Expanded form for 400,205

A computer can be classified as either cutting dash edge or ancient. Suppose that 86% of computers are classified as ancient.

Leno4ka [110]

Answer:

(a) The probability that both computers are ancient is 0.7396

(b) The probability that all seven computers are ancient is 0.3479

(c) The probability that at least one of seven randomly selected computers is cutting dash edge is 0.6520.

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

Step-by-step explanation:

We know that 86% of computers are classified as ancient. This means, if one computer is chosen at random, there is an 86% chance that it will be classified as ancient.

$P(ancient)=0.86$

(a) To find the probability that two computers are chosen at random and both are ancient you must,

The probability that the first computer is ancient is $P(ancient)=0.86$ and the probability that the second computer is ancient is $P(ancient)=0.86$

These events are independent; the selection of one computer does not affect the selection of another computer.

When calculating the probability that multiple independent events will all occur, the probabilities are multiplied, this is known as the rule of product.

Let A be the event "the first computer is ancient" and B the event "the second computer is ancient".

$P(A\:and \:B)=P(A)\cdot P(B)=0.86\cdot 0.86=0.86^2= 0.7396$

(b) To find the probability that seven computers are chosen at random and all are ancient you must,

Following the same logic in part (a) we have

Let A be the event "the first computer is ancient",

B the event "the second computer is ancient",

C the event "the third computer is ancient",

D the event "the fourth computer is ancient",

E the event "the fifth computer is ancient",

F the event "the sixth computer is ancient", and

G the event "the seventh computer is ancient"

$P(A\:and \:B\:and \:C\:and \:D\:and \:E\:and \:F\:and \:G)=\\P(A)\cdot P(B)\cdot P(C)\cdot P(D)\cdot P(E)\cdot P(F)\cdot P(G) =(0.86)^7=0.3479$

(c) To find the probability that at least one of seven randomly selected computers is cutting dash edge you must

Use the concept of complement. The complement of an event is the subset of outcomes in the sample space that are not in the event.

Let C the event "the computer is cutting dash edge".

Let A the event "the seven computers are ancient".

$P(C)=1-P(A)=1-0.3479=0.6520$

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

5 0

2 years ago

4. Which of the following equations represent

den301095 [7]

Answer:

Can you take picture, please?

7 0

2 years ago

What is the answer need it

denis-greek [22]

Answer:

$\frac{7}{10} |$

Step-by-step explanation:

STEP 1:

2/3 + 7/10 = ?

The fractions have unlike denominators. First, find the Least Common Denominator and rewrite the fractions with the common denominator.

LCD(2/3, 7/10) = 30

Multiply both the numerator and denominator of each fraction by the number that makes its denominator equal the LCD. This is basically multiplying each fraction by 1.

$(\frac{2}{3}$ * $\frac{10}{10})$ + $(\frac{7}{10} * \frac{3}{3})$ = ?

Complete the multiplication and the equation becomes

$\frac{20}{30} + \frac{21}{30}$

The two fractions now have like denominators so you can add the numerators.

Then:

$\frac{20+21}{30} = \frac{41}{30}$

This fraction cannot be reduced.

The fraction 41/30

is the same as

41 divided by 30

Convert to a mixed number using

long division for 41 ÷ 30 = 1R11, so

41/30 = 1 11/30

Therefore:

2/3+7/10= 1 11/30

STEP 2:

41/30 + -2/3

The fractions have unlike denominators. First, find the Least Common Denominator and rewrite the fractions with the common denominator.

LCD(41/30, -2/3) = 30

Multiply both the numerator and denominator of each fraction by the number that makes its denominator equal the LCD. This is basically multiplying each fraction by 1.

$(\frac{41}{30} *\frac{1}{1} ) + ( \frac{-2}{3} * \frac{10}{10} )$

The two fractions now have like denominators so you can add the numerators.

Then:

$\frac{41+-20}{30} = \frac{21}{30}$

This fraction can be reduced by dividing both the numerator and denominator by the Greatest Common Factor of 21 and 30 using

GCF(21,30) = 3

$\frac{21/3}{30/3} =\frac{7}{10}$

Therefore:

$\frac{41}{30} + \frac{-2}{3} =\frac{7}{10}$ |

8 0

2 years ago

Find the directional derivative of f(x,y,z)=2z2x+y3f(x,y,z)=2z2x+y3 at the point (−1,4,3)(−1,4,3) in the direction of the vector

Feliz [49]

$f(x,y,z)=2z^2x+y^3$

$f$ has gradient

$\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k$

which at the point (-1, 4, 3) has a value of

$\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k$

I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say $\vec u=15\,\vec\imath+25\,\vec\jmath$ , in which case we have

$\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}$

Then the derivative of $f$ at (-1, 4, 3) in the direction of $\vec u$ is

$D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}$

4 0

3 years ago

Using only whole numbers, how can you draw three rectangles with perimeter of 12 units

Assoli18 [71]

Perimeter means the sides add up to 12.
Rectangles have two pairs of equal sides.

2(one side) + 2(another side) = 12
2 (one side + another side) = 12
(one side + another side) = 6

To draw a rectangle with a perimeter of 12 units, the two adjacent sides must add up to 6.
1 + 5
2 + 4
3 + 3

3 0

2 years ago