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lana66690 [7]
3 years ago
11

The scalar product can be described as the magnitude of B times the component of A that is parallel to B. In terms of the positi

ve scalar quantities a, b, and d, what is the component of A that is parallel to B? Suppose that c = 0.

Mathematics
1 answer:
Mice21 [21]3 years ago
4 0

Answer:

See it in the pic.

Step-by-step explanation:

See it in the pic.

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riadik2000 [5.3K]

Answer:

The answer is 75 degrees.

Step-by-step explanation:

The explanation is that the the supplementary angle to 80 is 100 since a straight line has an angle of 180, now the 100 degree angle is congruent to the angle x + 25 which means that x + 25 is equal to 100 so x is 75

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yan [13]

Answer:

1. Pyramid= ll+l^2  2. Rectangular prism= 2 lw+2lh+2wh 3. Cylinder= 2pi pi rh+2pi pi r^2  4.Triangular Prism= 2 B + Ph  5. Cone= Pi rl+ pi r^2

Step-by-step explanation:

Pretty sure this is your answer. Hope it helps!

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2 years ago
big water canoeing company rents conoes for 75$ per hour. explain why the point (0,0) and (1, 75) are on the graph of the relati
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Compare each set of rational numbers.
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3 years ago
For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains
AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

f(x) = (1-p)^{x}p

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

\mu = \frac{1-p}{p}

The standard deviation of the geometric distribution is given by the following formula:

\sigma = \sqrt{\frac{1-p}{p^{2}}

In this problem, we have that:

p = 0.383

So

\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61

\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05

(a) What is the probability that a drought lasts exactly 3 intervals?

This is f(3)

f(x) = (1-p)^{x}p

f(3) = (1-0.383)^{3}*(0.383)

f(3) = 0.09

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is P = f(0) + f(1) + f(2) + f(3)

f(x) = (1-p)^{x}p

f(0) = (1-0.383)^{0}*(0.383) = 0.383

f(1) = (1-0.383)^{1}*(0.383) = 0.236

f(2) = (1-0.383)^{2}*(0.383) = 0.146

Previously in this exercise, we found that f(3) = 0.09

So

P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4).

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

P(X \leq 3) + P(X \geq 4) = 1

0.855 + P(X \geq 4) = 1

P(X \geq 4) = 0.145

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

8 0
3 years ago
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