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Anastaziya [24]

2 years ago

12

1. Keira earns $11.50 for each hour she babysits. The amount of depends on how many 1.50x ig Dependent variable: Independent var

iable:

1 answer:

dolphi86 [110]2 years ago

6 0

Dependent variable is the amount she earns
Independent variable is the amount of hours she works

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Identify a number that results in a 4-digit-number when 156,350 is subtracted from it

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160,000 - 156,350 =3,650

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Mary worked a total of 6 1/2 hours during 5 days of work per week. What was her average workday?

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Express 10500 in terms of its prime factors

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Step-by-step explanation:

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A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o

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Answer:

a

$P(X \ge 1) = 0.509$

b

$P(X \ge 1) = 0.6807$

Step-by-step explanation:

From the question we are told that

The number of students in the class is N = 20 (This is the population )

The number of student that will cheat is k = 3

The number of students that he is focused on is n = 4

Generally the probability distribution that defines this question is the Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally probability mass function is mathematically represented as

$P(X = x) = \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}$

Here C stands for combination , hence we will be making use of the combination functionality in our calculators

Generally the that he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

Here

$P(X \le 0) = \frac{ ^{3} C_0 * ^{20 - 3} C_{4- 0}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ ^{3} C_0 * ^{17} C_{4}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ 1 * 2380}{ 4845}$

$P(X \le 0) = 0.491$

Hence

$P(X \ge 1) = 1 - 0.491$

$P(X \ge 1) = 0.509$

Generally the that he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

$P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]$

Here n = 6

So

$P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{20 -3}C_{6-0}}{^{20}C_6}]$

$P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]$

$P(X \ge 1) =1- [ \frac{1 * 12376}{38760}]$

$P(X \ge 1) =1- 0.3193$

$P(X \ge 1) = 0.6807$

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Thirty two thousand six hundred and sixty seven in numbers

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