What is the word form of 0.604

Mathematics

1 answer:

Law Incorporation [45]2 years ago

6 0

Answer:

zero point six hundred and four

Step-by-step explanation:

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Hey can you please help me posted picture of question

JulijaS [17]

Answer: 32

Explanation:

The number of leaves in a tree diagram for probabilities is equal to the number of possible outcomes.

Tossing a coin 5 times leads to these results:

First,second, third and fourth tossing are the nodes; and fifth tossing has the leaves.

Root → 2 nodes → 2² nodes → 2³ nodes → 2⁴ → 2⁵ leaves = 32

Those 32 leaves is this sample space:

head-head-head-head-head
head-head-head-head-tail

head-head-head-tail-head
head-head-head-tail-tail

head-head-tail-head-head
head-head-tail-head-tail

head-head-tail-tail-head
head-head-tail-tail-tail

head-tail-head-head-head
head-tail-head-head-tail

head-tail-head-tail-head
head-tail-head-tail-tail

.......... (continue)

And so on until you have included all the possibilities, ending with tail-tail-tail-tail-tail. Those will be 32 outcomes.

7 0

2 years ago

Suppose you toss a fair coin 10 times, let X denote the number of heads. (a) What is the probability that X=5? (b) What is the p

zubka84 [21]

Answer: The required answers are

(a) 0.25, (b) 0.62, (c) 6.

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

We are to find

(a) the probability that X=5

(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

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