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djyliett [7]
2 years ago
12

Ben wants to have his birthday at the bowling alley with a few of his friends, but he can spend no more than $80. The bowling al

ley charges a flat fee of $45 for a private party and $5.50 per person for shoe rentals and unlimited bowling. If you had to write an inequality that represents the total cost of Ben's birthday for,p, people have given his budget, what number represents the coefficient?
Mathematics
1 answer:
Debora [2.8K]2 years ago
3 0

Answer:

He can have 6 people

Step-by-step explanation:

45 + 5.50p = 50.50p

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During the three weeks how many customers came from zones 3 and 4
Mama L [17]

Times Is your answer just times 3 by 4 and twelves' your answer don't ask for more.  
4 0
2 years ago
Given EP = FP and GQ = FQ, what is the perimeter of ΔEFG?
eduard

x is  \frac{2y + 1}{2y - 3}

<u>Explanation:</u>

Given:

EP = FP

GQ = FQ

Since the sides are equal, the ΔFPQ and ΔFEG are similar.

According to the property of similarity:

\frac{FP}{PE} =\frac{FQ}{QG}

On substituting the value:

\frac{2x}{4y+2} = \frac{3x-1}{4y + 4} \\\\2x ( 4y + 4) = (3x-1)(4y+2)\\\\8xy + 8x = 12xy + 6x - 4y - 2\\\\4xy - 6x - 4y - 2 = 0\\\\2xy - 3x - 2y - 1 = 0\\\\2xy - 3x = 2y + 1\\\\x(2y - 3) = 2y +1\\\\x = \frac{2y+1}{2y-3}

Perimeter will be sum of all the sides. If y is known then substitute the value and then find the perimeter.

5 0
2 years ago
A rational function can have infinitely many x-values at which it is not continuous. True or False. If false, explain
sattari [20]
<span>It is false since the rational function is discontinuous when the denominator is zero. But the denominator is a polynomial and a polynomial has only finitely many zeros. So the discontinuity points of a rational function is finite. </span>
3 0
2 years ago
A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe
xxMikexx [17]

Answer:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

Step-by-step explanation:

For the random variable N_1 we define the possible values for this variable on this case [1,2,3,4,5] . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable N_1 like this:

P(N_1 = a) = \frac{5-a C 1}{5C2}

For the distribution of N_2 we need to take in count that we are finding a conditional distribution. N_2 given N_1 =a, for this case we see that N_2 \in [1,2,...,5-a], so then exist 5-a C 1 ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}

And if we want to find the joint probability we just need to do this:

P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)

And if we multiply the probabilities founded we got:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

8 0
2 years ago
At a financial institution, a fraud detection system identifies suspicious transactions and sends them to a specialist for revie
labwork [276]

Answer:

a. E(X) = 54.4

b. E(X) = 2.5

c. P(Y=2) = .0116

Step-by-step explanation:

a.

    E(X) = np = .40 probability * 136 trials = 54.4 blocked transmissions

    To get the expected value, we simply multiply probability times number of trials. You can look at it in simple terms by thinking if there's a 50% chance of flipping heads and you flip a coin twice, in an ideal world you will have .5*2 = 1 head.

b.

    i. Let X represent the number of suspicious transmissions reviewed until finding the first blocked one. We will use a geometric distribution to model the "first" transmission. Whenever we're looking for the "first" time something happens, we use geometric.

   ii. E(X) = 1/p , according to the geometric model.

              = 1/.4 = 2.5.

       We expect that the specialist will review 2.5 suspicious transactions <em>on average </em>before finding the first transmission that will be blocked.

c.

    i. Let Y represent the exact number of blocked transmissions out of 10. We will use a binomial distribution to model the "fixed" number of transmissions. Whenever we're looking for a "fixed" number of times something happens, we use binomial.

    ii. P(Y=k) = (n choose k)(p^k)(q^n-k)

        P(Y=2) = (¹⁰₂)(.4^2)(.6^10-2)

                    = 45 (.4^2)(.6^10-2) = .0016

        As for calculator notation, the n choose k can be accessed on a TI-84 via MATH -> PRB -> nCr. It looks like 10 nCr 2 on the display.

        Hence the probability that two transactions out of ten will be blocked is .0016 by the binomial model.

5 0
2 years ago
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