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2 years ago

10

Ms. Casey grades 4 tests during her lunch. She grades 1/3 of the remainder after school. If she still has 16 tests to grade afte

r school, how many tests are there?

1 answer:

CaHeK987 [17]2 years ago

6 0

Answer:30 tests

Step-by-step explanation:

16=2/3 -4

20=2/3

10=1/3

30tests

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The given values are discrete. Use the continuity correction and describe the region of the normal distribution that corresponds

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Step-by-step explanation:

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which right triangle’s unknown leg measures is 43 square root units ? Assume the given length are in the same unit of measure.

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A study is being conducted in which the health of two independent groups of ten policyholders is being monitored over a one-year

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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).

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

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

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