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olga_2 [115]
3 years ago
12

James wants to pour a patio that is 18 feet by 20 feet. He wants the concrete to be 6 inches deep. How many cubic feet of concre

te mix are needed for the patio? [1 ft = 12 inches]
A) 45 cubic feet
B) 90 cubic feet
C) 180 cubic feet
D) 240 cubic feet
Mathematics
1 answer:
atroni [7]3 years ago
6 0
The correct answer would be C) 180 cubic feet.

WORK:  18 x 20 = 360, 6/12 = 0.5... then multiply 360 x 0.5 and you get 180
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<h3>Given:</h3>

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Y = 2

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The answer would be B. 
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Use the number line​ below, where RS=5y+5​, ST=2y+5, and RT=52. a. What is the value of​ y? b. Find RS and ST.
Alchen [17]

Answer:

y = 6, RS = 35 and ST = 17

Step-by-step explanation:

Given that,

RS = 5y+5

ST = 2y+5

RT = 52

If we consider a number line,

RT = RS + ST

52 = 5y+5 + 2y+5

52 = 7y + 10

7y = 52-10

7y = 42

y = 6

RS = 5y+5

= 5(6)+5

= 35

ST = 2y+5

= 2(6) +5

= 17

Hence, this is the required solution.

3 0
3 years ago
Which of the following is a true statement about functions?
chubhunter [2.5K]

Answer:

A

Step-by-step explanation:

6 0
3 years ago
A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company prod
tekilochka [14]

Answer:

C) 515 hours.

D) 500 hours

c) sample 3

Step-by-step explanation:

1. Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample Service Life (hours)

1                2              3

495      525            470

500         515           480

505        505            460

<u>500         515             470        </u>

<u>∑2000     2060         1880</u>

x1`= ∑x1/n1= 2000/4= 500 hours

x2`= ∑x2/n2= 2060/4= 515 hours

x3`= ∑x3/n3= 1880/4=  470 hours

2. The mean of the sampling distribution of sample means for whenever service life is in control is 500 hours . It is the given mean in the question and the limits are determined by using  μ ± σ , μ±2 σ  or μ ± 3 σ.

In this question the limits are determined by using  μ ± σ .

3. Upper control limit = UCL = 520 hours

Lower Control Limit= LCL = 480 Hours

Sample 1 mean = x1`= ∑x1/n1= 2000/4= 500 hours

Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample 3 mean = x3`= ∑x3/n3= 1880/4=  470 hours

This means that the sample mean must lie within the range 480-520 hours but sample 3 has a mean of 470 which is out of the given limit.

We see that the sample 3 mean is lower than the LCL. The other  two means are within the given UCL and LCL.

This can be shown by the diagram.

8 0
2 years ago
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