% probability that at least 2 of them choose the same quote.

More can be learned about the binomial distribution at brainly.com/question/24863377

6 0

2 years ago

A consulting firm has received 2 Super Bowl playoff tickets from one of its clients. To be fair, the firm is randomly selecting

Soloha48 [4]

Answer:

Therefore the only statement that is not true is b.)

Step-by-step explanation:

There employees are 6 secretaries, 5 consultants and 4 partners in the firm.

a.) The probability that a secretary wins in the first draw

= $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secreataries}{total \hspace{0.1cm} number \hspace{0.1cm} of \hspace{0.1cm} employees} = \frac{6}{15}$

b.) The probability that a secretary wins a ticket on second draw. It has been given that a ticket was won on the first draw by a consultant.

p(secretary wins on second draw | consultant wins on first draw)

= $\frac{p((consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)\cap( secretary\hspace{0.1cm} wins\hspace{0.1cm} on second \hspace{0.1cm}draw))}{p(consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)}$

= $\frac{\frac{5}{15} \times \frac{6}{14}}{\frac{5}{15} } = \frac{6}{14}$ .

The probability that a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw = $\frac{6}{15}$ is not true.

The probability that a secretary wins on the second draw = $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secretaries \hspace{0.1cm} remaining } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} remaining} = \frac{6 - 1}{15 - 1} = \frac{5}{14}$

c.) The probability that a consultant wins on the first draw =

$\frac{number \hspace{0.1cm} of \hspace{0.1cm} consultants \hspace{0.1cm} } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} } = \frac{5 }{15} = \frac{1}{3}$