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Romashka-Z-Leto [24]
4 years ago
14

27.2 is 68% of what number

Mathematics
2 answers:
FinnZ [79.3K]4 years ago
8 0
68% = 27.2
Divide by 68 to find 1%
1% = 0.4
Multiply by 100 t find 100%
0.4 x 100 = 40
stealth61 [152]4 years ago
6 0
27.2 is 68 percent of 40
to find this answer I divided 27.2 by 68 to find out how much 1 percent was; 1%=0.4then i just multiplied 0.4 by 100 to find out how much 100% was, and my answer was 40

hope this helps
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Help!!!!!!!!!!!!!!<br><br> 2x+3=x-7
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Answer:

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Step-by-step explanation:

2x+3=x-7

=> 2x - x = - 7 - 3

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3 years ago
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4 years ago
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KATRIN_1 [288]
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8 0
3 years ago
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7 0
3 years ago
) Determine the probability that a bit string of length 10 contains exactly 4 or 5 ones.
yanalaym [24]

Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the  0 and 1 area equally likely to any place.

i.e. P(0)= P(1)= \dfrac{1}{2}

The length of bits : n = 10

Let X = Number of getting ones.

Then , X \sim Bin(n=10,\ p=\dfrac{1}{2})

Binomial distribution formula : P(X=x)=^nC_x p^x q^{n-x} , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , p=q=\dfrac{1}{2}

Then ,The probability that a bit string of length 10 contains exactly 4 or 5 ones.

P(X= 4\ or\ 5)=P(x=4)+P(x=5)\\\\=^{10}C_4(\dfrac{1}{2})^{10}+^{10}C_4(\dfrac{1}{2})^{10}

=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}

=(\dfrac{1}{2})^{10}(\dfrac{10!}{4!6!}+\dfrac{10!}{5!5!})

=(\dfrac{1}{2})^{10}(210+252)

=(0.0009765625)(462)

=0.451171875\approx0.4512

Hence, the  probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.

3 0
4 years ago
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