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marishachu [46]
3 years ago
6

Jenna goes fishing every Saturday morning. She is only allowed to catch a maximum of 222 fish each trip. The table below display

s the probability distribution of XXX, the number of fish that Jenna catches in a trip. X= \# \text{ of fish}X=# of fishX, equals, \#, start text, space, o, f, space, f, i, s, h, end text 000 111 222 P(X)P(X)P, left parenthesis, X, right parenthesis 0.040.040, point, 04 0.320.320, point, 32 0.640.640, point, 64 Given that \mu_X=1.6μ X ​ =1.6mu, start subscript, X, end subscript, equals, 1, point, 6 fish, find the standard deviation of the number of fish that Jenna catches. Round your answer to two decimal places.
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
4 0

Answer: 0.57 fish. Khan academy told me.

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A length of string is 42 1/2 inches long.
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Answer:

5 5/16

Step-by-step explanation:

you just divide 42 1/2 by 8

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Heather has $45.71 in her savings account. She bought six packs of markers to donate to her school. If each pack of markers cost
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What is 40 meters in 8 seconds?
Basile [38]

5 meters/ 1 second because if you 40 and 8 you will get 5 and divided 8 by 8 and you get 1, you just divide the denominator by both numerator and denominator.

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Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

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And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

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5 0
3 years ago
Can anybody tell me how to solve this
Lyrx [107]
The answer to your question is:  Yes, someone undoubtedly can.

Although you haven't asked to be told or shown how to solve it, I'm here
already, so I may as well stick around and go through it with you.


The sheet is telling you to find the solutions to two equations, AND THEN
DO SOMETHING WITH THE TWO SOLUTIONS.  But you've cut off the
instructions in the pictures, so all we have are the two equations, and
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<u>First equation:</u>
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Divide each side by 2 :
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<u>Second equation:</u>
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Divide each side by -12 :
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5 0
3 years ago
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