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4 years ago

Section B

Mathematics

1 answer:

MA_775_DIABLO [31]4 years ago

5 0

Answer/Step-by-step explanation:

Using α as reference angle:

Perpendicular side = BC (one of the legs that adjoins to form a right angle. It is opposite to the reference angle.)

Base side = AB (The other leg that forms a right angle with the perpendicular side at the adjoining point. It is adjacent to the reference angle)

Hypotenuse = AC (the side opposite to the right angle, 90°)

Using β as reference angle:

Perpendicular side = AB (one of the legs that adjoins to form a right angle. It is opposite to the reference angle.)

Base side = BD (The other leg that forms a right angle with the perpendicular side at the adjoining point. It is adjacent to the reference angle)

Hypotenuse = AD (the side opposite to the right angle, 90°)

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6. Circle all the expressions that are equivalent to this expression: 6p+ 7.

-BARSIC- [3]

Number 4 and 6 are correct because you can only add variables with other variables so that is why 4 and 6 are correct

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4 years ago

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What is f(4) for the function f(x) = 7x + 6? Please help asap

Black_prince [1.1K]

Answer:

f(4) for the function $f(x) = 7x + 6$ is $\mathbf{f(4)=34}$

Step-by-step explanation:

We need to find f(4) for the function $f(x) = 7x + 6$

For finding f(4) we will put x= 4 in the given function

$f(x) = 7x + 6$

Put x = 4

$f(4)=7(4)+6\\f(4)=28+6\\f(4)=34$

SO, f(4) for the function $f(x) = 7x + 6$ is $\mathbf{f(4)=34}$

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3 years ago

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What is the question

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3 years ago

A total of 200 people attended a carnival. Children tickets to the carnival cost $2.00. Adult tickets to the carnival cost $4.00

Orlov [11]

Answer:

$15.02

Step-by-step explanation:

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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%.

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4 years ago

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