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BARSIC [14]
3 years ago
15

A 40% antifreeze solution is to be mixed with a 70% antifreeze

Mathematics
1 answer:
Alenkasestr [34]3 years ago
7 0

Answer: 160 liters of the

40% solution and 80 liters of the 70% solution will be used.

Step-by-step explanation:

Let x represent the number of liters of 40% antifreeze solution that should be used.

Let y represent the number of liters of 70% antifreeze solution that should be used.

The volume of the mixture to be mixed is 240 liters. It means that

x + y = 240

The 40% antifreeze solution is to be mixed with a 70% antifreeze

solution to get 240 liters of a 50% solution. This means that

0.4x + 0.7y = 0.5(240)

0.4x + 0.7y = 120 - - - - - - - - - - - -1

Substituting x = 240 - y into equation 1, it becomes

0.4(240 - y) + 0.7y = 120

96 - 0.4y + 0.7y = 120

- 0.4y + 0.7y = 120 - 96

0.3y = 24

y = 24/0.3

y = 80

x = 240 - y = 240 - 80

x = 160

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Consider the following expression and determine which statements are true. x^2+5yz−8
statuscvo [17]

Answer:

A.there are 3 terms.

B.the variables are x, y, and z.

Step-by-step explanation:

Complete question below:

Consider the following expression and determine which statements are true. x^2+5yz−8

CHOOSE 2 ANSWERS)

A.there are 3 terms.

B.the variables are x, y, and z.

C.The coefficient of x is 2.

D.the term 5yz is made up of 2 factors

The expression x^2+5yz-8

has three terms namely: x^2, 5yz and -8

The variables are x, y and z

The coefficient of x is 1 not 2, so option c is wrong

The term 5yz is made up of more than two factors namely 5, y, z, 5y, 5z, 5yz . So option d is also wrong.

The correct answers are option a and option b

A.there are 3 terms.

B.the variables are x, y, and z

8 0
3 years ago
On a standardized test with a normal distribution the mean score was 67.2. The standard deviation was 4.6. What percent of the d
Reika [66]

Answer:

P ( -1 < Z < 1 ) = 68%

Step-by-step explanation:

Given:-

- The given parameters for standardized test scores that follows normal distribution have mean (u) and standard deviation (s.d) :

                         u = 67.2

                         s.d = 4.6

- The random variable (X) that denotes standardized test scores following normal distribution:

                         X~ N ( 67.2 , 4.6^2 )

Find:-

What percent of the data fell between 62.6 and 71.8?

Solution:-

- We will first compute the Z-value for the given points 62.6 and 71.8:

                          P ( 62.6 < X < 71.8 )

                          P ( (62.6 - 67.2) / 4.6 < Z < (71.8 - 67.2) / 4.6 )

                          P ( -1 < Z < 1 )

- Using the The Empirical Rule or 68-95-99.7%. We need to find the percent of data that lies within 1 standard about mean value:

                          P ( -1 < Z < 1 ) = 68%

                          P ( -2 < Z < 2 ) = 95%

                          P ( -3 < Z < 3 ) = 99.7%

6 0
3 years ago
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