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svlad2 [7]
3 years ago
12

A box contains 1 plain pencil and 4 pens. A second box contains 6 color pencils and 2 crayons. One item from each box is chosen

at random. What is the probability that a pen from the first box and a crayon from the second box are selected?
Mathematics
1 answer:
FrozenT [24]3 years ago
8 0

Probability that a pen from the first box and a crayon from the second box are selected is 1/5.

<u>Step-by-step explanation</u>:

Step 1 :

  • First box contain = 1 pencil + 4 pens = 5 items
  • Second box contain = 6 color pencils + 2 crayons = 8 items

Step 2 :

Probability for a pen from 1st box, P(pen) = no. of pens / total items

P(pen) = 4/5

Probability for a crayon from 2nd box, P(crayon) = no. of crayons / total items

P(crayon) = 2/8 = 1/4

Step 3 :

P(pen) and P(crayon) = (4/5) \times (1/4)

                                   = 4/20

P(pen) and P(crayon) = 1/5

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Dmitriy789 [7]

Answer:

this is the answer

Step-by-step explanation:

a 65

b 650

c 52

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6 0
2 years ago
52m when m=105 explain this to me please
JulsSmile [24]
52m when m= 105
52(105)
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5 0
3 years ago
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Ira Lisetskai [31]

Answer:

A) 2.50

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the length of diagonal of a rectangular field is 23.7 m and one of its sides is 18.8 m. find the perimeter of the field.​
katen-ka-za [31]

Answer:

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Step-by-step explanation:

So we have a rectangle with a width of 18.8 meters and a diagonal with 23.7 meters. To find the perimeter, we need to find the length first. Since a rectangle has four right angles, we can use the Pythagorean Theorem, where the diagonal is the hypotenuse.

a^2+b^2=c^2

Plug in 18.8 for either <em>a </em>or <em>b. </em>Plug in the diagonal 23.7 for <em>c. </em>

<em />(18.8)^2+b^2=23.7^2\\b^2=23.7^2-18.8^2\\b=\sqrt{23.7^2-18.8^2} \\b\approx14.4 \text{ meters}

Therefore, the length is 14.4 meters. Now, find the perimeter:

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7 0
3 years ago
How many pounds of candy worth 70 cents a pound must be mixed with 30 pounds of candy worth 90 cents a pound to produce a mixtur
Svet_ta [14]
<h3>Answer:</h3>

B) 10 pounds

<h3>Explanation:</h3>

Let x represent the amount of 70¢ candy to be added. The value of the mixture can be written as ...

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... 150 = 15x . . . . . . add -70x-2550

... 10 = x . . . . . . . . . divide by the coefficient of x

10 pounds of candy at 70¢/lb must be added.

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