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zlopas [31]
2 years ago
14

The height of a baseball in feet can be found by the equation -4.982 – 20t + 1000. How far has the

Mathematics
1 answer:
disa [49]2 years ago
8 0

Answer:

875.018

Step-by-step explanation:

To solve this, we must plug in 6 for our t value. So, we have the equation:

-4.982 - 20(6) + 1000

Following the rules of PEMDAS, we have:

-4.982 - 120 + 1000 = 875.018

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A plane is approaching an aircraft carrier at 25° angle of elevation from 3.2 miles away how far above sea level is the plane
s2008m [1.1K]

Answer:

Height\approx 1.35\ miles

Step-by-step explanation:

The angle of elevation=25\textdegree

It is 3.2\ miles away from the point.

If we imagine a right  triangle having an angle 25\textdegree.

Hypotenuse=3.2, Opposite=Height from sea level(h)

\sin 25=\frac{opposite}{hypotenuse}\\\\\sin 25=\frac{h}{3.2}\\\\h=3.2\times \sin 25\\\\h=3.2\times 0.4226\\\\h=1.3524\approx 1.35\ miles

8 0
3 years ago
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Please help me with this graph!
trasher [3.6K]

Answer:

y=4/3x-1

Step-by-step explanation:

5 0
3 years ago
What is the solution to the equation 3/<br> x+4+3/2x+8 = 0?
mixas84 [53]

Answer:

X= -4

Step-by-step explanation:

^ you have to isolate the radical than raise each side to the power of its index which gets you

x= -4

5 0
3 years ago
Please help me ASAP!!!<br> Solve for y.<br> −4.2y+2.1&gt;−2.52
Maru [420]

Answer:

hope it helps you

Step-by-step explanation:

https://mathsolver.microsoft.com/en/solve-problem/-4.2y%2B2.1%20%3E%20%20-2.52

5 0
2 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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