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loris [4]
2 years ago
8

Walter used the iterative process to determine that √13 is between 3.61 and 3.62.Analyze Walter’s estimation. Is he correct? If

not, what was his mistake?
A-Yes, Walter is correct.
B-No, 3.612 is less than 13.
C-No, both 3.612 and 3.622 are greater than 13.
D-No, both 3.612 and 3.622 are less than 13.
i need the right answer
Mathematics
1 answer:
Shkiper50 [21]2 years ago
8 0

The correct option is C no, both 3.61^2 and 3,62^2 are greater than 13.

<h3>Is he correct? If not, what was his mistake?</h3>

To check this, we can just square both of the bounds that he found, and see if it makes sense.

Walter says hat:

3.61 < √13 < 3.62

Then we must have:

(3.61)^2 < 13 < (3.62)^2

The lower bound is 3.61, if we square it we get:

3.61*3.61 = (3 + 0.61)*(3 + 0.61) = 9 + 6*0.61 + 0.61*0.61 = 13.0321

Now, this is larger than 13, so the above inequality is false, which means that Walter is incorrect, because both 3.61 and 3.62 are larger than √13 , then the correct option is C.

If you want to learn more about square roots:

brainly.com/question/98314

#SPJ1

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A particular model of walkie-talkie can broadcast in a circular area. The radius of the broadcast area is 7,000 feet. Find the a
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153938040.0259 ft2

Step-by-step explanation:

R= Square root of A/π

   

7 0
3 years ago
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Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the
kodGreya [7K]

Answer:

  • p=0.7103 (4-game series)
  • p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters  n=4, p=0.6:

P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\  i=4,5,6,7

#We find probabilities for the different values of i:

P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659

Hence, probability of the stronger team winning overall is:

=P(X=4)+P(X=5)+P(X=6)+P(X=7)\\\\=0.7103

#Define Y as the random variable for winning 2/3 games.:

P(Y=2)={1\choose 1}0.6^2=0.3600\\\\P(Y=3)={2\choose3}0.6^20.4=0.2880\\\\P(win)=0.2880+0.3600=0.6480

Hence, probability of the stronger team winning in 2 out 3 game series is 0.6480

The stronger team has a higher chance of winning in a 4-game series(0.7103>0.6480)

8 0
3 years ago
Is this right ? Im honestly confused
fgiga [73]

Answer:

In step by step

Step-by-step explanation:

For #1, everything is perfect.

For #2, you wrote your functions out incorrectly, but your x input and y output are correct

#3 is a little messed up.

Your function is f(x)=-3x+1. Your given x's are -3, -1, 0, 1, 2, and 3. Plug all of these in.

1. f(-3)=-3(-3)+1

  • f(-3)=9+1
  • f(-3)=1

2. f(-1)=-3(-1)+1

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3. f(0)=-3(0)+1

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4. f(1)=-3(1)+1

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5. f(2)=-3(2)+1

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3 0
3 years ago
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asambeis [7]

Answer:

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Step-by-step explanation:

I square rooted it by using a calculator, and by working it out on paper.

5 0
2 years ago
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