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diamong [38]
2 years ago
9

How to write 12430000 in expanded form

Mathematics
2 answers:
Marrrta [24]2 years ago
8 0
10,000,000+2,000,000+400,000+30,000
Georgia [21]2 years ago
5 0
10000000+2000000+400000+30000
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The lengths of trout in a lake are normally distributed with a mean of 30 inches and a standard deviation of 4.5 inches. Enter t
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The answer is -0.4....
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Which equation combines with the given equation to form a system of equations with the solution x = 3 and y = 9? x + 2y = 21
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Yes. x=3 and y =9 are the solutions.
 Because if you change x=3 and y =9 ,you will have 
3+2* 9 =21
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A dairy milker's gross pay per week is $546. If she works 30 hours per week, how much does she earn per hour?
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$18.20

Explanation:
$546/30hours=$18.20

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Brody had his computer repaired at Store A. His bill was:
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Answer:

$190.80 gratuity tip.

Step-by-step explanation:

Adding up the initial cost and tax, it adds up to $1,272. Subtracting 85% from the total is $190.80.

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2 years ago
A study is being conducted in which the health of two independent groups of ten policyholders is being monitored over a one-year
uysha [10]

Answer:

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

Step-by-step explanation:

We use two binomial trials 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.

Probability of at least nine participants finishing the study in a group.

0.2 probability of a students dropping out. So 1 - 0.2 = 0.8 probability of a student finishing the study. This means that p = 0.8.

10 students, so n = 10

We have to find:

P(X \geq 9) = P(X = 9) + P(X = 10)

Then

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

P(X = 9) = C_{10,9}.(0.8)^{9}.(0.2)^{1} = 0.2684

P(X = 10) = C_{10,10}.(0.8)^{10}.(0.2)^{0} = 0.1074

P(X \geq 9) = P(X = 9) + P(X = 10) = 0.2684 + 0.1074 = 0.3758

0.3758 probability that at least nine participants complete the study in a group.

Calculate the probability that at least nine participants complete the study in one of the two groups, but not in both groups?

0.3758 probability that at least nine participants complete the study in a group. This means that p = 0.3758

Two groups, so n = 2

We have to find P(X = 1).

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

P(X = 2) = C_{2,1}.(0.3758)^{1}.(0.6242)^{1} = 0.4691

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

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2 years ago
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