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igor_vitrenko [27]
3 years ago
7

What is the total number of digits required in numbering the pages of a book, which has 1,150 pages?

Mathematics
1 answer:
zhuklara [117]3 years ago
5 0

Answer:

1150 numbers of digits since to every page is a digit

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Find the gradients of lines A and B.
Tanya [424]

Let's see

For line A

  • (2,2)
  • (3,6)

Slope

  • m=6-2/3-2
  • m=4/1
  • m=4

For line B

  • (1,6)
  • (2,4)

Slope

  • m=4-6/2-1
  • m=-2/1
  • m=-2
3 0
2 years ago
You need to hang seven pictures in a straight line, in how many ways can this be done?
Furkat [3]
This is simple. It is 7 factorial, or 7!. This just means 7*6*5*4*3*2*1, which is 5040.
3 0
3 years ago
Scar travels 350 miles on 20 gallons of gasoline. How many gallons will be used to travel 875 miles under the same conditions?
Hatshy [7]
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6 0
3 years ago
There are 4 green marbles and 2 red marbles in the jar. You just randomly draw one by one without replacement and stop when you
zalisa [80]

Answer:

Then the probability distribution is:

P(0) = 1/3

P(1) = 4/15

P(2) = 1/5

P(3)  = 2/15

P(4) = 1/15

The expected value for X is:

EV = 1.33...

Step-by-step explanation:

We have a total of 6 marbles in the jar.

The probability of getting a red marble in the first try  (X = 0) is equal to the quotient between the number of red marbles and the total number of marbles, this is:

P(0) = 2/6 = 1/3

The probability of drawing one green marble (X = 1)

is:

First, you draw a green marble with a probability of 4/6

Then you draw the red one, but now there are 5 marbles in the jar (2 red ones and 3 green ones), then the probability is 2/5

The joint probability is:

P(1) = (4/6)*(2/5) = (2/3)*(2/5) = 4/15

The probability of drawing two green marbles (X  = 2)

Again, first we draw a green marble with a probability of 4/6

Now we draw again a green marble, now there are 3 green marbles and 5 total marbles in the jar, so this time the probability is 3/5

Now we draw the red marble (there are 2 red marbles and 4 total marbles in the jar), with a probability of 2/4

The joint probability is:

P(2) = (4/6)*(3/5)*(2/4) = (2/6)*(3/5) = 1/5

The probability of drawing 3 green marbles (X = 3)

At this point you may already understand the pattern:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

finally, we draw a red marble with a probability 2/3

The joint probability is:

P(3) = (4/6)*(3/5)*(2/4)*(2/3) = (2/6)*(3/5)*(2/3) = (1/5)*(2/3) = (2/15)

Finally, the probability of drawing four green marbles (X = 4) is given by:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

fourth, we draw a green marble with a probability 1/3

Finally, we draw a red marble with a probability 2/2 = 1

The joint probability is:

P(4) = (4/6)*(3/5)*(2/4)*(1/3)*1 = (1/5)*(1/3) = 1/15

Then the probability distribution is:

P(0) = 1/3

P(1) = 4/15

P(2) = 1/5

P(3)  = 2/15

P(4) = 1/15

The expected value will be:

EV = 0*P(0) + 1*P(1) + 2*P(2) + 3*P(3) + 4*P(4)

EV = 1*(4/15) + 2*( 1/5) + 3*( 2/15) + 4*(1/15 ) = 1.33

So we can expect to draw 1.33 green marbles in this experiment.

5 0
3 years ago
Five oranges were selected at random from a crate of 53 oranges. Their diameters were measured, in millimeters, and then recorde
Olin [163]
Ok so mean means average so exmple
mean of 1,2,3+5 would be (1+2+3+5)/4
(sum of terms)/(number of terms)=mean so

number of terms=5 since 5 oranges
sum=(46+39+53+61+49)

so mean=(46+39+53+61+49)/5=248/5=49.6

so the answer would be 49.6 (estimateed to be 50)
5 0
3 years ago
Read 2 more answers
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