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denis-greek [22]
3 years ago
7

The mass of a block of stone is 6400 kg. If the block has a volume of 0.8 m cubed, what is its density? Remember, density is mas

s per unit of volume.
Please help ... due in less than 12 hours.
Mathematics
1 answer:
vovangra [49]3 years ago
7 0
The question is asking for mass/unit of volume. 
Therefore the density of the material would be 6400kg/0.8m^3. You then want to change the bottom into a 1 by multiplying the top and bottom both by 10/8. You now have 8000kg/m^3, which is the solution. 

8000  \frac{kg}{m^3}
You might be interested in
A vertical right circular cylindrical tank has height h=8 feet high and diameter d=6 feet. It is full of kerosene weighing 50 po
Mariana [72]

Answer:

The work done to  pump all of the kerosene from the tank to an outlet is W=45238.9\: J  

Step-by-step explanation:

The work is defined by:

W=\int dFdx (1)    

The force here will be the product between the volume and the kerosene weighing, so we have :

dF=\pi R^{2}dy*50

This force will be in-lbs.

Where R is the radius (3 feet)                    

Then using (1), we have:

W=\int \pi R^{2}dy*50(8-y)  

Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.

W=\int_{0}^{8} \pi 3^{2}dy*50(8-y)

W=450\pi \int_{0}^{8}dy(8-y)

W=450\pi(\int_{0}^{8} 8dy-\int_{0}^{8} ydy)

W=450\pi(8y|_{0}^{8} -\frac{y^{2}}{2}|_{0}^{8})  

W=450\pi(8*8 -\frac{8^{2}}{2})

W=450\pi(64 -\frac{64}{2})

Therefore, the work done to  pump all of the kerosene from the tank to an outlet is W=45238.9\: J  

I hope it helps you!  

6 0
3 years ago
Type the correct answer in each box. Use numerals instead of words. Any non-integer answers in this problem should be entered as
ladessa [460]

Answer:

The interquartile range of the data set is 5

The mean absolute deviation of the data set is 3.6

Step-by-step explanation:

* Lets explain how to find the interquartile range and the mean absolute

  deviation (MAD)

- The steps to find the interquartile range is:

1- Arrange the values from the smallest to the largest

∴ The values are 5 , 6 , 10 , 11 , 12 , 13 , 14 , 15 , 18 , 20

2- Find the median

- The median is the middle value after arrange them

* If there are two values in the middle take their average

∵ The values are 10 then the 5th and the 6th are the values

∵ The 5th is 12 and the 6th is 13

∴ The median = \frac{12+13}{2}=12.5

∴ The median is 12.5

3- Calculate the median of the lower quartile

- The lower quartile is the median of the first half data values

∵ There are 10 values

∴ The first half is the first five values

∴ The first half values are 5 , 6 , 10 , 11 , 12

∵ The middle value is 10

∴ The median of lower quartile = 10

- Similar find the median of the upper quartile

- The upper quartile is the median of the second half data values

∵ There are 10 numbers

∴ The second half is the last five values

∴ The second half values are 13 , 14 , 15 , 18 , 20

∵ The middle value is 15

∴ The median of upper quartile = 15

4- The interquartile range (IQR) is the difference between the upper

    and the lower medians

∴ The interquartile range = 15 - 10 = 5

* The interquartile range of the data set is 5

* Lets talk about the mean absolute deviation

- Mean absolute deviation (MAD) of a data set is the average distance  

 between each data value and the mean

- To find the mean absolute deviation of the data, start by finding

   the mean of the data set.  

1- Find the sum of the data values, and divide the sum by the  

   number of data values.  

∵ The data set is 5 , 6 , 10 , 11 , 12 , 13 , 14 , 15 , 18 , 20

∵ Its sum = 5 + 6 + 10 + 11 + 12 + 13 + 14 + 15 + 18 + 20 = 124

∵ The mean = the sum of the data values/the number of the data

∵ The set has 10 numbers

∴ The mean = 124/10 = 12.4

2- Find the absolute value of the difference between each data value  

   and the mean ⇒ |data value – mean|

# I5 - 12.4I = 7.4

# I6 - 12.4I = 6.4

# I10 - 12.4I = 2.4

# I11 - 12.4I = 1.4

# I12 - 12.4I = 0.4

# I13 - 12.4I = 0.6

# I14 - 12.4I = 1.6

# I15 - 12.4I = 2.6

# I18 - 12.4I = 5.6

# I20 - 12.4I = 7.6

3- Find the sum of the absolute values of the differences.

∵ Their sum = 7.4 + 6.4 + 2.4 + 1.4 + 0.4 + 0.6 + 1.6 + 2.6 + 5.6 + 7.6

∴ Their sum = 36

4- Divide the sum of the absolute values of the differences by the

    number of data values  to find MAD

∴ MAD = The sum of the absolute values/number of the values

∵ The sum = 36

∵ The data set has 10 values

∴ MAD = 36/10 = 3.6

* The mean absolute deviation of the data set is 3.6

6 0
3 years ago
Based on Pythagorean identities, which equation is true? A. Sin^2 theta -1= cos^2 theta B. Sec^2 theta-tan^2 theta= -1 C. -cos^2
Arturiano [62]

Answer:

D

Step-by-step explanation:

our basic Pythagorean identity is cos²(x) + sin²(x) = 1

we can derive the 2 other using the listed above.

1. (cos²(x) + sin²(x))/cos²(x) = 1/cos²(x)

1 + tan²(x) = sec²(x)

2.(cos²(x) + sin²(x))/sin²(x) = 1/sin²(x)

cot²(x) + 1 = csc²(x)

A. sin^2 theta -1= cos^2 theta

this is false

cos²(x) + sin²(x) = 1

isolating cos²(x)

cos²(x) = 1-sin²(x), not equal to sin²(x)-1

B. Sec^2 theta-tan^2 theta= -1

1 + tan²(x) = sec²(x)

sec²(x)-tan(x) = 1, not -1

false

C. -cos^2 theta-1= sin^2

cos²(x) + sin²(x) = 1

sin²(x) = 1-cos²(x), our 1 is positive not negative, so false

D. Cot^2 theta - csc^2 theta=-1

cot²(x) + 1 = csc²(x)

isolating 1

1 = csc²(x) - cot²(x)

multiplying both sides by -1

-1 = cot²(x) - csc²(x)

TRUE

3 0
3 years ago
Find the area of the shape below, giving your answer to 1 decimal place
Andre45 [30]

Answer:the answer is 150

Step-by-step explanation:multiply 10 and 15 and you get 150

7 0
3 years ago
Read 2 more answers
A series of three? separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers. Each
Nadya [2.5K]

Answer:

312.5\pi \text{ km}^3\approx 981.75\text{ km}^3

Step-by-step explanation:

We have been given that a series of 3 separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers.

Each of the three tunnels is shaped like a half-cylinder with a radius of 5 meters.

Since we know that volume of a semicircular or a half cylinder is half the volume of a circular cylinder.

\text{Volume of a semicircular cylinder}=\frac{\pi r^2h}{2}, where,

r = Radius of cylinder,

h = height of the cylinder.

Upon substituting our given values in volume formula we will get,

\text{Volume of a semicircular cylinder}=\frac{\pi (5\text{ km})^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*25\text{ km}^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*625\text{ km}^3}{2}

\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3

\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3

\text{Volume of a semicircular cylinder}=981.74770\text{ km}^3

Therefore, the volume of earth removed to build the three tunnels is 312.5\pi \text{ km}^3\approx 981.75\text{ km}^3.


4 0
3 years ago
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