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natta225 [31]
3 years ago
14

If you toss four fair coins, in how many ways can you obtain at least one head? Group of answer choicesA. 1/16B. 1/2C. 3/16D. 11

/16E. 3/8
Mathematics
1 answer:
LenKa [72]3 years ago
3 0

Answer:

The correct answer is:

15/16

Step-by-step explanation:

In a toss of four fair coin, there are possibility of 16 occurrence because a coin have an head and a tail (2).

In 4 coins, we will have : 2⁴ = 16

Let S be the sample space, then:

S = { HHHH, HHHT, HHTT, HTTT, TTTH, TTHH, TTTH, TTTT, HTHT, THTH, HHTH, HTHH, TTHT,THHT, HTTH, THTT}

Let E₁ be the event that at least one head appears:

n(E₁) = 15

∴ Pr (E₁) = n(E₁)/ n(S)

             = 15/16

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The expression i/pt represents the rate of instrest being charged if a loan of p dollars for t years required i dollars in intre
zysi [14]

Answer:

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Step-by-step explanation:

Given:

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Solution:

Putting the given values in the formula:

R=\dfrac{151.11}{690 \times 3}\\\Rightarrow R=\dfrac{151.11}{2070}\\\Rightarrow R=0.073

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8 0
3 years ago
Given the following observed and expected data (total of 1000), using chi-squared calculate a p-value that corresponds with this
finlep [7]

Answer:

The answer is "0.90>p>0.75."

Step-by-step explanation:

\text{Cinnabar vestigial} \ \ \ \ \ \ \ \ \ \ \ 384 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \  -6 36 \ \ \ \ \ \ \ \ \ \ \  0.092308\\\\

roof \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \  408 \ \ \ \ \ \ \ \ \ \ \  390  \ \ \ \ \ \ \ \ \ \ \  18 \ \ \ \ \ \ \ \ \ \ \  324  \ \ \ \ \ \ \ \ \ \ \ 0.830769\\\\\text{Cinnabar vestigial roof} \ \ \ \ \ \ \ \ \ \ \ \ \ 63 \ \ \ \ \ \ \ \ \ \ \ \ \  70\ \ \ \ \ \ \ \ \ \ \ \ -7 \ \ \ \ \ \ \ \ \ \ \  49 \ \ \ \ \ \ \ \ \ \ \  0.7\\\\\text{wild type} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 72  \ \ \ \ \ \ \ \ \ \ \  70  \ \ \ \ \ \ \ \ \ \ \ 2  \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \  0.057143\\\\

vestigial \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 32 \ \ \  \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  0.257143\\\\

\text{Cinnabar roof} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 34\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  -1 \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \  1  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  0.028571\\\\\text{roof vestigial}  \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  0.2\\\\

\text{cinnabar}  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  3 \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \  5 \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \  -2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  0.8 \\\\

Total \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1000 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ 2.965934

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\to df = 8-1= 7

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The p-value of 0.75 is 4.5. Chi-square

Chi-sqaure value is observed at 2.965.

That means 0.90>p-value>0.75.

7 0
3 years ago
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son4ous [18]
The remainder (which is what Blake says is the remainder), can never be more than the number you are dividing by (the dividend), which in this problem is three.

Divide 136 by 3 and you will prove that answer is incorrect.
3 0
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