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andreyandreev [35.5K]
3 years ago
5

Select all of the situations which can be represented by the expression 25.75x + 10.

Mathematics
2 answers:
Kobotan [32]3 years ago
6 0

Answer:

The situation that says Sasha sells a scarf for $10.00 and xjackets for 25.75 each

Step-by-step explanation:

We know thatt she is only selling one scarf for $10.00 and we know we are multiplying X jackets times 25.75

so 25.75x + 10

GarryVolchara [31]3 years ago
4 0
25.75x+10 tht the answer
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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.

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for the given license plate configuration, determine how many different plates are possinle if letters and digits (a) can ba rep
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Answer:

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Step-by-step explanation:

The permutations formula is important to solve this question.

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

(a) can be repeated:

There are 3 letters, each one with 26 possible outcomes.

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So

T = 26*26*26*10*10 = 1757600

1,757,600 different plates possible.

(b) cannot be repeated:

Here the permutations formula is used.

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Two digits, from a set of 10. So

T = P_{(26,3)}*P_{(10,2)} = \frac{26!}{(26-3)!}*\frac{10!}{(10-2)!} = 1404000

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