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4 years ago

Name the value of each 2 in 222,222?

1 answer:

3 0

The first is one hundred thousand
the second is ten thousand
the third is one thousand
the fourth is one hundred
the fifth is ten
the sixth is one

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How do divided 220 into 3

Zina [86]

I put 220 divided 3 and The answer is 7.3

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3 years ago

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You’ve bought a half-dozen (six) eggs from the store but you forgot to check them first! The probability that no eggs are broken

GREYUIT [131]

Answer:

$P(X = 0) = C_{6,0}.(0.1416)^{0}.(0.8584)^{6} = 0.4$

$P(X = 1) = C_{6,1}.(0.1416)^{1}.(8584)^{5} = 0.3960$

$P(X = 2) = C_{6,2}.(0.1416)^{2}.(8584)^{4} = 0.1633$

$P(X = 3) = C_{6,3}.(0.1416)^{3}.(8584)^{3} = 0.0359$

$P(X = 4) = C_{6,4}.(0.1416)^{4}.(8584)^{2} = 0.0044$

$P(X = 5) = C_{6,5}.(0.1416)^{5}.(8584)^{1} = 0.0003$

$P(X = 6) = C_{6,6}.(0.1416)^{6}.(8584)^{0} = 0.00001$

b) 56.77% probability that an even number of eggs is broken.

Expectation: 0.8496

Variance: 0.7293

Step-by-step explanation:

For each egg, there are only two possible outcomes. Either it is broken, or it is not. The probability of an egg being broken is independent from other eggs. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The variance of the binomial distribution is:

$V(X) = np(1-p)$

There are 6 eggs

So $n = 6$

The probability that no eggs are broken is 0.4.

This means that $P(X = 0) = 0.4$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.p^{0}.(1-p)^{6}$

$(1 - p)^{6} = 0.4$

Taking the sixth root from both sides of the equality

$(1 - p) = 0.8584$

p = 0.1416

Each egg has a 0.1416 probability of being broken

(a) Write out the pmf of X.

Probability of each value, from 0 to 6

$P(X = 0) = C_{6,0}.(0.1416)^{0}.(0.8584)^{6} = 0.4$

$P(X = 1) = C_{6,1}.(0.1416)^{1}.(8584)^{5} = 0.3960$

$P(X = 2) = C_{6,2}.(0.1416)^{2}.(8584)^{4} = 0.1633$

$P(X = 3) = C_{6,3}.(0.1416)^{3}.(8584)^{3} = 0.0359$

$P(X = 4) = C_{6,4}.(0.1416)^{4}.(8584)^{2} = 0.0044$

$P(X = 5) = C_{6,5}.(0.1416)^{5}.(8584)^{1} = 0.0003$

$P(X = 6) = C_{6,6}.(0.1416)^{6}.(8584)^{0} = 0.00001$

(b) Compute the probability that an even number of eggs is broken.

0, 2, 4 or 6

$P = P(X = 0) + P(X = 2) + P(X = 4) + P(X = 6) = 0.4 + 0.1633 + 0.0044 + 0.00001 = 0.5677$

56.77% probability that an even number of eggs is broken.

(c) Compute the expectation and variance of X.

Expectation:

$E(X) = np = 6*0.1416 = 0.8496$

Variance:

$V(X) = np(1-p) = 6*0.1416*0.8584 = 0.7293$

7 0

4 years ago

I need this answer quickly

iragen [17]

X = 23 is the answer

7 0

3 years ago

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Which decimal is equivalent to \dfrac{4}{3} 3 4 start fraction, 4, divided by, 3, end fraction? Choose 1 answer: Choose 1 answ

Otrada [13]

Answer:

The correct option is option D.

Step-by-step explanation:

Any fraction number $\frac mn$ (m>n ) can be written as $a\frac bn$

where a = quotient

b= reminder.

Given number is $\frac{4}{3}$ .

3)4( 1 ← quotient

- 3

____

1 ← reminder

Here quotient= 1, reminder=1 and n=3

$\therefore \frac43=1\frac13$ .

7 0

3 years ago

Arbitron Media Research Inc. conducted a study of the iPod listening habits of men and women. One facet of the study involved th

trasher [3.6K]

Answer:

The null and alternative hypotheses are:

$H_{0}:\sigma_{men} = \sigma_{women}$

$H_{a}:\sigma_{men} \neq \sigma_{women}$

Under the null hypothesis, the test statistic is:

$F=\frac{\sigma^{2}_{men}}{\sigma^{2}_{women}}$

$=\frac{19^{2}}{9^{2}}$

$=\frac{361}{81}$

$=4.46$

Therefore, the test-statistic is $F=4.46$

Now the F critical value at 0.02 significance level for df1 = 10- 1 =9 and df2 = 16 - 1 =15 is:

$F_{critical} = 3.303$

Since F statistic is greater than the F critical value, we therefore, reject the null hypothesis and conclude that there is sufficient evidence to support the claim that there is a difference in the variation in the listening times for men and women.

7 0

3 years ago