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Viktor [21]
3 years ago
10

Please give Examples for real life examples for expressing very small number sin standard form.

Mathematics
1 answer:
Digiron [165]3 years ago
3 0
Scientific notation is a system for expressing very large or very small numbers in a compact manner. It uses the idea that such numbers can be rewritten as a simple number multiplied by 10 raised to a certain exponent, or power.
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The system of linear equations 5 x + 3 y = 3 and x + y = negative 1 is graphed below.
Thepotemich [5.8K]

Answer:

I think the answer is (4,-3)

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brian is building a wood frame around a window in his house. if the window is 4 feet by 5 feet, how much wood does he need for t
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18 eighteen feet long
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If f(x) =x/2 -2 and g(x) = 2x2 + x - 3, find (8 +g)(x),
vesna_86 [32]
I would go with D. 2x^2+3/2x-5
3 0
3 years ago
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Please help urgently ​
Andreyy89

Hey there!

2ab + 2 / 2ab - 2

= 2(1)(3) + 2 / 2(1)(3) - 2

= 2(3) + 2 / 2(3) - 2

= 6 + 2 / 6 - 2

= 8 / 4

= 2


Therefore, your answer should be: 2


Good luck on your assignment & enjoy your day!


~Amphitrite1040:)

6 0
1 year ago
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Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

3 0
3 years ago
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