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UkoKoshka [18]
2 years ago
7

Photon Lighting Company determines that the supply and demand functions for its most popular lamp are as follows: S(p) = 400 - 4

p + 0.00002p4 and D(p) = 2800 - 0.0012p3, where p is the price. Determine the price for which the supply equals the demand.
A) $93.24
B) $100.24
C) $96.24
D) $99.24
Mathematics
1 answer:
BlackZzzverrR [31]2 years ago
6 0
S(p) = 400 - 4p + 0.00002p^4
D(p) = 2800 - 0.0012p^3

S(p) = D(p)
400 - 4p + 0.00002p^4 = 2800 - 0.0012p^3
0.00002p^4 + 0.0012p^3 - 4p - 2400 = 0
p = $96.24
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(8.8 × 106)(5 × 102)

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(A) The weight of cans of vegetables is normally distributed with a mean of 1380 grams and a standard deviation of 80 grams. Wha
pychu [463]

Answer:

7.35%

Step-by-step explanation:

μ = 1380

σ = 80  

n = 15  

P(x>1410)  

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= 1.4524  

P(z>1.4524) = 0.4265  (from the graph)

P(z>1.4524) = 0.5 - 0.4265 = 0.0735

7 0
1 year ago
On a coordinate plane, a triangle has points (negative 5, 1), (2, 1), (2, negative 1).
daser333 [38]

Answer:

Step-by-step explanation:

Given a triangle has points:

(-5,1),(2,1), (2,-1)

Let us label the points:

A(2,1),

B(-5,1) and  

C(2,-1)

To find:

Distance between (−5, 1) and (2, −1) i.e. BC.

Horizontal leg AB and

Vertical leg, AC.

Solution:

Please refer to the attached diagram for the labeling of the points on xy coordinate plane.

We can simply use Distance formula here, to find the distance between two coordinates.

Distance formula :

D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

For BC:

x_2 = 2\\x_1 = -5\\y_2 = -1\\y_1 = 1

BC = \sqrt{(2--5)^2+(-1-1)^2} = \sqrt{63}

Horizontal leg, AC:

x_2 = -5\\x_1 = 2\\y_2 = 1\\y_1 = 1

AC = \sqrt{(2-(-5))^2+(1-1)^2} = 7

Vertical Leg,  AB:

x_2 = 2\\x_1 = 2\\y_2 = -1\\y_1 = 1

AB = \sqrt{(2-2)^2+(-1-1)^2} = 2

5 0
2 years ago
Read 2 more answers
At what points does the helix r(t) = sin t, cos t, t intersect the sphere x2 + y2 + z2 = 65? (round your answers to three decima
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7 0
2 years ago
A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F
Dima020 [189]

Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

(a) exactly 5

(b) 2 or fewer

(c) more than 1.

Let <em>X</em> = number of unspoiled eggs in the bad carton of eggs.

Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

The probability of selecting an unspoiled egg is:

P(X)=p=\frac{10}{18}=0.556

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.556.

The success is defined as the selection of an unspoiled egg.

The probability mass function of <em>X</em> is given by:

P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531

Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

              =\sum\imits^{2}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=0.0173+0.1080+0.2706\\=0.3959

Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

              = 1 - P (X = 0) - P (X = 1)

              =1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

6 0
2 years ago
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