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3 years ago

There are 83 apples. Each bag can hold 4 apples. How many full bags are there? How many are left over?

2 answers:

8 0

First, You have to divide 4 from 83,
Result would be: Quotient = 20, Remainder = 3

So, There are 20 full bags [ equal to quotient ]
There are 3 leftovers [ equal to remainder ]

Hope this helps!

klio [65]3 years ago

3 0

20 full bags with 3 left over.

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Which postulate or theorem justifies this statement: ∠ 11 + ∠ 10 = 180°

Trava [24]

Answer:

Same-side Interior Angles theorem justifies ∠ 11 + ∠ 10 =180°

Step-by-step explanation:

To Prove:

∠ 11 + ∠ 10 =180°

Proof:

Consider lines are Parallel,then

Corresponding Angles are Equal

∴ ∠ 10 = ∠ 12 say (equation 1)

Now by Linear Pair postulate we have,

∴ ∠ 11 + ∠ 12 = 180° say (equation 2)

Now by replacing ∠12 by ∠ 10 from equation 1 in equation 2 we get

∴ ∠ 11 + ∠ 10 = 180°

Hence proved the above statement.

7 0

3 years ago

How many steel rods are in a staircase frame with 12 steps?

kolbaska11 [484]

180 steel rods to be exact

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3 years ago

Write an equation in slope-intercept form of the line having the given slope and y-intercept.

pychu [463]

Answer:

The answer to your question is y = 4/6x - 4

Step-by-step explanation:

Data

m = 4/6

P (0, -4)

slope-intercept form = ? y = mx + b where m is the slope and

b is the y-intercept

Formula

y - y1 = m(x - x1)

Substitution

y - (-4) = 4/6(x - 0)

Simplification

y + 4 = 4/6x

Result

y = 4/6x - 4

slope = 4/6 y-intercept = -4

7 0

3 years ago

Calculate the sum. 10+ (-42) O A. 52 O B. -32 O C. 32 D. -52

kotykmax [81]

Answer:

Step-by-step explanation:

10 + (-42)

10 - 42

-32

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

$(A \cup B) + C = 1$