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stiks02 [169]
3 years ago
5

ill give brainliest!! Solve these equations for the variable. Check your solutions. please put the answers to this

Mathematics
1 answer:
natima [27]3 years ago
4 0

Answer:

I have done one question for you

hope it helps

plz Mark as brainlist

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Write 10 × 10 × 10 with an exponent. Explain how you decided which exponent to write.
n200080 [17]
10^3 is the answer to this
3 0
3 years ago
Rose bought 5 pencils for $0.69 each, 3 notebooks for $ 2.75 each, and a pocket dictionary for
Marysya12 [62]

Answer: $2.81

Step-by-step explanation:

Pencils: 5 * 0.69 = 3.45

Notebooks: 3 * 2.75 = 8.25
pocket dictionary: 5.49

3.45 + 8.25 + 5.49 = 17.19

20 - 17.19 = 2.81

Therefore, she will get $2.81 change back.

3 0
2 years ago
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8. In an examination 40% of the students failed. The number of students that passed was 180. How many students failed?
EastWind [94]

Answer:

Neither. 72 students failed.

Step-by-step explanation:

40% of 180 is 72

8 0
3 years ago
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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
3 years ago
Kerri drives 150 miles to visit her grandmother. She starts off driving 60 miles per hour, but then encounters road construction
QveST [7]
I think it would be 2 and a half hours/ 2 hours and 30 minutes.
8 0
3 years ago
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