All categories

3 years ago

I just wanted to see if this was 8 or negative 8 :)

Mathematics

1 answer:

nirvana33 [79]3 years ago

4 0

Answer:

It's a positive 8.

Step-by-step explanation:

You can't have a negative amount of money.

You might be interested in

The length of a radius of a circle, measured in centimeters, is represented by the expression x + 1.5. The diameter of the circl

Tomtit [17]

The value of x of a circle with radius x + 1.5 and diameter of 9 2 / 5 cm is 7.9 cm

<h3>Radius of a circle</h3>

Radius of a circle is half of the diameter of a circle. The radius extend from the centre of the circle to the circumference.

Mathematically,

radius = 1 / 2 (diameter)

Therefore,

radius = (x + 1.5) cm

diameter = 9 2 / 5 cm = 47 / 5 cm

let's find x with the relationship above.

Therefore,

x + 1.5 = 47 / 5

subtract 1.5 from both sides

x + 1.5 - 1.5 = 9.4 - 1.5

x = 7.9 cm

learn more on radius here: brainly.com/question/12183018

6 0

2 years ago

At these numbers functions: (1, 2), (2, 3), (3, 4), (4, 3)

Leona [35]

Answer:

(Add one then subtract one)

Step-by-step explanation:

8 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

3 years ago

Ronald invested $68160 for 8 years and earned 6.2% interest compounded annually

erastova [34]

Answer:

He earned 110287.3536 dollars

6 0

3 years ago

3/4 m = 66<br> What is the answer?

omeli [17]

Answer:

m=88

Step-by-step explanation:

First we have to isolate the variable. In this case, it would be by multiplying the reciprocal of the fraction the both sides. M will be alone and the other side should come out to be 88. Assuming you know how to multiply with fractions, of course.

5 0

3 years ago

I just wanted to see if this was 8 or negative 8 :)​

I just wanted to see if this was 8 or negative 8 :)