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konstantin123 [22]
4 years ago
12

What is the shape of this distribution?

Mathematics
1 answer:
PolarNik [594]4 years ago
8 0

Answer:

A

Step-by-step explanation:

It is not E because it is not symmetric. It is not uniform because that is a rectangular looking distributiom. (not D)

It is not C or B because it is not unimodal. Unimodal means there is one mode (or one column that is the highest) I beleieve this one is A because it has to modes at 1 and 10

You might be interested in
Are these Equivalent expressions 4(2 + 3) and 4(2) + 4(3)
slava [35]

Answer:

Yes, they are equivalent.

Step-by-step explanation:

  1. To see if they're equivalent, you have to solve them. Start with the first one. Steps are below.
  2. Apply the distributed property: 8 + 12
  3. 8 + 12 = 20

For second expression:

  1. 4(2) = 4 × 2 = 8
  2. 4(3) = 4 × 3 = 12
  3. Put them together: 8 + 12
  4. 8 + 12 = 20

Making a conclusion:

Because they both have the same answer, they are equivalent expressions.

I hope this helps!

4 0
3 years ago
Read 2 more answers
Plz answer and thank you
icang [17]
A . It has a value GREATER than 8
4 0
4 years ago
Read 2 more answers
2(p + 1) = 18<br> what is p?
marshall27 [118]
It would be: 2(p + 1) = 18
= 2p + 2 = 18
= 2p = 18 - 2
= 2p = 16
= p = 16/2
= p = 8

In short, Your Answer would be 8

Hope this helps!
5 0
3 years ago
Read 2 more answers
A sphere and a cylinder have the same radius and height. The volume of the cylinder is 18 cm
Naily [24]

Answer:

12 cm³

Step-by-step explanation:

the volume of the sphere = 4/3 πr³

the volume of the cylinder (h=2r)= πr².2r

= 2πr³

the volume ratio of S : C =

4/3 πr³ : 2 πr³

= 4/3 : 2

= 4 : 6

= 2 : 3

so, the volume of the Sphere =

2/3 × 18 = 12 cm³

4 0
3 years ago
Read 2 more answers
Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin
ozzi

I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta  \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt

by using the integration formula

we get,

\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c

now put the value of t=\sin\theta in the above equation

we get,

\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c

hence proved

7 0
3 years ago
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