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2 years ago

15

Sia sells large candles for $3 each and small candles for $2 each. She sold 17 candles for $46.00. How many of each size candle

did she sell?

1 answer:

kipiarov [429]2 years ago

8 0

Answer:

The number of large size candles sells are 12 and the number of small size candles are 5 .

Step-by-step explanation:

As given

Sia sells large candles for $3 each and small candles for $2 each.

She sold 17 candles for $46.00.

Let us assume that the large size candle sells are x .

Let us assume that the small size candle sells are y.

Equation becomes

x + y = 17

3x + 2y = 46

Multiply x + y = 17 by 3 and subtracted from 3x + 2y = 46 .

3x - 3x + 2y - 3y = 46 - 51

-y = - 5

y = 5

Put in the equation x + y = 17 .

x + 5 = 17

x = 17 - 5

x = 12

Therefore the number of large size candles sells are 12 and the number of small size candles are 5 .

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Select all the pairs that could be reasonable approximations for the diameter and circumference of a cirlce 5 meters and 22 mete

bazaltina [42]

Answer:

Option a) circle 5 meters and 22 meters

Step-by-step explanation:

We are given the following information in the question:

A pair of diameter and the circumference is given. We have to find a correct approximations for the diameter and circumference.

a) circle 5 meters and 22 meters

$\text{Diameter} = 5\text{ meters}\\\text{Circumference} = \pi d = 3.14\times 5 = 15.7\text{ meters}$

b) 19 inches and 50 inches

$\text{Diameter} = 19\text{ inches}\\\text{Circumference} = \pi d = 3.14\times 19 = 59.66\text{ inches}$

c) 33 centimeters and 80 centimeters

$\text{Diameter} = 33\text{ centimeters}\\\text{Circumference} = \pi d = 3.14\times 33 =103.62\text{ centimeters}$

Thus, no pair gives a reasonable approximation. Only the circle with diameter 5 and circumference 22 meters have closest approximation.

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2 years ago

SAT Writing scores are normally distributed with a mean of 491 and a standard deviation of 113.A university plans to send letter

Sholpan [36]

Answer:

$z=1.405$

And if we solve for a we got

$a=491 +1.405*113=649.765$

So the value of height that separates the bottom 92% of data from the top 8% is 649.765.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(491,113)$

Where $\mu=491$ and $\sigma=113$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.08$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.92 of the area on the left and 0.08 of the area on the right it's z=1.405. On this case P(Z<1.405)=0.92 and P(z>0.92)=0.08

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.405$

And if we solve for a we got

$a=491 +1.405*113=649.765$

So the value of height that separates the bottom 92% of data from the top 8% is 649.765.

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2 years ago

A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th

QveST [7]

we know that

the volume of a cone is equal to

$V= \frac{1}{3} \pi r^{2}h$

in this problem

the radius is equal to

$r= \frac{112}{2}= 56ft$

1) <u>Find the height of the cone before the storm</u>

Applying the Pythagorean Theorem find the height

$h^{2} = l^{2}-r^{2}$

$l=65 ft$

$h^{2} = 65^{2}-56^{2}$

$h^{2} = 1,089$

$h=33 ft$

2) <u>Find the volume before the storm</u>

$V= \frac{1}{3}*\pi* 56^{2}*33$

$V=34,496\pi\ ft^{3}$

3) <u>Find the volume after the storm</u>

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

so

r=(56/3) ft

h=(33/3)=11 ft

$V= \frac{1}{3}*\pi* (56/3)^{2}*11$

$V= 1,277.63\pi\ ft^{3}$

<u>4) Find how this change affect the volume of the pile</u>

Divide the volume after the storm by the volume before the storm

$\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}$

therefore

<u>the answer part a) is</u>

The volume of the pile after the storm is $\frac{1}{27}$ times the original volume

<u>Part b)</u> Estimate the number of lane miles that were covered with salt

5) <u>Find the amount of salt that was used during the storm</u>

$=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}$

6) <u>Find the pounds of road salt used</u>

$104,358.59*80=8,348,687.2\ pounds$

7) <u>Find the number of lane miles that were covered with salt</u>

$8,348,687.2/350=23,853.39 \ lane\ miles$

therefore

<u>the answer part b) is</u>

the number of lane miles that were covered with salt is $23,853.39 \ lane\ miles$

<u>Part c) </u>How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to $1,277.63\pi\ ft^{3}$

$1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}$

8) <u>Find the pounds of road salt </u>

$4,013.79*80=321,103.20\ pounds$

9) <u>Find the number of lane miles </u>

$321,103.20/350=917.44 \ lane\ miles$

therefore

<u>the answer part c) is</u>

the number of lane miles is $917 \ lane\ miles$

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2 years ago

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Olenka [21]

Answer:

Conclusion

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Step-by-step explanation:

From the question we are told that

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The sample mean for Ascension parish is $\= x_2 = \$246,000$

The p-value is $p-value = 0.045$

The level of significance is $\alpha = 0.01$

The null hypothesis is $H_o : \mu_2 = \mu_1$

The alternative hypothesis is $H_a : \mu_2 > \mu_1$

Here $\mu_2$ is the population mean for Ascension parish

From the data given values we see that

$p-value > \alpha$

So we fail to reject the null hypothesis

So we conclude that there is no sufficient evidence to conclude that the mean of the home prices from Ascension parish is higher than the EBR mean

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ZanzabumX [31]

Answer:

dav d

Step-by-step explanation:

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