What is <br><img src="https://tex.z-dn.net/?f=%20-%20%20%5Csqrt%7B36%7D%20" id="TexFormula1" title=" - \sqrt{36} " alt="

4x-3=5x-21 what does x=

torisob [31]

U subtract 4x by 5x and then you add 3 by -21 and the answer is x=18

3 0

3 years ago

May someone help me with this equation, please <br> 95=20x + 5

arsen [322]

Answer:

The answer is 9/2 or decimal form is 4.5 i gave you the work in the screenshot below hope it helps :)

Step-by-step explanation:

6 0

2 years ago

The table below shows the scores of a group of students on a 10-point quiz.Test ScoreFrequency3 54 15 16 27 28 49 310 1The mean

Katarina [22]

Explanation

Part A

We can use the formula below to find the mean.

$\text{Mean}=\sum ^{}_{}(\frac{fx}{f})$

We will then have;

$\begin{gathered} \text{Mean = }\frac{5\times3+4\times1+5\times1+6\times2+7\times2+8\times4+9\times3+10\times1}{19}=\frac{15+4+5+12+14+32+27+10}{19} \\ =\frac{119}{19}=6.2632 \end{gathered}$

Mean = 6.2632

Part B

Since the data set is odd.

$\begin{gathered} \text{Median}=\frac{(n+1)}{2}^{th}\text{observation} \\ =\text{Value of }\frac{\text{(19+1)}}{2}^{th}\text{observation} \\ =\text{value of 10th observation} \end{gathered}$

Answer: From the column of cumulative frequency cf, the median is 7

5 0

1 year ago

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

balu736 [363]

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

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

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