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Yuki888 [10]
3 years ago
15

Find the cross product (7,9,6) x (-4,1,5). Is the resulting vector perpendicular to the given vectors

Mathematics
1 answer:
melisa1 [442]3 years ago
6 0

Expand each vector as linear combinations of the standard basis vectors:

(7, 9, 6) = 7(1, 0, 0) + 9(0, 1, 0) + 6(0, 0, 1)

(-4, 1, 5) = -4(1, 0, 0) + (0, 1, 0) + 5(0, 0, 1)

For brevity, write

i = (1, 0, 0)

j = (0, 1, 0)

k = (0, 0, 1)

Then by definition of the cross product,

i x i = j x j = k x k = (0, 0, 0)

i x j = k

j x k = i

k x i = j

and for any two vectors a and b, we have a x b = - b x a.

Now compute the product:

(7i + 9j + 6k) x (-4i + j + 5k)

= -28 (i x i) - 36 (j x i) - 24 (k x i)

...   + 7 (i x j) + 9 (j x j) + 6 (k x j)

...   + 35 (i x k) + 45 (j x k) + 30 (k x k)

= -36 (-k) - 24 j + 7 k + 6 (-i) + 35 (-j) + 45 i

= 39 i - 59 j + 43 k

which is the same as the vector

(39, -59, 43)

And yes, this vector is perpendicular to both given vectors.

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Part B:

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c

Step-by-step explanation

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A different species of cockroach has weights that are approximately Normally distributed with a mean of 50 grams. After measurin
Ratling [72]

Answer:

The standard deviation of weight for this species of cockroaches is 4.62.

Step-by-step explanation:

Given : A different species of cockroach has weights that are approximately Normally distributed with a mean of 50 grams. After measuring the weights of many of these cockroaches, a lab assistant reports that 14% of the cockroaches weigh more than 55 grams.

To find : What is the approximate standard deviation of weight for this species of cockroaches?

Solution :

We have given,

Mean \mu=50

The sample mean x=55

A lab assistant reports that 14% of the cockroaches weigh more than 55 grams.

i.e. P(X>55)=14%=0.14

The total probability needs to sum up to 1,

P(X\leq 55)=1-P(X>55)

P(X\leq 55)=1-0.14

P(X\leq 55)=0.86

The z-score value of 0.86 using z-score table is z=1.08.

Applying z-score formula,

z=\frac{x-\mu}{\sigma}

Where, \sigma is standard deviation

Substitute the values,

z=\frac{x-\mu}{\sigma}

1.08=\frac{55-50}{\sigma}

1.08=\frac{5}{\sigma}

\sigma=\frac{5}{1.08}

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The standard deviation of weight for this species of cockroaches is 4.62.

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