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2 years ago

729 divided 10 to the 3rd power

Mathematics

2 answers:

vladimir2022 [97]2 years ago

8 0

$\huge\text{Hey there!}$

$\large\boxed{\mathsf{729\div10^3}}\\\\\\\large\boxed{\mathsf{\rightarrow \dfrac{729}{10^3}}}\\\\\\\\\large\boxed{\rm{10^3}}\\\large\boxed{\rm{= 10\times10\times10}}\\\large\boxed{\rm{= 100 \times 10}}\\\large\boxed{\rm{= 1,000}}\\\\\\\\\large\boxed{=\mathsf{\dfrac{729}{1,000}}}\\\\\\\large\boxed{\mathsf{= 729\div 1,000}}}\\\\\\\large\boxed{= \rm{0.729}}\\\\\huge\boxed{\textsf{Therefore, your answer is: 0.729}}}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

grin007 [14]2 years ago

6 0

Answer:

0.729

Step-by-step explanation:

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Step-by-step explanation: Given that we toss a fair coin 10 times and X denote the number of heads.

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(b) the probability that X greater or equal than 5

(c) the minimum value of a such that P(X ≤ a) > 0.8.

We know that the probability of getting r heads out of n tosses in a toss of coin is given by the formula of binomial distribution as follows :

$P(X=r)=^nC_r\left(\dfrac{1}{2}\right)^r\left(\dfrac{1}{2}\right)^{n-r}.$

(a) The probability of getting 5 heads is given by

$P(X=5)\\\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}\\\\\\=\dfrac{10!}{5!(10-5)!}\dfrac{1}{2^{10}}\\\\\\=0.24609\\\\\sim0.25.$

(b) The probability of getting 5 or more than 5 heads is

$P(X\geq 5)\\\\=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}+^{10}C_6\left(\dfrac{1}{2}\right)^6\left(\dfrac{1}{2}\right)^{10-6}+^{10}C_7\left(\dfrac{1}{2}\right)^7\left(\dfrac{1}{2}\right)^{10-7}+^{10}C_8\left(\dfrac{1}{2}\right)^8\left(\dfrac{1}{2}\right)^{10-8}+^{10}C_9\left(\dfrac{1}{2}\right)^9\left(\dfrac{1}{2}\right)^{10-9}+^{10}C_{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{10-10}\\\\\\=0.24609+0.20507+0.11718+0.04394+0.0097+0.00097\\\\=0.62295\\\\\sim 0.62.$

(c) Proceeding as in parts (a) and (b), we see that

if a = 10, then

$P(X\leq 0)=0.00097,\\\\P(X\leq 1)=0.01067,\\\\P(X\leq 2)=0.05461,\\\\P(X\leq 3)=0.17179,\\\\P(X\leq 4)=0.37686,\\\\P(X\leq 5)=0.62295,\\\\P(X\leq 6)=0.82802.$

Therefore, the minimum value of a is 6.

Hence, all the questions are answered.

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