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Bogdan [553]
2 years ago
6

A new centrifugal pump is being considered for an application involving the pumping of ammonia. The specification is that the fl

ow rate be more than 3 gallons per minute (gpm). In an initial study, eight runs were made. The average flow rate was 6.3 gpm and the standard deviation was 1.9 gpm. If the mean flow rate is found to meet the specification, the pump will be put into service.
a) State the appropriate null and alternate hypotheses.
b) Find the P-value.
c) Should the pump be put into service? Explain.
Mathematics
1 answer:
Dovator [93]2 years ago
5 0
Your answer would be a because of the state of the appropriate bill and alternate hypotheses makes the most sense out of B and C
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The graphs of the equations $y=3x-20$ and $y=-2x+10$ intersect at the point $(6,-2)$. Without solving, find the solution of the
Maru [420]

Answer:

The solution is

\boxed{x=6,y=-2}

Step-by-step explanation:


The given equations are


y=3x-20


and

y=-2x+10


The graph of the two equations intersects at (6,-2) as show in the graph in the attachment.


The point of intersection of the two graphs is the solution to system of equations.

Hence the solution is

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Solve the equation.<br> 25.14 = d + (-15.3)<br> 0 40.44<br> 0 -9.84<br> O 9.84<br> -40.44
Gnom [1K]

Answer:

d = 40.44

Step-by-step explanation:

Simplifying

25.14 = d + (-15.3)

25.14 = d + -15.3

Reorder the terms:

25.14 = -15.3 + d

Solving

25.14 = -15.3 + d

Solving for variable 'd'.

Move all terms containing d to the left, all other terms to the right.

Add '-1d' to each side of the equation.

25.14 + -1d = -15.3 + d + -1d

Combine like terms: d + -1d = 0

25.14 + -1d = -15.3 + 0

25.14 + -1d = -15.3

Add '-25.14' to each side of the equation.

25.14 + -25.14 + -1d = -15.3 + -25.14

Combine like terms: 25.14 + -25.14 = 0.00

0.00 + -1d = -15.3 + -25.14

-1d = -15.3 + -25.14

Combine like terms: -15.3 + -25.14 = -40.44

-1d = -40.44

Divide each side by '-1'.

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Simplifying

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Angels Inscribed and circumscribed.
zhenek [66]

Remark

The diameter of the circle is AC. The hypotenuse of the right triangle is also AC. This assumes that A,B and C all lie on the circumference of the circle.

Answers

One

The center of the circle is on the diameter of the circle. In fact, the center bisects the diameter. So the center lies on AC and if the center is labeled O then AO = BO.

<em><u>Answer 1: </u></em> AC

Two

The answer is the single word diameter.

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