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2 years ago

Which equation accurately represents this statement? Select three options.

Mathematics

2 answers:

N76 [4]2 years ago

5 0

Answer:

X = 2

Step-by-step explanation:

4.9x - (-3) = 12.8

4.9x = 12.8 + (-3)

4.8x - 12.8 = -3

Can be three different forms

To solve:

4.9x = 12.8 - 3

4.9x = 9.8

x = 2

atroni [7]2 years ago

3 0

Complete question:

Which equation accurately represents this statement? Check all that apply.

Negative 3 less than 4.9 times a number, x, is the same as 12.8

A: -3-4.9x=12.8

B: 4.9x-(-3)=12.8

C: 3+4.9x=12.8

D: (4.9-3)x=12.8

E: 12.8=4.9x+3

Answer:

B: 4.9x-(-3)=12.8

C: 3+4.9x=12.8

E: 12.8=4.9x+3

Step-by-step explanation:

The word equation can be expressed mathematically as follows

negative 3 → -3

less than 4.9 times a number , x → 4.9 x - (-3)

is same as 12.8 → 4.9 x - (-3) = 12.8

The equation can be expressed as

4.9 x - (-3) = 12.8

when you open the bracket, you multiply -3 time - 1 . MInus times minus is equals to plus. so you get

4.9 x + 3 = 12.8

12.8 = 4.9x+3

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How can we find how many inches wide the coat box is?

Alex777 [14]

Answer:

First we found the sum of 15.438 and 3.18. The subtract the answer from 32.476.

Option A is correct.

Step-by-step explanation:

The total width of box is = 32.476

Width of box of T-shirts = 15.438

Width of box of Socks = 3.18

We need to find width of coat box.

The width of coat box can be found as:

Total Width = Width of box of T-shirts + Width of box of Socks + width of coat box

32.476 = 15.438 + 3.18 + width of coat box

32.476 = 18.618 + width of coat box

width of coat box= 32.476 - 18.618

width of coat box = 13.858

So, The Width of coat box = 13.858 inches

First we found the sum of 15.438 and 3.18. The subtract the answer from 32.476.

Option A is correct.

5 0

1 year ago

Problem 5 (4+4+4=12) We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely. 1) Fin

tekilochka [14]

Answer:

$p(b) = \frac{1}{6}$

$p(k) = \frac{1}{3}$

$P(a) = \frac{1}{3}$

Step-by-step explanation:

Generally when two fair 6-sided dice is rolled the doubles are

(1 1) , ( 2 2) , (3 3) , (4 4) , ( 5 5 ), (6 6)

The total outcome of doubles is N = 6

The total outcome of the rolling the two fair 6-sided dice is

n = 36

Generally the probability that doubles (i.e., having an equal number on the two dice) were rolled is mathematically evaluated as

$p(b) = \frac{N}{n}$

$p(b) = \frac{6}{36}$

$p(b) = \frac{1}{6}$

Generally when two fair 6-sided dice is rolled the outcome whose sum is 4 or less is

(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)

Looking at this outcome we see that there are two doubles present

The conditional probability that doubles were rolled is mathematically represented as

$p(k) = \frac{2}{6}$

$p(k) = \frac{1}{3}$

Generally when two fair 6-sided dice is rolled the number of outcomes that would land on different numbers is L = 30

And the number of outcomes that at least one die is a 1 is W = 10

The conditional probability that at least one die is a 1 is mathematically represented as

$P(a) = \frac{W}{L}$

=> $P(a) = \frac{10}{30}$

=> $P(a) = \frac{1}{3}$

3 0

2 years ago

A pound of oranges costs $2 less than a pound of pears. Together, ten pounds of oranges and eight pounds of pears cost $61. How

Thepotemich [5.8K]

Let us say that:

o = cost of oranges per pound

p = cost of pears per pound

so that:

o = p – 2

Therefore:

10o + 8p = 61

10 (p – 2) + 8p = 61

10p – 20 + 8p = 61

18p = 81

p = 4.5

p = $4.5 per pound

So 3 pounds of pears would cost:

total cost = 3 * 4.5

total cost = $13.5

3 0

2 years ago

A restaurant serves custom-made omelets, where guests select meat, cheese, and vegetables to be added to their omelet. There are

kondor19780726 [428]

Answer: The number of different combinations of 2 vegetables are possible = 15 .

Step-by-step explanation:

In Mathematics , the number of combinations of selecting r values out of n values = $^nC_r=\dfrac{n!}{r!(n-r)!}$

Given : Number of available vegetables = 6

Then, the number of different combinations of 2 vegetables are possible will be :

$^6C_2=\dfrac{6!}{2!(6-2)!}=\dfrac{6\times5\times4!}{2\times4!}=15$

Hence , the number of different combinations of 2 vegetables are possible = 15 .

5 0

2 years ago

Find the product. (2n + 2)(2n – 2)

vekshin1

<span>(2n + 2)(2n – 2)
= (2n)^2 -2^2
= 4n^2 - 4

hope it helps</span>

3 0

2 years ago

Read 2 more answers