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3 years ago

There are 20 machines in a factory. 7 of the machines are defective.

Mathematics

1 answer:

adelina 88 [10]3 years ago

4 0

Answer:

0.1225

Step-by-step explanation:

Given

Number of Machines = 20

Defective Machines = 7

Required

Probability that two selected (with replacement) are defective.

The first step is to define an event that a machine will be defective.

Let M represent the selected machine sis defective.

P(M) = 7/20

Provided that the two selected machines are replaced;

The probability is calculated as thus

P(Both) = P(First Defect) * P(Second Defect)

From tge question, we understand that each selection is replaced before another selection is made.

This means that the probability of first selection and the probability of second selection are independent.

And as such;

P(First Defect) = P (Second Defect) = P(M) = 7/20

So;

P(Both) = P(First Defect) * P(Second Defect)

PBoth) = 7/20 * 7/20

P(Both) = 49/400

P(Both) = 0.1225

Hence, the probability that both choices will be defective machines is 0.1225

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X = ? y = ? 16 45 degrees

mezya [45]

Let's put more details in the figure to better understand the problem:

Let's first recall the three main trigonometric functions:

$\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}$ $\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}$ $\text{ Tangent }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Adjacent Side}}$

For x, we will be using the Cosine Function:

$\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}$ $Cosine(45^{\circ})\text{ = }\frac{\text{ x}}{\text{ 1}6}$ $(16)Cosine(45^{\circ})\text{ = x}$ $(16)(\frac{1}{\sqrt[]{2}})\text{ = x}$ $\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}$ $\text{ 8}\sqrt[]{2}\text{ = x}$

Therefore, x = 8√2.

For y, we will be using the Sine Function.

$\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}$ $\text{ Sine }(45^{\circ})\text{ = }\frac{\text{ y}}{\text{ 1}6}$ $\text{ (16)Sine }(45^{\circ})\text{ = y}$ $\text{ (16)(}\frac{1}{\sqrt[]{2}})\text{ = y}$ $\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}$ $\text{ 8}\sqrt[]{2}\text{ = y}$

Therefore, y = 8√2.

5 0

1 year ago

Help mw with this one

Alex787 [66]

1/4 in fraction form, or 0.4 in decimal

5 0

3 years ago

Solve for x by completing the square: x2 - 12x + 11 = 0

sineoko [7]

Answer:

Step-by-step explanation:

x² - 12x + 11 = 0 ( subtract 11 from both sides )

x² - 12x = - 11

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 6)x + 36 = - 11 + 36

(x - 6)² = 25 ( take square root of both sides )

x - 6 = ± $\sqrt{25}$ = ± 5 ( add 6 to both sides )

x = 6 ± 5

Then

x = 6 - 5 = 1 ⇒ (1, 0 )

x = 6 + 5 = 11 ⇒ (11, 0 )

6 0

2 years ago

F(x)=(5x+3)(x−2)(3x+7)(x+5) has zeros at x = -5, x = -7/3, x = 2

musickatia [10]

The sign of f on the interval -7/3 < x < -3/5 is always positive.

<h3>How to solve for the sign on the interval</h3>

We have the equation

(5x+3)(x−2)(3x+7)(x+5) > 0

Now when f(x) > 0

Then -7/3 < x < -3/5

This would tell us that the sign would become positive when it changes from the less than to greater than sign

<h3>Complete Question</h3>

f(x)=(5x+3)(x-2)(3x+7)(x+5) has zeros at x=-5, x=-7/3, x=-3/5, and x=2

What is the sign of f on the interval -7/3<x<-3/5?

answer choices

f is always positive on the interval

f is always negative on the interval

f is sometimes positive and sometimes negative on the interval

f is never positive or negative on the interval