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2 years ago

Help with question 5,9, and 13. It would be a big help.

Mathematics

1 answer:

solniwko [45]2 years ago

3 0

I am using "x" but alphabet letter does not matter.

#9

Words: x is greater than or equal to -4
Algebra: x ≥ -4
Possible solutions: 0,1,2

#13

Words: x is greater than 0
Algebra: x > 0
Possible solutions: 1,2,3

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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:

$p = 0.383$

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

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

(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 $f(3) = 0.09$

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

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

Matrix multiplication was used to encode a message using the given encoding matrix: [i 47 A=1 |-1 -3] The original message was c

nadya68 [22]

Answer:

$A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right]$ message SHOW_ME_THE_MONEY_

Step-by-step explanation:

The matrix

$A=\left[\begin{array}{cc}1&4\\-1&-3\end{array}\right]\rightarrow |A|=(1 \times -3)-(-1\times 4)=1\\\rightarrow A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right] \\$

We can check that in fact A*A^⁻1=I_2 the identity matrix of size 2 x 2.

Now the message was divided in 1 x 2 matrices, then we have that the sequence given is the result of multiplying m by A, so to get m again we multiply now by A^⁻1. and we get the next table

Encoded message Decoded message message in letters by association

11 52 19 8 S H

-8 -9 15 23 O W

-13 -39 0 13 _ M

5 20 5 0 E _

12 56 20 8 T H

5 20 5 0 E _

-2 7 13 15 M O

9 41 14 5 N E

25 100 25 0 Y _

Then the message decoded is SHOW_ME_THE_MONEY_

3 0

2 years ago

Pls help me with this problem

Alchen [17]

Answer:

Angle A: 90

Angle B: 48

Angle C: 42

Side length A (Hypotinuse): 23.9

SIde length B (Opposite): 17.8

Side length C (Adjacent): 16

Step-by-step explanation:

https://www.calculator.net/triangle-calculator.html?vc=42&vx=&vy=&va=90&vz=16&vb=&angleunits=d&x=0&y=0

6 0

2 years ago

Anyone know the answer to this algebra question?

lana [24]

Answer:

possibly the third maybe

4 0

2 years ago

Help anyone pleaseeee

AlexFokin [52]

Answer:

36 <127

(36,127°)

36 [ cos (127) + i sin (127)]

Step-by-step explanation:

w1 = 2 < 95

w2 = 18 < 32

w1 * w2

We multiply the magnitude and add the angles

w1 * w2 = 2*18 < (95+32)

=36 < 127

36 [ cos (127) + i sin (127)]

4 0

3 years ago