Which is True?????/<br> Value of x???

Solve by factorization<br> 2x^2 + 3x -20 = 0

saul85 [17]

Answer:

x=2.5 or x=-4

Step-by-step explanation:

factorisation= factors:1, 2, 4, 5, 10, 20 (only the positive factors)

We will use the factors 4 and 5.

=(2x-5)(x+4)=0

2x-5=0 || x+4=0

2x=+5 || x=-4

x=2.5 or x=-4

4 0

2 years ago

Read 2 more answers

Round 4382 to 1 significant figure

Nataly [62]

I think is 4000 because when it's sad 1 singnificant figure you have to round the first name which 4

8 0

2 years ago

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End Behaviors

Harlamova29_29 [7]

<h3>#End behaviour:-</h3>

#1

Left:-Rise
Right:-Fall

#2

Left:-Rise
Right:-Fall

<h3>#Degree:-</h3>

Find nodes

#1

2
Even degree function

#2

It's a parabola so it's the graph of a quadratic equation.

Even degree function

<h3>Real zeros</h3>

#1

3

#2

2

3 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

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

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