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bearhunter [10]
3 years ago
10

When you flip a baised coin the probability of getting tail is 0.75 How many times would you expect to get tails uf you flip the

coin 320 times?
Mathematics
2 answers:
Elis [28]3 years ago
7 0

Answer: 60

How?: Well, we know that we have a 0.75 chance for tails. And on a coin, there is heads and tails. So, we multiply 0.75 (tails) by 320 and then divide that by 4. We do this because there is a 0.75 and 1 = 1.00. 1.00 - 0.75 = 0.25. 0.25 = 60

Ierofanga [76]3 years ago
3 0

Answer:240

Step-by-step explanation:

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3 0
4 years ago
What is the first step when solving this equation for t?
Bogdan [553]

Answer:

D. subtract 31 from both sides

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Step-by-step explanation:

hope this helped a little bit :)

7 0
4 years ago
Read 2 more answers
Tyrone factored the polynomial completely. What is the value of b? 12x^4+30x^3+4x^2+10x
Arlecino [84]
The value of B = 2x (3x^2 + 1) (2x + 5)

Given: 12x^4 + 30x^3 + 4x^2 + 10x

Req'd: B = ?

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*use elimination

solving for B:

= 2x (6x^3 +15x^2 + 2x +5)
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3 0
3 years ago
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A box of crayons has 12 different coloured crayons in it. 4 crayons are given to one
k0ka [10]

Answer:   34,560

Step-by-step explanation:

Total crayons of different color = 12

Total children = 3

Each child will get 4 crayons.

Number of combinations of selecting <em>r</em> things out of <em>n</em> things = ^nC_r=\dfrac{n!}{r!(n-r)!}

The number of ways of selecting the first 4 crayons for first child = ^{12}C_4=\dfrac{12!}{4!(12-4)!}=\dfrac{12\times11\times10\times9\times8!}{(24)\times 8!}\\\\=495

Colors left = 12-4 =8

Now, the again selecting 4 colors out of 8 for second child = ^{8}C_4=\dfrac{8!}{4!(8-4)!}=\dfrac{8\times7\times6\times5\times4!}{(24)\times4!}=70

Colors left =4

Number of ways of selecting 4 colors out of 4 for third child  =1

Total number of ways = 495 x 70 x 1 = 34650   [By fundamental principle of counting]

Hence, the total number of ways = 34,560.

7 0
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