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Maru [420]
3 years ago
9

Solve by substitution. Please show work -x+9y=-5 x-5y=1

Mathematics
1 answer:
kkurt [141]3 years ago
3 0

Let's solve for x.

−x+9y=−5

Step 1: Add -9y to both sides.

−x+9y+−9y=−5+−9y

−x=−9y−5

Step 2: Divide both sides by -1.

−x ÷ −1 = −9y−5 ÷ −1

x=9y+5

THE ANSWER FOR THE OTHER EQUATION: (x-5y=1)

Let's solve for x.

x−5y=1

Step 1: Add 5y to both sides.

x−5y+5y=1+5y

x=5y+1

Answer:

x=5y+1

I hope this helped I was a little confused on what your problem meant...so if this is not what you asked for just lmk so I can fix it for you :)

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Step-by-step explanation:

2.88/16 = 0.18

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Answer:

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Please help me on this question!
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-8 to 8

Step-by-step explanation:

f(x) is the same as y when discussing functions. Therefore we look for the places where the line is above or touching the x axis

4 0
3 years ago
I cant wrap my head around this
vazorg [7]

Answer:

B is the answer

Step-by-step explanation:

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7 0
3 years ago
is flipped ten times, and he is asked to predict in advance the outcome. Our individual gets seven out of ten correct. What is t
egoroff_w [7]

The required probability is 0.171.

What is probability?

Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.

We can find required Probability as shown:

From 10 coins we choose 7 he guessed correct with probability of each 1/2. For three remaining coins he failed to predict the outcomes of the flip with the probability of 1/2 for each. Similarly for 8,9 or 10 successive trials. Let X denote number of successive trials

P(X\geq 7)= \tbinom{10}{7}.\frac{1}{2^{10}}+\tbinom{10}{8}.\frac{1}{2^{10}} +\tbinom{10}{9}.\frac{1}{2^{10}}+\tbinom{10}{10}.\frac{1}{2^{10}}

=120.\frac{1}{2^{10}} +45.\frac{1}{2^{10}} +10.\frac{1}{2^{10}} +.\frac{1}{2^{10}}

=\frac{120+45+10+1}{2^{10}}

=\frac{176}{1024}

=0.171

Hence, the required probability is 0.171.

Learn more about Probability here:

brainly.com/question/6649771

#SPJ4

Given question is incomplete. This is complete question:

A man claims to have extrasensory perception. As a test, a fair coin is flipped 10 times and the man is asked to predict the outcome in advance. He gets 7 out of 10 correct. What is the probability that he would have done at least this well if he had no ESP?

3 0
2 years ago
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