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2 years ago

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A twelve inch candle and an 18 inch candle are lit at 6pm. The 12-in. candle burns 0.5 inches every hour. The 18 inch candle bur

ns two inches every hour. At what time will the two candles be the same height? Let h represent the number of hours.

2 answers:

svp [43]2 years ago

8 0

Answer:

At 10 PM both candles will be same height.

Step-by-step explanation:

Let h represent the number of hours.

We have twelve inch candle and an 18 inch candle are lit at 6pm and the 12-in. candle burns 0.5 inches every hour. The 18 inch candle burns 2 inch every hour.

Height 18 inch candle after h hours, = 18 - 2h

Height 12 inch candle after h hours, = 12 - 0.5h

At same height we have

18 - 2h = 12 - 0.5h

1.5 h = 18 - 12

h = 4 hours.

Time = 6 PM + 4 hours = 10 PM

At 10 PM both candles will be same height.

torisob [31]2 years ago

5 0

12 - 0.5x = 18 - 2x

12 + 1.5h = 18
1.5h = 6
h = 4
4 hours

or to work it out the other way ..

18 - 2t = 12 - 0.5t
6 = 1.5t
t= 6/1.5

t= 4 hours , at time of 10 p.m.

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Answer:

Step-by-step explanation:

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2 years ago

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

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collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

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The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

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2 years ago

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reasonable domain of h
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2 years ago

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vivado [14]

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2 years ago

Read 2 more answers

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DiKsa [7]

Answer:

Tuesday z-score was 3.26.

Tuesday was a significantly good day.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

A score is said to be significantly high if it has a z-score higher than 1.64, that is, it is at least in the 95th percentile.

In this problem, we have that:

$\mu = 28372.72, \sigma = 2000$

On Tuesday, the store sold $34,885.21 worth of goods. Find Tuesday's z-score.

This is Z when $X = 34885.21$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{34885.21 - 28372.72}{2000}$

$Z = 3.26$

Tuesday z-score was 3.26.

Was Tuesday a significantly good day?

A z-score of 3.26 has a pvalue of 0.9994. So only 1-0.9994 = 0.0006 = 0.06% of the day are better than Tuesday.

So yes, Tuesday was a significantly good day.

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2 years ago