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3 years ago

A 3-yard roll of brown paper costs $1.14.What is the unit price HELP PLEASE

Mathematics

2 answers:

Lapatulllka [165]3 years ago

8 0

Per 1 yard the cost would be $0.38. Not sure what unit you need it in, but thats the answer in yards.

Butoxors [25]3 years ago

8 0

Answer:

0.38 or 38 cents

Step-by-Step:

1.14 divided by 3 is 0.38

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What is the volume of the composite figure? Explain your work. A complete answer should include how you broke up the figure, whi

Sphinxa [80]

Answer:

$15,000\:\mathrm{mm^3}$

Step-by-step explanation:

The composite figure consists of a square prism and a trapezoidal prism. By adding the volume of each, we obtain the volume of the composite figure.

The volume of the square prism is given by $V=s^2\cdot h$ , where $s$ is the base length and $h$ is the height. Substituting given values, we have: $V=14^2\cdot 30=196\cdot 30=5,880\:\mathrm{mm^3}$

The volume of a trapezoidal prism is given by $V=\frac{b_1+b_2}{2}\cdot l\cdot h$ , where $b_1$ and $b_2$ are bases of the trapezoid, $l$ is the length of the height of the trapezoid and $h$ is the height. This may look very confusing, but to break it down, we're finding the area of the trapezoid (base) and multiplying it by the height. The area of a trapezoid is given by the average of the bases ( $\frac{b_1+b_2}{2}$ ) multiplied by the trapezoid's height ( $l$ ).

Substituting given values, we get:

$V=\frac{14+24}{2}\cdot (30-14)\cdot 30,\\V=19\cdot 16\cdot 30=9,120\:\mathrm{mm^3}}$

Therefore, the total volume of the composite figure is $5,880+9,120=\boxed{15,000\:\mathrm{mm^3}}$ (ah, perfect)

Alternatively, we can break the figure into a larger square prism and a triangular prism to verify the same answer:

$V=30^2\cdot 14+\frac{1}{2}\cdot10\cdot 16\cdot 30=\boxed{15,000\:\mathrm{mm^3}}\checkmark$

8 0

2 years ago

Which would you be most likely to measure using yards?

kvasek [131]

A) i would say inches
B) miles
C) i would say inches
D) i would say is the correct answer!!!!

ANSWER IS D

6 0

3 years ago

Read 2 more answers

The managers of 21 supermarkets counted the number of cars in their parking lots on the same day. The results are shown in the l

mixer [17]

The IQR is 42.5

Step-by-step explanation:

Interquartile range is the difference of third and first quartile.

First of all we have to find the median for that purpose the data has to be arranged in ascending order. The data is already in ascending order.

As the number of values are odd

n=21

The median will be: $(\frac{n+1}{2}) th\ term$

Putting n=21

$(\frac{21+1}{2})th\ term\\=(\frac{22}{2})th\ term\\= 11th\ term$

The 11th term is 133

So median = 133

Now the data is divided into two halves

One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132

2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q1 will be the median of first half and Q3 will be the median of 2nd half.

As now the halves contain even number of values, the medians will be the average of middle two values

For First Half:

98, 100, 101, 102, 108, 109, 111,118, 129, 132

$Q_1 = \frac{108+109}{2}\\Q_1 = \frac{217}{2}\\Q_1 = 108.5$

For Second Half:

135, 135, 145, 146, 146, 156, 170 176, 180, 180

$Q_2 = \frac{146+156}{2}\\Q_2 = \frac{302}{2}\\Q_2 = 151$

Now

Interquartile Range:

$IQR = Q_3-Q_1\\= 151-108.5\\=42.5$

Hence,

The IQR is 42.5

Keywords: Median, IQR

Learn more about median at:

#LearnwithBrianly

7 0

3 years ago

Find the length of -- CD.

Sunny_sXe [5.5K]

D. About 3.2 units is the correct answer

6 0

3 years ago

The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6.

Wittaler [7]

Answer:

a) There is a 10.75% probability of observing less than 60 hits in an hour.

b) The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) There is a 24% probability of observing between 80 and 90 hits an hour

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem, we have that

The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6, so $\mu = 75, \sigma = 8.6$ .