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3 years ago

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What is the absolute value of 2.3

1 answer:

Hoochie [10]3 years ago

8 0

<h3>Answer: 2.3</h3>

The absolute value of any positive number is that number itself.

If you take the absolute value of a negative number, then just erase the negative sign and you have your answer.

The absolute value is never negative. It represents distance from that value to 0 on the number line.

So saying |2.3| = 2.3 means the number 2.3 is exactly 2.3 units away from 0 on the number line.

Another example would be |-7| = 7 showing that from 0 to -7 is 7 spaces.

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Use the alternative form of the dot product to find u · v.

vodka [1.7K]

Answer:

Step-by-step explanation:

u = 45

v = 30

Angle between them, θ = 5π / 6 = 150°

(a) The formula for the dot product is given by

$\overrightarrow{u}.\overrightarrow{v}= u v Cos\theta$

$\overrightarrow{u}.\overrightarrow{v}= 45 \times 30 \times Cos150$

$\overrightarrow{u}.\overrightarrow{v}= -1169.13$

(b) $\overrightarrow{u}=Cos\frac{\pi }{3}\widehat{i}+Sin\frac{\pi }{3}\widehat{j}$

$\overrightarrow{u}=0.5\widehat{i}+0.866\widehat{j}$

$\overrightarrow{v}=Cos\frac{2\pi }{3}\widehat{i}+Sin\frac{2\pi }{3}\widehat{j}$

$\overrightarrow{v}=-0.707\widehat{i}+0.707\widehat{j}$

Let the angle between them is θ

The formula in terms of the dot product is given by

$Cos\theta =\frac{\overrightarrow{u}.\overrightarrow{v}}{u v}$

$u=\sqrt{0.5^{2}+0.866^{2}}=1$

$v=\sqrt{-0.707^{2}+0.707^{2}}=1$

$Cos\theta =\frac{0.5 \times (-0.707) + 0.866 \times 0.707)}{1\times 1}$

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(c) $\overrightarrow{u}=1\widehat{i}+4\widehat{j}+8\widehat{k}$

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3 years ago

A Confidence interval is desired for the true average stray-load loss mu (watts) for a certain type of induction motor when the

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Answer:

A sample size of 35 is needed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

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Now, find the width W as such

$W = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large must the sample size be if the width of the 95% interval for mu is to be 1.0:

We need to find n for which W = 1.

We have that $\sigma^{2} = 9$ , then $\sigma = \sqrt{\sigma^{2}} = \sqrt{9} = 3$ . So

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$(\sqrt{n})^2 = (1.96*3)^{2}$

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Rounding up

A sample size of 35 is needed.

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Step-by-step explanation:

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