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2 years ago

Write the repeating decimal 0.63333 as afraction

Mathematics

1 answer:

sergeinik [125]2 years ago

5 0

Step-by-step explanation:

just count off the positions after the decimal point.

they go with decreasing negative exponents of ten.

the first position is 1/10.

the second 1/100.

the third 1/1000.

the fourth 1/10000

the fifth 1/100000

and that is how we can easily convert such decimals into fractions.

0.63333 = 63333 / 100000

or were you looking for the infinite decimal ?

that would be 1.9/3 = 19/10 / 3 = 19/30

I found this by simply suspecting that there is a division by 3 involved (due to the 3333333 repeating pattern).

so, by multiplying 0.63333 by 3 we get 1.9.

1.9 = 19/10

and then dividing by 3 gives us 19/30

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53% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

faltersainse [42]

Answer:

$P(X = 5) = 0.2423$

$P( X \ge 6 )= 0.4516$

$P( X < 4 )= 0.12694$

Step-by-step explanation:

From the question we are told that

The population proportion is p = 0.53

The sample size is n = 10

Generally the distribution of the confidence of US adults in newspapers follows a binomial distribution

i.e

$X \~ \ \ \ B(n , p)$

and the probability distribution function for binomial distribution is

$P(X = x) = ^{n}C_x * p^x * (1- p)^{n-x}$

Here C stands for combination hence we are going to be making use of the combination function in our calculators

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is exactly five is mathematically represented as

$P(X = 5) = ^{10}C_5 * 0.53^5 * (1- 0.53)^{10-5}$

=> $P(X = 5) = 252 * 0.04182 * 0.023$

=> $P(X = 5) = 0.2423$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is at least six is mathematically represented as

$P( X \ge 6 )= P( X = 6) + P( X = 7) + P( X = 8) + P( X = 9) + P( X = 10)$

=> $P( X \ge 6 )= [^{10}C_6 * [0.53]^6 * (1- 0.53)^{10-6}] + [^{10}C_7 * [0.53]^7 * (1- 0.53)^{10-7}] + [^{10}C_8 * [0.53]^8 * (1- 0.53)^{10-8}] + [^{10}C_9 * [0.53]^9 * (1- 0.53)^{10-9}] + [^{10}C_{10} * [0.53]^{10} * (1- 0.53)^{10-10}]$

=> $P( X \ge 6 )= [0.227] + [0.1464] + [0.0619] + [0.0155] + [0.00082]$

=> $P( X \ge 6 )= 0.4516$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is less than four is mathematically represented as

$P( X < 4 )= P( X = 3) + P( X = 2) + P( X = 1) + P( X = 0)$

=> $P( X < 4 )= [^{10}C_3 * 0.53^3 * (1- 0.53)^{10-3}] + [^{10}C_2 * 0.53^2 * (1- 0.53)^{10-2}] + [^{10}C_1 * 0.53^1 * (1- 0.53)^{10-1}] + [^{10}C_0 * 0.53^0 * (1- 0.53)^{10-0}]$