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3 years ago

Can you please solve the following equation?

Mathematics

1 answer:

AlladinOne [14]3 years ago

7 0

Answer:

x=3.903

Step-by-step explanation:

$2^{x-3}.5^{x-1}=200\\2^x.2^{-3}.5^x.5^{-1}=200\\\frac{2^x}{2^3}.\frac{5^x}{5^1}=200\\\frac{2^x.5^x}{2^3.5}=200\\\frac{(2.5)^x}{8.5}=200\\10^x=200 \times 40\\10^x=8000\\log 10^x=log8000\\x=\frac {log8000}{log10}=log8000=log 8\times log1000=log 8+log 1000=log2^3+log10^3=3log2+3log10=3log2+3 \approx0.903+3 \approx3.903$

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Answer:

$P(X \ge 1) = 0.509$

$P(X \ge 1) = 0.6807$

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

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

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