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2 years ago

88511, 16351, ?, 10251 missing number in sequence

Mathematics

1 answer:

Mrac [35]2 years ago

8 0

Step-by-step explanation:

this is really complicated and tricky.

to create the next number, you have to add the first 2 digits of the previous number, then subtract the 3rd from the 2nd digit, multiply the 3rd and the 4th digit, and finally divide the 4th digit by the 5th.

so,

8+8 = 16

8-5 = 3

5×1 = 5

1/1 = 1

giving 16351 as next number.

the missing number we get then

1+6 = 7

6-3 = 3

3×5 = 15

5/1 = 5

giving 73155 as the next (missing) number.

control :

7+3 = 10

3-1 = 2

1×5 = 5

5/5 = 1

gives us 10251 as next number - correct.

but then it will end

1+0 = 1

0-2 = -2 ???? take the absolute value ?

2×5 = 10

5/1 = 5

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In a survey, three out of four students said that courts show too much concern for criminals. Of seventy randomly selected stude

vladimir1956 [14]

Answer:

B. Yes, because P(X ≥ 68) ≤ .05

Step-by-step explanation:

The probability of a score X occuring is said to be unusually high if the probability of a score of X or larger is lower than 5%.

For each student, there are only two possible outcomes. Either they think that the court shows too much concern for criminals, or they do not think this. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

Three out of four students said that courts show too much concern for criminals. This means that $p = \frac{3}{4} = 0.75$

Would 68 be considered an unusually high number of students who feel that courts are partial to criminals?

We have to find $P(X \geq 68)$ .

$P(X \geq 68) = P(X = 68) + P(X = 69) + P(X = 70)$

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

$P(X = 68) = C_{70,68}.(0.75)^{68}.(0.25)^{2} < 0.000001$

$P(X = 69) = C_{70,69}.(0.75)^{69}.(0.25)^{1} < 0.000001$

$P(X = 70) = C_{70,70}.(0.75)^{70}.(0.25)^{0} < 0.000001$

$P(X \geq 68) = P(X = 68) + P(X = 69) + P(X = 70) < 0.000003$

$P(X \geq 68)$ is lower than 0.05, so this is an unusually high number.

So the correct answer is:

B. Yes, because P(X ≥ 68) ≤ .05

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You use new over old
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Answer:

Answer

4.0/5

Americ

Ambitious

1.5K answers

627.6K people helped

x + 4/9= 3/4

subtract 4/9 from both sides

x= 3/4 - 4/9

change to common denominators

x= 27/36 - 16/36

subtract numerators only

x= 11/36

ANSWER: (A) 11/36

Hopes this helps :)

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