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3 years ago

I need help with this problem

Mathematics

1 answer:

maxonik [38]3 years ago

6 0

Answer:

Sorry i really need points you can report me but i just want you to knw that I REALLY need points

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F(x) = x2+3x+2 x2+5x+4 Why is there no zero at x = –1?

dmitriy555 [2]

You could rewrite $F(x)$ as

$\dfrac{x^2+3x+2}{x^2+5x+4}=\dfrac{(x+1)(x+2)}{(x+1)(x+4)}$

and be tempted to cancel out the factors of $x+1$ . But this cancellation is only valid when $x\neq-1$ .

When $x=-1$ , you end up with the indeterminate form $\dfrac00$ , which is why $-1$ is not a zero.

7 0

3 years ago

Help ASAP please please 14 and 15

Yuliya22 [10]

I think 14 is B (idrk) and 15 is definitely C

8 0

3 years ago

Can someone offer help to this ratio question

AlekseyPX

Answer:

Cherry Pink has not been made ( $25:17\neq 4:3$ )

Step-by-step explanation:

Given that she already has 60litres of Rose Pink paint, we use the ratios to determine the volumes of red/white used:

$Rose Pink=60l\\\\Red=\frac{1}{3}\times 60\\\\=20l\\\\White=\frac{2}{3}\times 60\\\\=40l$

#Add these proportions to the additional red and white paints to determine if Cherry Pink has been made:

$V_r=20l+80l\\\\=100l\\\\V_w=40l+28l\\\\=68l\\\\\therefore r:w=100l:68l \ \ \ \ \ \ \ \ \ \ \# simplify\\\\r:w=25:17$

We equate this proportion to the Cherry Pink proportion to verify if Cherry pink has been made:

$25:17\neq 4:3$

Hence, Cherry pink has not been made.

3 0

3 years 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:

$P(X \ge 1) = 0.509$

$P(X \ge 1) = 0.6807$

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

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

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

Here n = 6

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

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

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

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

$P(X \ge 1) = 0.6807$

5 0

3 years ago

7.92 8.055 7.9 7.98 8.072 8.027 Least to greatest

ioda

Answer:

First, we compare the given numbers using a place value chart. Since I don't have one of those, we won't be using a guide.

7.92

8.055

7.9

7.98

8.072

8.027

Comparing those, this order is the correct arrangement from least to greatest:

7.9

7.92

7.98

8.027

8.055

8.072

~Hope this helps~

8 0

3 years ago

I need help with this problem​

I need help with this problem