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3 years ago

What place value is the 5 in the following number, 123.465? earn 100 points!!!

Mathematics

2 answers:

GenaCL600 [577]3 years ago

6 0

Answer:

The place value of 5 in "123.465" is the thousandths

Step-by-step explanation:

The reason for this is because the "5" is 3 points behind the decimal, making it in the thousanths place.

Papessa [141]3 years ago

5 0

Answer:

it will be in the thousands place for sure

Step-by-step explanation:

hope that helps =) i forgot there were a decimal point

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-(5a+6)=2(3a+8) what is the answer

ValentinkaMS [17]

Answer:

a = -2

Step-by-step explanation:

-(5a+6)=2(3a+8)

-5a -6 = 2*3a +2*8

-5a -6 = 6a +16

-5a -6a = 16+6

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a= -22/11

a = -2

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3 years ago

It is an unfair coin with unknown probability of head and tail. Can you generate even odds using this coin?

Vladimir79 [104]

Answer:

Yes, it can be generated if $p = \frac{n}{n+1}$ , where n is a positive even integer.

Step-by-step explanation:

An odds for an event, is the probability or chance that this event will occur, divided by the probability that it will not occur.

Assume that the probability of getting tail when throwing the coin is $p$ , therefore, the probability of getting a head is $(1-p)$ and the odds for the event of getting tail is $\frac{p}{1-p}$ .

If $k$ is a positive integer, and we assume that the odds of getting a tail are even, then:

$\frac{p}{1-p} = 2k$

$p = 2k - 2kp$

$p = \frac{2k}{2k+1}$

Conclusion: if $p = \frac{2k}{2k+1}$ , for some positive integer $k$ , even odds are generated. For example: $p = 2/3, 4/5, 6/7, ..$ .

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3 years ago

41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

lys-0071 [83]

Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used 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.

41% of U.S. adults have very little confidence in newspapers.

This means that $p = 0.41$

You randomly select 10 U.S. adults.

This means that $n = 10$

(a) exactly five

This is $P(X = 5)$ . So

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

$P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087$

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

(b) at least six

This is:

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

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

$P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209$

$P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480$

$P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125$

$P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019$

$P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001$

Then

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834$

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

(c) less than four.

This is:

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

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

$P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051$

$P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355$

$P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111$

$P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575$

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

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3 years ago

Find the measure of 7

Scorpion4ik [409]

Answer:

96° because it is congruent to angle 7

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3 years ago