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Kipish [7]
3 years ago
8

Simplify the problem: Sin theta Sec Theta

Mathematics
1 answer:
kherson [118]3 years ago
3 0

Answer:

\tan \theta

Step-by-step explanation:

We have to simplify the following expression as given by

\sin \theta \times \sec \theta

= \frac{\sin \theta}{\cos \theta}

= \tan \theta ( Answer )

Because, we know that \sec \theta =\frac{1}{\cos \theta} and \tan \theta = \frac{\sin \theta}{\cos \theta}

If we consider \sin \theta = \frac{Perpendicular}{Hypotenuse} and \sec \theta = \frac{Hypotenuse}{Base},

then \sin \theta \times \sec \theta = \frac{Perpendicular}{Hypotenuse}\times \frac{Hypotenuse}{Base} = \frac{Perpendicular}{Base}= \tan \theta

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The equivalent fraction for 5/15 is 1/5

Option B is correct

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Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 141 millimeters,
tiny-mole [99]

Answer:

Probability that the sample mean would be greater than 141.4 millimetres is 0.3594.

Step-by-step explanation:

We are given that Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 141 millimetres, and a standard deviation of 7.

A random sample of 39 steel bolts is selected.

Let \bar X = <u><em>sample mean diameter</em></u>

The z score probability distribution for sample mean is given by;

                            Z  =  \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} }  } }  ~ N(0,1)

where, \mu = population mean diameter = 141 millimetres

           \sigma = standard deviation = 7 millimetres

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Now, Percentage the sample mean would be greater than 141.4 millimetres is given by = P(\bar X > 141.4 millimetres)

      P(\bar X > 141.4) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} }  } } > \frac{141.4-141}{\frac{7}{\sqrt{39} }  } } ) = P(Z > 0.36) = 1 - P(Z \leq 0.36)

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The above probability is calculated by looking at the value of x = 0.36 in the z table which has an area of 0.6406.

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Find the quotient. 492 ÷ 4 = ? HELP PLEASE
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Answer:

123

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492 divided by four equals 123

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