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Vaselesa [24]
2 years ago
5

What is the range of the function f(x) = 3x - 5, for the domain { -1, 2, 4}

Mathematics
1 answer:
morpeh [17]2 years ago
7 0
Hope this helps!! Solved it step by step:)

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One month Abdul rented 3 movies and 5 video games for a total of . The next month he rented 6 movies and 2 video games for a tot
MrMuchimi

Answer:

Video Games = $5.25

Movies = $1.25

Step-by-step explanation:

(3x + 5y = 30)2

6x + 2y = 18

                                           Video Games (y) = 5.25

  6x + 10y = 60

<u>-  6x + 2y = 18   </u>

      (8y)/8 = (42)/8

<u>                                                                                </u>

6x + 2(5.25) = 18

6x + 10.5 = 18

6x = 7.5

x = 1.25

Double Check by plugging in the numbers

7 0
3 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.

6 0
2 years ago
Can someone help me please i don’t understand it i really need too finish it by tonight
guajiro [1.7K]

Don't mind me, I just need points haha

6 0
3 years ago
In order to estimate the average time spent on the computer terminals per student at a local university, data were collected fro
VladimirAG [237]

Answer:

Margin of error  for a 95% of confidence intervals is 0.261

Step-by-step explanation:

<u>Step1:-</u>

 Sample n = 81 business students over a one-week period.

 Given the population standard deviation is 1.2 hours

 Confidence level of significance = 0.95

 Zₐ = 1.96

Margin of error (M.E) = \frac{Z_{\alpha  }S.D }{\sqrt{n} }

Given n=81 , σ =1.2 and  Zₐ = 1.96

<u>Step2:-</u>

<u />Margin of error (M.E) = \frac{Z_{\alpha  }S.D }{\sqrt{n} }<u />

<u />Margin of error (M.E) = \frac{1.96(1.2) }{\sqrt{81} }<u />

On calculating , we get

Margin of error = 0.261

<u>Conclusion:-</u>

Margin of error  for a 95% of confidence intervals is 0.261

<u />

4 0
3 years ago
Simplify the expression.
snow_tiger [21]
The answer to the question is d.
8 0
3 years ago
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