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sergejj [24]
3 years ago
5

Evaluate the expression. (6–2)[(2+3)+4] a. 9 b. 13 c. 24 d. 36

Mathematics
1 answer:
Ulleksa [173]3 years ago
4 0
(6-2)[(2+3)+4] \\ 4 [(2+3)+4] \\ 4 [5+4] \\ 4 (9) \\ 36

Your answer is D. 36
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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
Write words to match the expression.<br><br> 3+(4x12)
IgorLugansk [536]
Add three to the product of 4 times twelve
3 0
2 years ago
Read 2 more answers
Thomas invested $40,000. He put part of the money in an account paying 2% interest and the remainder
Rainbow [258]

Answer:

18,000 at 2%

22,000 at 3%

Step-by-step explanation:

x: the amount in account paying 2% interest

y: the amount in stock paying 3% interest

x + y = 40,000

2%. x + 3%. y = 1020

⇒ x = 18,000

y = 22,000

5 0
2 years ago
In a certain​ country, the mean birth weight for boys is 3.27 ​kg, with a standard deviation of 0.51 kg. Assuming that the distr
Degger [83]

Answer:

0.066

Step-by-step explanation:

We solve this question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation

For the question

mean birth weight for boys is 3.27 ​kg, with a standard deviation of 0.51 kg.

x = 2.5

Hence:

z = 2.5 - 3.27/0.51

z = -1.5098

Probability value from Z-Table:

P(x ≤ 2.5) =P(x < 2.5) = P(x = 2.5) =

=0.065547

Therefore, the proportion of baby boys that are born with a low birth weight is 0.066

5 0
3 years ago
Question: "If y &gt; 3, what is the value of n ?"
iris [78.8K]

Answer:

y-3

Problem:

What is the remainder when the dividend is xy-3, the divisor is y, and the quotient is x-1. ?

Step-by-step explanation:

Dividend=quotient×divisor+remainder

So we have

xy-3=(x-1)×(y)+remainder

xy-3=(xy-y)+remainder *distributive property

Now we just need to figure out what polynomial goes in for the remainder so this will be a true identity.

We need to get rid of minus y so we need plus y in the remainder.

We also need minus 3 in the remainder.

So the remainder is y-3.

Let's try it out:

xy-3=(xy-y)+remainder

xy-3=(xy-y)+(y-3)

xy-3=xy-3 is what we wanted so we are done here.

7 0
2 years ago
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