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aleksandrvk [35]
3 years ago
15

Can someone please help me with this ?!!

Mathematics
1 answer:
bija089 [108]3 years ago
8 0

Answer:

I believe the answer is B

Step-by-step explanation:

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A. G(x) = 7x^2

D. No, because one x-value corresponds to two different y-values.

Step-by-step explanation:

If you were to vertically stretch the quadratic parent function, the correct answer would be "A," because you are stretching upwards, which in turn, also changes the y-values.

As you can see in the table, the x-value of 3 repeats, which cannot occur. Each input (x-value) can go to only one output (y-value). As a result, the table does not represent a function.

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Can an octagon have interior angles that measure 100 degrees, 156 degrees, 125 degrees, 90 degrees, 175 degrees, 160 degrees, an
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Step-by-step explanation:

The 7 angles listed have a total of 946 degrees. The total interior angle measure in an octagon is 1080 degrees. So, the remaining angle must have a measure of 134°. The figure is entirely feasible.

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2 years ago
All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be
Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

A = a + (A \cap B)

In which a is the probability that a student is proficient in reading but not mathematics and A \cap B is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

B = b + (A \cap B)

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

(A \cup B) + C = 1

In which

(A \cup B) = a + b + (A \cap B)

65% were found to be proficient in both reading and mathematics.

This means that A \cap B = 0.65

78% were found to be proficient in mathematics

This means that B = 0.78

B = b + (A \cap B)

0.78 = b + 0.65

b = 0.13

85% of the students were found to be proficient in reading

This means that A = 0.85

A = a + (A \cap B)

0.85 = a + 0.65

a = 0.20

Proficient in at least one:

(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98

What is the probability that the student is proficient in neither reading nor mathematics?

(A \cup B) + C = 1

C = 1 - (A \cup B) = 1 - 0.98 = 0.02

There is a 2% probability that the student is proficient in neither reading nor mathematics.

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