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Nitella [24]
3 years ago
11

two sides of a triangle measures 20 centimeters and 30 centimeters which of the following could be the measure of the third side

Mathematics
1 answer:
brilliants [131]3 years ago
6 0
To find the third side of a triangle (there are two answers you could get), you need to use the Pythagorean formula: a^2+b^2=c^2.  First answer, 20^2+30^2=c^2.  Solve for c.  c=36.06.  The second answer, 20^2+b^2=30^2.  Solve for b.  b=26.36.
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You have already run 4 miles. If you run at a speed of 8 miles per hour, how many total miles will you run in 2 more hours? Choo
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Answer:

20 miles

Step-by-step explanation:

I'm not sure if that is exactly how you solve it but

If its

8x+4 as the equation and x is the number of hours run

the total number of miles run should be 20 miles

8(2)+4=20

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3 years ago
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What is the surface area of the cylinder with the height of 6ft and the radius of 6ft. round your answer to the nearest thousand
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Devaughn is 15 years younger than Sydney. The sum of their ages is 73. What is Sydney’s age?
Olegator [25]

Answer:

Sydney is 44 years old

Step-by-step explanation:

<u>Step 1:  Make a system of equations </u>

d = s - 15

<em>d + s = 73 </em>

<u>Step 2:  Plug in s - 15 for d in the second equation and solve </u>

s - 15 + s = 73

2s - 15 = 73 + 15

2s  / 2= 88 / 2

<em>s = 44 </em>

Answer:  Sydney is 44 years old

6 0
3 years ago
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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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31 between is 32 then 33
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