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3 years ago

Please help solve much appreciated

Mathematics

1 answer:

DIA [1.3K]3 years ago

7 0

Step-by-step explanation:

For the Cuboid, the formula will be:

Width x Height x Length

For the cylinder the volume formula will be:

$\pi \: \times \: {radius}^{2} \times \: length$

You might be interested in

In a three-dimensional coordinate system, which of the following generally points upward?

tatiyna

In a three-dimensional coordinate system, the following generally points upward
the positive y-axis
b is the right option
hope this helps

5 0

3 years ago

If anyone could solve this question for me while showing the steps. It would be GREATLY appreciated

Blizzard [7]

this is a very annoying problem, with lots of steps. it is on a next level of dragging on and on

Step-by-step explanation:

You can split this into different shapes. A triangle, a rectangle, and a void of a quarter circle. The total area is rectangle+triangle-quarter circle.

We find the second side of the triangle by doing 65-48=17

We then use the pythagorean theorem to find the third length. This is about 49.1426.

We can then use whatever method you'd like for finding the area of the triangle. it turns out to be about 417.7125

Next, we find the rectangle, easy enough, 49.1426*48=2358.8448

which seems a bit scuffed until you realize that the diagram itself is scuffed as well. Great, quarter circle next

16*16*pi/4=201.0619

Alright, now to combine it to find the area.

417.7125+2358.8448-201.0619=2575.4954

great, now let's find the perimeter, simple enough. Curved perimeter+straight perimeter-that one corner in the bottom right

52+48+49.1426+65-32+25.1327=207.2753

and now we convert to meters

207.2753feet=63.1775meters

We are almost done! Now let's find the cost.

Fencing first:

63.1775*8.50=537.01

Then the cement

2575.4954*5.50=14165.2247

Then we add it up and apply tax:

537.01+14165.2247=14702.2347

14702.23*1.13=$16613.53

if you find anything wrong message me, I may have made errors because its 10pm and i need sleep. I recommend you work through my steps on your own as well. There might be complications with rounding as well, so yea. Hope this helps!

3 0

2 years ago

Which lines in the graph have a slope greater than 1 but less than 2?

lapo4ka [179]

Answer:

Lines 3 and 4

Step-by-step explanation:

Let's examine the slope of each of the given lines to have the exact value and be able to answer the question. We use the definition of: $slope=\frac{rise}{run}$

Line 1: $slope=\frac{rise}{run}=\frac{6}{2} = 3$

Line 2: $slope=\frac{rise}{run}=\frac{6}{3} = 2$

Line 3: $slope=\frac{rise}{run}=\frac{6}{4} = 1.5$

Line 4: $slope=\frac{rise}{run}=\frac{6}{5} = 1.2$

Line 5: $slope=\frac{rise}{run}=\frac{6}{6} = 1$

Therefore the lines that have slope strictly greater than 1 and less than 2 are:

Lines 3 and 4

3 0

3 years ago

Read 2 more answers

To avoid modeling the physical characteristics of wheel encoders such the square wave outputs, it is desirable to generate encod

grandymaker [24]

Answer:

The probability distribution is Normal continuous

Step-by-step explanation:

The basic idea is that velocity values have different noise level and an important thing regarding continuous probability distributions is that the probability of the random variable is equal to a specific outcome is 0

In other words, is practically impossible that one value of velocity could be the same as others.

5 0

3 years ago

Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.

adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0

3 years ago