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soldier1979 [14.2K]
3 years ago
5

The cost of a loaf of bread increases by $0.50 for every $1 increase in the cost of 1 pound of flour. Which equation relates the

cost of one loaf of bread to the cost of one pound of flour?
Mathematics
2 answers:
lions [1.4K]3 years ago
7 0

Answer:

y=0.5x+C

Step-by-step explanation:

Given that the cost of a loaf of bread increases by $0.50 for every $1 increase in the cost of 1 pound of flour.

Let x be the cost of flour and y the cost of bread.

Rate of change of y with respect to x is constant as 0.50

This can be represented as

\frac{dy}{dx} =0.5

This can be solved as

y = 0.5x+C where C is the cost of bread even when x=0

Klio2033 [76]3 years ago
5 0
.5x=new price. X= dollars increased per loaf
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(1) The probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

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Let <em>X</em> = number of computers to be tested before the 1st defect is found.

Then the random variable X\sim Geo(p).

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P (X ≥ 5) = 1 - P (X < 5)

              = 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]

              =1 -[(1-0.40)^{0}0.40+(1-0.40)^{1}0.40+(1-0.40)^{2}0.40\\+(1-0.40)^{3}0.40+(1-0.40)^{4}0.40]\\=1-[0.40+0.24+0.144+0.0864+0.05184]\\=0.07776\\\approx0.078

Thus, the probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2)

Let <em>Y</em> = number of computers infected.

The number of computers in the company is, <em>n</em> = 20.

Then the random variable Y\sim Bin(20,0.40).

The probability function of a binomial distribution is:

P(Y=y)={n\choose y}p^{y}(1-p)^{n-y};\ y=0,1,2,...

Compute the probability at least 5 computers are infected as follows:

P (Y ≥ 5) = 1 - P (Y < 5)

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Thus, the probability at least 5 computers are infected is 0.949.

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