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arlik [135]
3 years ago
7

If someone could help me out with this, I would greatly appreciate it!!

Mathematics
1 answer:
kati45 [8]3 years ago
7 0

We are given that revenue of Tacos is given by the mathematical expression -7x^{2}+32x+240.

(A) The constant term in this revenue function is 240 and it represents the revenue when price per Taco is $4. That is, 240 dollars is the revenue without making any incremental increase in the price.

(B) Let us factor the given revenue expression.

-7x^{2}+32x+240=-7x^{2}+60x-28x+240\\-7x^{2}+32x+240=x(-7x+60)+4(-7x+60)\\-7x^{2}+32x+240=(-7x+60)(x+4)\\

Therefore, correct option for part (B) is the third option.

(C) The factor (-7x+60) represents the number of Tacos sold per day after increasing the price x times. Factor (4+x) represents the new price after making x increments of 1 dollar.

(D) Writing the polynomial in factored form gives us the expression for new price as well as the expression for number of Tacos sold per day after making x increments of 1 dollar to the price.

(E) The table is attached.

Since revenue is maximum when price is 6 dollars. Therefore, optimal price is 6 dollars.


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a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles  carry just one person.    

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

\mu = np = 92(0.64) = 58.88

\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60

We have to evaluate the probability that more than 47 cars carry just one person.

P(x \geq 47)

After continuity correction, we will evaluate

P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)

= 1 - P(z \leq -2.6913)

Calculation the value from standard normal z table, we have,  

P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0
3 years ago
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