Please answer ASAP. The question is down below

Mathematics

1 answer:

belka [17]3 years ago

4 0

Answer: A

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

Notes: Dividing by a fraction means to multiply by its reciprocal.

The denominator cannot equal zero.

$\dfrac{5a^3bc}{8ab^3}\div\dfrac{-ab^2}{6a^5b}\cdot \dfrac{2a^2b^3}{3b}\qquad \rightarrow a\neq 0,b\neq 0\\\\\\=\dfrac{5a^3bc}{8ab^3}\cdot\dfrac{6a^5b}{-ab^2}\cdot \dfrac{2a^2b^3}{3b}\\\\\\=\dfrac{5\cdot 6\cdot 2\quad a^3\cdot a^5\cdot a^2\quad b\cdot b\cdot b^3\quad c}{8\cdot -1 \cdot 3\quad a\cdot a\qquad b^3\cdot b^2\cdot b \quad}\\\\\\=\dfrac{-60a^{10}b^5c}{-24a^2b^6}\\\\\\=\dfrac{-5a^8c}{2b}$

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The chance that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses is P=0.0488.

Step-by-step explanation:

To solve this problem we divide the tossing in two: the first 5 tosses and the last 5 tosses.

Both heads and tails have an individual probability p=0.5.

Then, both group of five tosses have the same binomial distribution: n=5, p=0.5.

The probability that k heads are in the sample is:

$P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{5}{k}\cdot0.5^k\cdot0.5^{5-k}$

Then, the probability that exactly 2 heads are among the first five tosses can be calculated as:

$P(x=2)=\dbinom{5}{2}\cdot0.5^{2}\cdot0.5^{3}=10\cdot0.25\cdot0.125=0.3125\\\\\\$

For the last five tosses, the probability that are exactly 4 heads is:

$P(x=4)=\dbinom{5}{4}\cdot0.5^{4}\cdot0.5^{1}=5\cdot0.0625\cdot0.5=0.1563\\\\\\$

Then, the probability that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses can be calculated multypling the probabilities of these two independent events:

$P(H_1=2;H_2=4)=P(H_1=2)\cdot P(H_2=4)=0.3125\cdot0.1563=0.0488$