All categories

3 years ago

Please help me!!!!! I don't want to get a bad grade and CAN YOU PLEASE SHOW YOUR WORK!!!!

Mathematics

2 answers:

Veseljchak [2.6K]3 years ago

4 0

E. 53
D. 2
B. -42
A.-92

Igoryamba3 years ago

4 0

A.) -92
B.) -42
D.) 2
E.) 53
I hope this helps I think you solve these by PEMDAS if that helps any PEMDAS is Parentheses Exponents Multiplication Division Addison Subtraction

You might be interested in

In the term 4p, the variable p is raised to the power of 0 or 1 ?

Svetlanka [38]

Answer:

Step-by-step explanation:

anything raised to the power of 0 would equal 1 and p doesn't equal 1 here.

3 0

3 years ago

Read 2 more answers

17 plus what equals 79

Tasya [4]

The answer would be 62

4 0

3 years ago

Read 2 more answers

A bag contains blue marbles and red marbles, 75 in total. The number of blue marbles is 5 more than 4 times the number of red ma

horrorfan [7]

There are 61 blue marbles

<em><u>Solution:</u></em>

Let "b" be the number of blue marbles

Let "r" be the number of red marbles

Given that bag contains blue marbles and red marbles, 75 in total

number of blue marbles + number of red marbles = 75

b + r = 75 -------- eqn 1

The number of blue marbles is 5 more than 4 times the number of red marbles

number of blue marbles = 5 + 4(number of red marbles)

b = 5 + 4r ------- eqn 2

Let us solve eqn 1 and eqn 2

Substitute eqn 2 in eqn 1

5 + 4r + r = 75

5 + 5r = 75

5r = 75 - 5

5r = 70

r = 14

From eqn 2,

b = 5 + 4(14) = 5 + 56 = 61

b = 61

Thus there are 61 blue marbles

3 0

3 years ago

Which of the following is a true statement?

zlopas [31]

Answer:

A. |5| < |–8| is the correct option.

5 0

2 years ago

From a bag containing 5 nickels, 8 dimes, and 7 quarters, 5 coins are drawn at random and all at once. What is the probability o

murzikaleks [220]

Answer:

0.13

Step-by-step explanation:

From a bag containing 5 nickels, 8 dimes, and 7 quarters, 5 coins are drawn at random and all at once

we need to select 2 nickels from 5 nickels and select 2 dimes from 8 dimes and 1 quarter from 7 quarters

There are total of 5+8+7=20 coins

select 5 coins from total of 20 coins

2 nickels can be selected from 5 nickels in 5C2 ways

$5C2=\frac{5!}{2!(5-2)!} =\frac{5!}{2!(3!)!}=10$

2 dimes selected from 8 dimes

$8C2=\frac{8!}{2!(8-2)!} =\frac{6!}{2!(6!)}=28$

1 quarter selected from 7 quarter

$7C1=\frac{7!}{1!(7-1)!} =\frac{7!}{1!(6!)}=7$

5 coins selected from 20 coins

$20C5=\frac{20!}{5!(15)!} =15504$

probability of getting 2 nickels, 2 dimes, and 1 quarter

$\frac{10 \cdot 28 \cdot 7}{15504} =\frac{1960}{15504} =0.13$

5 0

3 years ago