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3 years ago

X-y+z=-7 x+y+z=1 x+y-z=9 How would you solve this equation?

1 answer:

3 0

X - y + z = -7
x + y + z = 1
-------------------add
2x + 2z = -6

x + y + z = 1
x + y - z = 9....multiply by -1
-------------------
x + y + z = 1
-x - y + z = -9 (result of multiplying by -1)
------------------add
2z = -8
z = -8/2
z = -4

2x + 2z = -6
2x + 2(-4) = -6
2x - 8 = -6
2x = -6 + 8
2x = 2
x = 1

x - y + z = -7
1 - y - 4 = -7
-y -3 = -7
-y = -7 + 3
-y = -4
y = 4

so x = 1, y = 4, and z = -4

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Answer:

a) There is a 59.87% probability that none of the LED light bulbs are defective.

b) There is a 31.51% probability that exactly one of the light bulbs is defective.

c) There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) There is a 100% probability that three or more of the LED light bulbs are not defective.

Step-by-step explanation:

For each light bulb, there are only two possible outcomes. Either it fails, or it does not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 10, p = 0.05$

a) None of the LED light bulbs are defective?

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}*(0.05)^{0}*(0.95)^{10} = 0.5987$

There is a 59.87% probability that none of the LED light bulbs are defective.

b) Exactly one of the LED light bulbs is defective?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{10,1}*(0.05)^{1}*(0.95)^{9} = 0.3151$

There is a 31.51% probability that exactly one of the light bulbs is defective.

c) Two or fewer of the LED light bulbs are defective?

This is

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 2) = C_{10,2}*(0.05)^{2}*(0.95)^{8} = 0.0746$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5987 + 0.3151 + 0.0746 0.9884$

There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) Three or more of the LED light bulbs are not defective?

Now we use p = 0.95.

Either two or fewer are not defective, or three or more are not defective. The sum of these probabilities is decimal 1.

$P(X \leq 2) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 0) = C_{10,0}*(0.95)^{0}*(0.05)^{10}\cong 0$

$P(X = 1) = C_{10,1}*(0.95)^{1}*(0.05)^{9} \cong 0$

$P(X = 2) = C_{10,1}*(0.95)^{2}*(0.05)^{8} \cong 0$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0$

$P(X \geq 3) = 1 - P(X \leq 2) = 1$

There is a 100% probability that three or more of the LED light bulbs are not defective.

8 0

3 years ago

A collection of quarters and dimes has a total value of$5.00 and contains 32 coins. How many of each kind of coins are there in

vaieri [72.5K]

9514 1404 393

Answer:

12 quarters

20 dimes

Step-by-step explanation:

If all were dimes the value would be $3.20. It is $1.80 more than that. Each replacement of a dime with a quarter adds $0.15 to the value, so there must be $1.80/$0.15 = 12 such replacements.

There are 12 quarters and 20 dimes.

_____

Let q represent the number of quarters.

0.25q +0.10(32 -q) = 5.00 . . . . . total value

0.15q +3.20 = 5.00 . . . . . . . . . simplify

q = (5.00 -3.20)/0.15 . . . . . . subtract 3.20 and divide by 0.15

q = 12

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2 years ago

Two planes leave an airport at the same time. Thier speeds are mph 130 and 150 mph, and the angle between thier courses is 36 de

Reil [10]

Answer:

132.89 miles

Step-by-step explanation:

Let plane A travel at 130 mph and plane B travel at 150 mph.

After 1.5 hours, the total distance traveled by each plane is:

$A = 1.5 *130\\A = 195\ miles\\B = 1.5 *150\\B = 225\ miles$

The law of cosines can be used to find the distance (D) between both planes as follows:

$D^2 = A^2+B^2-2AB*cos(36)\\D=\sqrt{195^2+225^2-(2*195*225*cos(36))}\\D=132.89\ miles$

They are 132.89 miles apart after 1.5 hours.

5 0

3 years ago