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Marta_Voda [28]
3 years ago
9

Two identical blue boxes together weigh the same as three identical red boxes together. Each red box weighs 15.2 ounces. How muc

h does one blue box weigh in ounces?
Mathematics
2 answers:
pshichka [43]3 years ago
6 0

Answer:

Each blue box weighs 22.8 ounces

Step-by-step explanation:

stich3 [128]3 years ago
6 0

Answer:

22.8 ounces

Step-by-step explanation:

Since one red box weighs 15.2 ounces, three red boxes weigh 45.6 ounces. This is the same as two blue boxes, so we have the equation 2b=45.6 where b represents the weight of the blue box. Solving the equation by multiplying both sides by 1/2 gives us 22.8 ounces.

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Help!!! Please will give points
galben [10]

Option D:

The solution to the equation is y = 27.

Solution:

Given equation:

$\frac{1}{3} y=9

To solve this equation:

$\frac{1}{3} y=9

Multiply by 3 on both sides of the equation.

$3 \times \frac{1}{3} y=9 \times 3

3 in the numerator and denominator get canceled.

y = 9 × 3

y = 27

The solution to the equation is y = 27.

Option D is the correct answer.

3 0
3 years ago
This equation has one solution. 5(x – 1) + 3x = 7(x + 1) What is the solution?
bogdanovich [222]
Open up the brackets
5x - 5 + 3x = 7x + 7
Collect like terms
5x + 3x - 7x = 7 + 5
Simplify
8x -7x = 12
x = 12
Therefore, your answer is 12.
5 0
3 years ago
Read 2 more answers
I need help with this question.This is due in like 7 mins :(​
Stells [14]

Answer:

the answer is the last option! 68/5 - 22/5 = 9 and 1/5

Step-by-step explanation:

6 0
2 years ago
Choose the surface area of a rectangular
olganol [36]

Answer:

592

Step-by-step explanation:

7 0
3 years ago
Suppose that 20% of the employees of a given corporation engage in physical exercise activities during the lunch hour. Moreover,
Harlamova29_29 [7]

Answer:

a) 0.152 = 15.2% probability that this person is a female who engages in physical exercise activities during the lunch hour.

b) 0.248 = 24.8% probability that this person is a female who does not engage in physical exercise activities during the lunch hour.

Step-by-step explanation:

Question a:

20% of employees engage in physical exercise.

This 20% is composed by:

8% of 60%(males)

x% of 100 - 60 = 40%(females).

Then, x is given by:

0.08*0.6 + 0.4x = 0.2

0.4x = 0.2 - 0.08*0.6

x = \frac{0.2 - 0.08*0.6}{0.4}

x = 0.38

0.38 = 38%

Probability of being a female who engages in exercise:

40% are female, 38% of 40% engage in exercise. So

0.38*0.4 = 0.152

0.152 = 15.2% probability that this person is a female who engages in physical exercise activities during the lunch hour.

B. If we choose an employee at random from this corporation,what is the probability that this person is a female who does not engage in physical exercise activities during the lunch hour?

40% are female, 100% - 38% = 62% of 40% do not engage in exercise. So

0.62*0.4 = 0.248

0.248 = 24.8% probability that this person is a female who does not engage in physical exercise activities during the lunch hour.

3 0
3 years ago
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