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Goshia [24]
3 years ago
6

Four numbers 1-6 you roll a six-sided die whose sides are numbered from 1 through 6. Find the probability of rolling the followi

ng. NO LINKS!!!!
​

Mathematics
1 answer:
mel-nik [20]3 years ago
5 0

Answer:

#1

<u>Since half of the numbers are even, 2, 4, 6</u>

  • P(even) = 3/6 = 1/2

#2

<u>Multiple of 3 are 3 and 6:</u>

  • P(multiple of 3) = 2/6 = 1/3

#3

<u>The numbers ≤ 4 are: 1, 2, 3, 4</u>

  • P(≤4) = 4/6 = 2/3

#4

<u>Prime numbers are 2, 3, 5</u>

  • P(prime) = 3/6 = 1/2

#5

<u>Numbers > 1 are 2, 3, 4, 5, 6</u>

  • P(>1) = 5/6

#6

<u>A four, there is one option</u>

  • P(4) = 1/6
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Fofino [41]

She gives away n books per week, so If w =week number , then she has (100-nw) books left at any time.

this will be the explicit formula.

so we need to write in terms of recrusive.

to be recrusive each term must be expressed in terms of previous term.

in this represent a_{w[tex] +1} [/tex] in terms of a_{w}

a_{w} =100-n(w-1)

a_{w+1} =100-n(w+1-1)

replace w+1 in place of w .

a_{W+1} =100-nwa_{w} =a_{w+1} +n.

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3 years ago
All my points to whoever can solve this
tankabanditka [31]

Answer:

1/5 + 2/5 i sqrt(6) = .2 + .98i

1/5 - 2/5 i sqrt(6) = .2 - .98i

Step-by-step explanation:

5z^2−9z=−7z−5

We need to get all the terms on one side (set the right side equal to zero)

Add 7z to each side

5z^2−9z+7z=−7z+7z−5

5z^2−2z=−5

Add 5 to each side

5z^2−2z+5=−5 +5

5z^2−2z+5=0

This is in the form

az^2 +bz+c = 0 so we can use the quadratic formula

where a = 5   b = -2  and c = 5

-b± sqrt(b^2-4ac)

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-(-2)± sqrt((-2)^2-4(5)5)

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2± sqrt(4-100)

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2± sqrt(-96)

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2± sqrt(16)sqrt(-1) sqrt(6)

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2± 4i sqrt(6)

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1/5 ± 2/5 i sqrt(6)

Splitting the ±

1/5 + 2/5 i sqrt(6) = .2 + .98i

1/5 - 2/5 i sqrt(6) = .2 - .98i


3 0
3 years ago
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ludmilkaskok [199]
The answer is sometimes!
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3 years ago
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You are thinking of making your mansion more energy efficient by replacing some of the light bulbs with compact fluorescent bulb
Lunna [17]

Answer:

1) The number of compact fluorescent light bulbs to buy = 120 bulbs

The number of square feet of insulation to be purchased = 2,800 sqft

2) The amount of savings in energy cost per year = $800

Step-by-step explanation:

The cost of each compact fluorescent bulb = $4

The average amount saved by each bulb per year = $ 2

The cost of each square foot of wall insulation = $ 1

The amount saved by each square foot of wall insulation = $ 0.20

The number of light fittings in the mansion = 300 fittings

The area of uninsulated exterior wall of the mansion = 2,800 ft.²

The maximum amount saved in energy cost = $800

1) Let 'x' represent the number of compact fluorescent bulb used and let 'y' represent the number of square feet of insulation purchased

We have;

2·x + 0.2·y ≤ 800

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Total Cost, P = 4·x + y

From the combined graph of the total cost and cost savings, we have the highest total cost is given when the wall insulation area, y is maximum, that is y = 2,800 and x = 120

Therefore, we have;

Total Cost, P = 4·x + y

∴ P = 4 × 120 + 2,800 = 3,280

Therefore, the number of compact fluorescent light bulbs and the number of square feet of insulation to be purchased are 120 bulbs and 2,800 square feet of insulation respectively

2) The amount of savings in energy cost per year, C_s = 2·x + 0.2·y

Where;

x = 120, and y = 2,800

∴ C_s = 2 × 120 + 0.2 × 2,800 = 800

The amount of savings in energy cost per year, C_s = $800

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Semmy [17]

Step-by-step explanation:

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