All categories

3 years ago

In the problem below, what is the SECOND operation that should be performed: 42 - (28 + 7) + 111

Mathematics

2 answers:

Nata [24]3 years ago

6 0

Answer:

Add -28 and -7 (2nd operation)

Step-by-step explanation:

42-(28+7)+111

Step 1: Remove the brackets

42-28-7-11

Step 2: Add 42 and 111( 1st operation ) you can also rewrite to avoid confusion

42+111-28-7

= 153-28-7

Step 3: Add -28 and -7 (2nd operation)

153-28-7

=153-35

Step 4 : Subtract

153-35

=118

labwork [276]3 years ago

4 0

42-35+111
7+111
118
Hope you got the answer

You might be interested in

What is negative 5/12 plus negative 1/4 plus 3/12

shtirl [24]

-5/12 + (-1/4) + 3/12
Find the least common denominator.
-5/12 - 3/12 + 3/12
Combine like terms.
-5/12 is your final answer.

6 0

3 years ago

Read 2 more answers

What is 189% as a decimal

Dovator [93]

Answer:1.89

Step-by-step explanation:

u take 189 percent and u divide it by 100 or

189/100= 1.89

hope this helps

4 0

3 years ago

Read 2 more answers

Please explain how you got your answer!!! The dot plots below show the ages of students belonging to two groups of salsa classes

Rudik [331]

Answer:

Group b most likely has a lower mean age of salsa students

Step-by-step explanation:

Arithmetic Mean of the data is the average of a set of numerical values, calculated by adding them together and dividing by the number of terms in the set.

Here we are given with two groups that are Group A and Group B

both having total number of students = 20

Here the mean age of the data is addition of the all ages of different students divided by total number of students.

For group a

total age of the group = 3 × 5 + 4 × 10 + 6 × 17 + 4 × 24 + 3 × 29

= 15 + 40 + 102 + 96 + 87

=340

The mean age of salsa students= 340 ÷ 20 = 17

For group b

total age of the group = 6 × 7 + 3 × 10 + 4 × 14 + 5 × 16 + 2 × 21

= 42 + 30 + 56 + 80 + 42

=250

The mean age of salsa students= 250 ÷ 20 = 12.5

So the group b most likely has a lower mean age of salsa students

Learn more about Arithmetic Mean here - brainly.com/question/24688366

#SPJ10

7 0

2 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

We used Total - 1 because it is a probability without replacement

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$