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3 years ago

The product of a number and 9 less than the number is 90. find the number.

1 answer:

5 0

Factors of 36 that differ by 9 are 12 and 3;
12*3 = 36 
Or you could write x(x - 9) - 36 = (x - 12)(x + 3) = 0 
The number was 12

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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

3 years ago

Write the given trinomial if possible as a square of a binomial or as an expression opposite to a square of a binomial: 15ab-9a^

MrMuchimi

Answer:

$- \bigg(3a - \frac{5}{2} b) \bigg)^{2}$

Step-by-step explanation:

$15ab-9a^2-6 \frac{1}{4} b^2 \\ \\ = 15ab-(3a)^2-\frac{6 \times 4 + 1}{4} b^2 \\ \\ = 15ab-(3a)^2-\frac{24+ 1}{4} b^2 \\ \\ = 15ab-(3a)^2-\frac{25}{4} b^2 \\ \\ = 15ab- (3a)^2- \bigg(\frac{5}{2} b \bigg)^2 \\ \\ = - \{ - 15ab + (3a)^2 + \bigg(\frac{5}{2} b \bigg)^2 \} \\ \\ = - \{ (3a)^2 + \bigg(\frac{5}{2} b \bigg)^2 - 15ab \} \\ \\ = - \bigg(3a - \frac{5}{2} b \bigg)^{2}$

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2 years ago