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melamori03 [73]
2 years ago
10

A 20-ounce candle is expected to burn for 60 hours. A 12-ounce candle is expected to burn for 36 hours. Assuming the variables a

re directly related, how many hours would a 9-ounce candle be expected to burn? 18
Mathematics
2 answers:
ANTONII [103]2 years ago
5 0
Because the ratio for both candles is 20/60 or 1/3 then take the number of ounces the candle has and divide by 1/3. In expression: 9 / 1/3 => 9 * 3 = 18
alina1380 [7]2 years ago
3 0

Answer:  A 9-ounce candle is expected to burn for 27 hours.

Step-by-step explanation:  Given that a 20-ounce candle is expected to burn for 60 hours and a 12-ounce candle is expected to burn for 36 hours.

If the variables are directly related, we are to find the number of hours that a 9-ounce candle is expected to burn.

Let, x represents the number of ounces of the candle and y represents the corresponding number of hours for which it burns.

Then, since the variables are directly related, the graph will be a straight line.

And, the two points (x, y) = (20, 60) and (12, 36) lies on the line.

So, the slope of the line will be

m=\dfrac{36-60}{12-20}\\\\\\\Rightarrow m=\dfrac{-24}{-8}\\\\\Rightarrow m=3.

Therefore, the equation of the line is

y-36=m(x-12)\\\\\Rightarrow y-36=3(x-12)\\\\\Rightarrow y=3x.

So, if x = 9, then

y=3\times9=27.

Thus, a 9-ounce candle is expected to burn for 27 hours.

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Complete the square to rewrite y = x2 - 6x + 15 in vertex form. Then state
love history [14]

Since x^2 is the square of x and 6x is twice the product between x and 3, the second square must be 3 squared, i.e. 9.

So, if we think of 15 as 9+6, we have

x^2-6x+9+6 = (x-3)^2+6

Which is the required vertex form. This form tells us imediately that the vertex is the point (3,6).

Since the leading coefficient is 1, the parabola is facing upwards (it's U shaped), so the vertex is a minimum.

5 0
2 years ago
The average sea level in 1900 at London bridge was 33 feet. In 1990 it was 33.08 feet. Use linear interpolation or extrapolation
eduard

Answer:

In 1956, the height will be 33.05 ft

Step-by-step explanation:

Let +x+ = number of years after 1900

Calll 1900 +x+=+0+

Then 1990 is +x+=+90+

Let +y+ = height in feet

--------------------------

You are gven the points

( 0, 33 )

( 90, 33.08 )

Use general point-slope formula

+%28+y+-+33+%29+%2F+%28+x+-+0+%29+=+%28+33.08+-+33+%29+%2F+%28+90+-+0+%29+

Multiply both sides by +90x+

+90%2A%28+y+-+33+%29+=+.08x+

+90y+-+2970+=+.08x+

+90y+=+.08x+%2B+2970+

+y+=+.000889x+%2B+33+ ( equation to use )

--------------------------

Check:

does it go through ( 90, 33.08 ) ?

+33.08+=+.000889%2A90+%2B+33+

+33.08+=+.08001+%2B+33+

+33.08+=+33.08001+

close enough

-------------------------------

(a)

+x+=+61%0D%0A%7B%7B%7B+y+=+.000889x+%2B+33+

+y+=+.000889%2A61+%2B+33+

+y+=+.0542+%2B+33+

+y+=+33.0542+

In 1961, the height will be 33.05 ft

8 0
2 years ago
Suppose A = {x |x ≤ −2, x ≥ 3} and B = {x | −1 ≤ x < 5}. Which statement is true?
vagabundo [1.1K]
A.A∪ B is real numbers

3 0
2 years ago
Suppose a certain airline uses passenger seats that are 16.2 inches wide. Assume that adult men have hip breadths that are norma
Pachacha [2.7K]

Answer:

Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.

There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

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}

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem

Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that \mu = 14.4, \sigma = 1.1.

What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?

For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:

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

Z = \frac{16.2 - 14.4}{1.1}

Z = 1.64

Z = 1.64 has a pvalue of 0.9495.

This means that each male has a 1-0.9495 = 0.0505 = 5.05% probability of having a hip width greater than 16.2 inches.

For the average of the sample

What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches?

Now, we need to find the standard deviation of the sample before using the zscore formula. That is:

s = \frac{\sigma}{\sqrt{110}} = 0.1.

Now

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

Z = \frac{16.2 - 14.4}{0.1}

Z = 18

Z = 18 has a pvalue of 0.9999.

This means that there is a 1-0.9999 = 0.0001 = 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

7 0
2 years ago
Let d(t) be the total number of miles Joanna has cycled, and let t represent the number of hours after stopping for a break duri
Naddik [55]

Answer:

d(4)=?

where

d= distance

t= hours

In this Linear function given by the law d(t) = 12t +20 and the graph below. To find the d(4), means to find out how many miles she cycled in the instant 4 hours.

By this function, after a break , the instant t=0, she had already cycled 20 miles.

Let's calculate

d(4) = 12(4) +20

d(4) =48+20

d(4)= 68 miles

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
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