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igor_vitrenko [27]
3 years ago
6

Use parentheses to make this statement true: 8 times 4 - 2 times 3 + 8 divided by 2

Mathematics
1 answer:
PIT_PIT [208]3 years ago
7 0
8x4(-2x3)+8/2

I hope this is right!
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Solve-1/8<34 then graph it
Vlada [557]

Answer:

This isn’t really a valid equation, so I’m guessing you forgot to type something.

Step-by-step explanation:

6 0
3 years ago
Find the area of the figure.<br><br> answer choices:<br> 31 ft², 80 ft², 40 ft², 60 ft²
IrinaVladis [17]

Answer:

60 ft²

Step-by-step explanation:

You can divide the figure into 2 parts.

First, solve for the rectangle: Area of a rectangle = b*h = 4ft x 10ft = 40ft²

Then, solve for the triangle:

Height = 10ft - 5ft = 5ft

Base = 12ft - 4ft = 8ft

Area of a triangle = (b*h)/2 = (5ft x 8ft)/2 = 20ft²

Just add them together: 20ft² + 40ft² = 60ft²

:)

3 0
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.

6 0
3 years ago
What is 34567890p x 23er56y
Zielflug [23.3K]
 their is no answer
to that problem
6 0
3 years ago
A new pencil is 19 centimeters long. Levi uses it for a month. The pencil now measures 8 centimeters.
pogonyaev

Answer:

11 cm

Step-by-step explanation:

Given:

Length of new pencil = 19 cm

Length of pencil after using a month = 8 cm

Question asked:

The pencil is centimeters shorter now than when it was new = ?

Solution:

Length of new pencil = 19 cm

Length of pencil after using a month =  8 cm

The pencil is centimeters shorter now than when it was new = 19 cm -  8 cm

                                                                                                     = 11 cm

Length of pencil has been used during a month = 11 cm

8 0
3 years ago
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