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3 years ago

Find the volume of a cylinder with a height of 16ft and diameter of 7 feet. Write answer in pie

Mathematics

1 answer:

finlep [7]3 years ago

5 0

Answer:

Step-by-step explanation:

volume of cylinder=V=πr^2h

given diameter=7 feet

radius=7/2 ft

height=16ft

v=π×7/2×7/2×16

=12.25×16×π

=196π

hope this helps

plzz mark me brainliest

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2 years ago

The diameter of a spindle in a small motor is supposed to be 5.8 millimeters (mm) with a standard deviation of 0.14 mm. If the s

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Answer: A. H0: Mean= 5 and Ha: Mean is not equal to 5.

Step-by-step explanation:

Definitions:

Null hypothesis $(H_0)$ is a statement describing population parameters according to the objective of the study. It takes "≤,≥,=" signs.
Alternative hypothesis $(H_a)$ is a statement opposite to the null hypothesis elaborating population parameters according to the objective of the study. It takes ">, <, ≠" signs.

Given: The diameter of a spindle in a small motor is supposed to be 5.8 millimeters.

If the spindle is either too small or too large, the motor will not work properly. so, Mean diameter ≠ 5.8 to work correctly.

i.e. $H_a: \text{Mean}\neq 5.8$

So, $H_a: \text{Mean =5.8}$

The required null and alternative hypotheses:

H0: Mean= 5 and Ha: Mean is not equal to 5.

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2 years ago

Please someone help me to prove this..

Pachacha [2.7K]

Answer: see proof below

<u>Step-by-step explanation:</u>

Use the following Sum to Product Identities:

$\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)$

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}$

$\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}$

$\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}$ $\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}$

$\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}$

$\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}$

$\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10$

LHS = RHS: cot 10 = cot 10 $\checkmark$

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2 years ago