Lin wants to bike 9/10 of a mile .She has already biked 4/10 of a mile How much further does she have to bike

In the first equation in the system of equations, y represents the money collected from selling sweatshirts. In the second equat

NeTakaya

Answer:

The solution x = 229.89 ≅ 230 (since we cannot have a decimal number of sweatshirts)implies that, the money collected from selling sweatshirts equals the money used in producing sweatshirts when 230 sweatshirts have been sold or produced.

Step-by-step explanation:

Here is the complete question

In the first equation in the system of equations, y represents the money collected from selling sweatshirts. in the second equation, y represents the money spent to produce x sweatshirts with team logos on them for a professional sports league.

y=35x; y=-0.05(x-400)²+9,492

what does the solution of the system represent in this context?

Solution

The solution of the system is obtained when both equations are equal. That is y=35x = y=-0.05(x-400)²+9,492 .

So, 35x = -0.05(x-400)² + 9,492.

Expanding the right side, we have

35x = -0.05(x² - 800x + 160000) + 9492

35x = -0.05x² -800x × -0.05 + (-0.05) × 160000 + 9492

35x = -0.05x² +40x - 8000 + 9492

35x = -0.05x² +40x + 1492

Collecting all the terms to the left side, we have

0.05x² - 40x - 1492 + 35x = 0

0.05x² - 5x -1492 = 0

Using the quadratic formula,

$x = \frac{-(-5) +/- \sqrt{(-5)^{2} - 4 X 0.05 X - 1492} }{2X0.05} = \frac{5 +/- \sqrt{25 + 298.4} }{0.1} = \frac{5 +/- \sqrt{323.4} }{0.1}\\= \frac{5 +/- 17.983 }{0.1}\\= \frac{5 - 17.983 }{0.1} or \frac{5 + 17.983 }{0.1}\\= \frac{-12.983 }{0.1} or \frac{22.983 }{0.1}\\= -129.83 or 229.83$

We choose the positive answer x = 229.83. Since we cannot have a decimal number of sweatshirts, we approximate to the nearest whole number. So, x = 229.83 ≅ 230 sweatshirts.

The solution implies that the money collected from selling sweatshirts equals the money used in producing sweatshirts when 230 sweatshirts have been sold or produced.

We confirm that by plugging in x = 229.83 in both equations.

y = 35x = 35(229.83) = 8044.05 ≅ 8044.1

y=-0.05(x-400)²+9,492 = y=-0.05(229.83-400)²+9,492 = y=-0.05(-170.17)²+9,492 = y=-0.05(28,957.8289)+9,492 = -1447.8914 + 9492 = 8044.11 ≅ 8044.1

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Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

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Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

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Step-by-step explanation:

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