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zepelin [54]
4 years ago
7

claire mowed 5 lawns last week. show mowed each lawn in 7/12 hour. she mowed the same lawns this week in 5/12 hour each using he

r new lawn mower. How many times longer was Claire's time to now all the lawns last week than this week?
Mathematics
1 answer:
Zinaida [17]4 years ago
5 0

Answer:

\dfrac{5}{6} hour or 50 minutes

Step-by-step explanation:

<u>Last week:</u>

Claire mowed 5 lawns

She mowed each lawn in \frac{7}{12} hour.

She spent

5\cdot \dfrac{7}{12}=\dfrac{35}{12}=2\dfrac{11}{12}

hours to mow 5 lawns.

<u>This week:</u>

Claire mowed 5 lawns

She mowed each lawn in \frac{5}{12} hour.

She spent

5\cdot \dfrac{5}{12}=\dfrac{25}{12}=2\dfrac{1}{12}

hours to mow 5 lawns.

<u>Difference:</u>

2\dfrac{11}{12}-2\dfrac{1}{12}=\dfrac{11}{12}-\dfrac{1}{12}=\dfrac{10}{12}=\dfrac{5}{6}

hour or 50 minutes.

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Answer:

Step-by-step explanation:

Everything is correct. But you forgot to add

x = 6 - square root of 26. The answer is

x = 6 + square root of 26 or

x = 6 - square root of 26

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3 years ago
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mrs_skeptik [129]

9514 1404 393

Answer:

  10.49

Step-by-step explanation:

Since we know 110 = 10² +10, we can make a first approximation to the root as ...

  √10 ≈ 10 +10/21 . . . . . where 21 = 1 + 2×integer portion of root

This is a little outside the desired approximation accuracy, so we need to refine the estimate. There are a couple of simple ways to do this.

One of the best is to use the Babylonian method: average this value with the value obtained by dividing 110 by it.

  ((220/21) + (110/(220/21)))/2 = 110/21 +21/4 = 881/84 ≈ 10.49

An approximation of √110 accurate to hundredths is 10.49.

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The other simple way to refine the root estimate is to carry the continued fraction approximation to one more level.

For n = s² +r, the first approximation is ...

  √n = s +r/(2s+1)

An iterated approximation is ...

  s + r/(s +(s +r/(2s+1)))

The adds 's' to the approximate root to make the new fraction denominator.

For this root, the refined approximation is ...

  √110 ≈ 10 + 10/(10 +(10 +10/21)) = 10 +10/(430/21) = 10 +21/43 ≈ 10.49

_____

<em>Additional comment</em>

Any square root can be represented as a repeating continued fraction.

  \displaystyle\sqrt{n}=\sqrt{s^2+r}\approx s+\cfrac{r}{2s+\cfrac{r}{2s+\dots}}

If "f" represents the fractional part of the root, it can be refined by the iteration ...

  f'=\dfrac{r}{2s+f}

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The above continued fraction iteration <em>adds</em> 1+ good decimal places to the root with each iteration. The Babylonian method described above <em>doubles</em> the number of good decimal places with each iteration. It very quickly converges to a root limited only by the precision available in your calculator.

4 0
3 years ago
A process manufactures ball bearings with diameters that are normally distributed with mean 25.1 mm and standard deviation 0.08
marta [7]

Answer:

(a) The proportion of the diameters are less than 25.0 mm is 0.1056.

(b) The 10th percentile of the diameters is 24.99 mm.

(c) The ball bearing that has a diameter of 25.2 mm is at the 84th percentile.

(d) The proportion of the ball bearings meeting the specification is 0.8881.

Step-by-step explanation:

Let <em>X</em> = diameters of ball bearings.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 25.1 mm and standard deviation, <em>σ</em> = 0.08 mm.

To compute the probability of a Normally distributed random variable we need to first convert the raw scores to <em>z</em>-scores as follows:

<em>z</em> = (X - μ) ÷ σ

(a)

Compute the probability of <em>X</em> < 25.0 mm as follows:

P (X < 25.0) = P ((X - μ)/σ < (25.0-25.1)/0.08)

                    = P (Z < -1.25)

                    = 1 - P (Z < 1.25)

                    = 1 - 0.8944

                    = 0.1056

*Use a <em>z</em>-table for the probability.

Thus, the proportion of the diameters are less than 25.0 mm is 0.1056.

(b)

The 10th percentile implies that, P (X < x) = 0.10.

Compute the 10th percentile of the diameters as follows:

P (X < x) = 0.10

P ((X - μ)/σ < (x-25.1)/0.08) = 0.10

P (Z < z) = 0.10

<em>z</em> = -1.282

The value of <em>x</em> is:

z = (x - 25.1)/0.08

-1.282 = (x - 25.1)/0.08

x = 25.1 - (1.282 × 0.08)

  = 24.99744

  ≈ 24.99

Thus, the 10th percentile of the diameters is 24.99 mm.

(c)

Compute the value of P (X < 25.2) as follows:

P (X < 25.2) = P ((X - μ)/σ < (25.2-25.1)/0.08)

                    = P (Z < 1.25)

                    = 0.8944

                    ≈ 0.84

*Use a <em>z</em>-table for the probability.

Thus, the ball bearing that has a diameter of 25.2 mm is at the 84th percentile.

(d)

Compute the value of P (25.0 < X < 25.3) as follows:

P (25.0 < X < 25.3) = P ((25.0-25.1)/0.08 < (X - μ)/σ < (25.3-25.1)/0.08)

                    = P (-1.25 < Z < 2.50)

                    = P (Z < 2.50) - P (Z < -1.25)

                    = 0.99379 - 0.10565

                    = 0.88814

                    ≈ 0.8881

*Use a <em>z</em>-table for the probability.

Thus, the proportion of the ball bearings meeting the specification is 0.8881.

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