The question is the photo

When p2 – 4p is subtracted from p2 + p – 6, the result is

Galina-37 [17]

$\Large\maltese\underline{\textsf{Our problem:}}$

When $\bf{p^2-4p}$ is subtracted from $\bf{p^2+p-6}$ , the result is...?

$\Large\maltese\underline{\textsf{This problem has been solved!}}$

$\bf{Subtract:}$

$\bf{p^2+p-6-p^2-4p}$ . | combine like terms

$\bf{p^2-p^2+p-4p-6}$ | simplify

$\bf{0p^2-3p-6}$ | simplify

$\bf{-3p-6}$

$\rule{300}{1.7}$

$\bf{Result:}$

$\bf{-3p-6}$

$\boxed{\bf{aesthetic \not101}}$

5 0

2 years ago

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In a chemical plant, 24 holding tanks are used for final product storage. Suppose that six of the tanks contain high viscosity m

andriy [413]

Answer:

The probability is 0.461

Step-by-step explanation:

Since six tanks contain high viscosity material, it means that 18 does not contain since the total number of tanks is 24.

Now, here, we know that we are selecting a total of 4 tanks.

out of the tanks with high viscosity materials, we will be selecting one only while out of the others without viscosity materials we shall need selecting three.

Thus,

the number of ways we can select one tank with viscosity material from a total of six tanks will be 6C1 = 6 ways

Now, we want to select three from the remaining 18 tanks

The number of ways we can do this is 18C3 = 816 ways

Finally, we are selecting 4 out of 24 tanks.

The number of ways we can do this is 24C4 = 10,626 ways

Now, we proceed to calculate the probability.

Kindly note that we are selecting 1 from 4 and 3 from 18

The term ‘and’ in probability means we are multiplying the probabilities of these events together and thus, that means that we are going to combine our numbers of selections.

So the total number of ways in which we can select 1 out of 3 and 3 out of 18 will be = 6 * 816 = 4,896 ways

The probability would now be 4896/10626 = 0.461

5 0

4 years ago

3. In a game, you lose a turn six if you roll a six on a 6-sided number cube. (a) What is the probability that you roll a 6? Exp

Alex73 [517]

Answer: A) 1/6

B)5/36

C)5/6

Step-by-step explanation:

A) is 1/6 because there is only one number that you have to roll out of 6 numbers.

B) is 5/36 because the probability of not rolling a 6 is 5/6. The probability of rolling a 6 is 1/6. You multiply those and get 5/36.

C) is 5/6 because you don't roll a 6 is only one number you don't roll. There are 5 other numbers that you are trying to roll.( AKA: 1,2,3,4, and 5)

4 0

2 years ago

I need help on number 1

lesya [120]

Answer:

y = -9

Step-by-step explanation:

8 0

3 years ago

A veterinarian’s assistant made a table that shows the animals seen in the office in one week. What is the probability that the

horsena [70]

Answer:

The probability that the pet seen was sick is 53.57%.

Step-by-step explanation:

The probability of an event E is the ration of the number of favorable outcomes to the total number of outcomes.

$P(E)=\frac{n(E)}{N}$

Here,

n (E) = number of favorable outcomes

N = total number of outcomes

Denote the events as follows:

C = the pet is a cat

D = the pet is a dog

O = the pet is some other animal

S = the pet is sick.

The data provided is summarized as follows:

n (C) = 28

n (C ∩ S) = 3

n (D) = 42

n (B ∩ S) = 4

n (O) = 24

n (O ∩ S) = 8

Compute the probability that a cat was sick as follows:

$P(C\cap S)=\frac{n(C\cap S)}{n(C)}=\frac{3}{28}$

Compute the probability that a dog was sick as follows:

$P(D\cap S)=\frac{n(D\cap S)}{n(D)}=\frac{4}{42}=\frac{2}{21}$

Compute the probability that another animal was sick as follows:

$P(O\cap S)=\frac{n(O\cap S)}{n(O)}=\frac{8}{24}=\frac{1}{3}$

Compute the probability that the pet seen was sick as follows:

P (S) = P (C ∩ S) + P (D ∩ S) + P (O ∩ S)

$=\frac{3}{28}+\frac{2}{21}+\frac{1}{3}\\=\frac{9+8+28}{84}\\=\frac{45}{84}\\=0.5357$

Thus, the probability that the pet seen was sick is 53.57%.

8 0

4 years ago

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