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Maksim231197 [3]
3 years ago
14

The two-way frequency table below shows data from a random sample of 93 students on their favourite season and whether or not th

e student has allergies.
Which of the following statements is true about the students surveyed?
Choose 1 answer:

(Choice A)
A
A student with allergies is least likely to have spring as their favourite season.

(Choice B)
B
A student was most likely to have summer as their favourite season, whether or not they have allergies.

(Choice C)
C
A student with no allergies is more likely to have spring than fall as their favourite season.

(Choice D)
D
A student was more likely to have no allergies, no matter what their favourite season was.


Has allergies No allergies Total
Winter 5 9 14
Spring 6 7 13
Summer 13 23 36
Fall 14 16 30
Total 38 55 93
Mathematics
1 answer:
daser333 [38]3 years ago
8 0

Answer:

Choice D

Step-by-step explanation:

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Dafna1 [17]

22 pancakes

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7 0
3 years ago
How many diagonals can be constructed from one vertex of an n-gon? State your answer in terms of n and, of course, justify your
natulia [17]

Answer:

The \ total \ number \ of \ distinct \ diagonals \ in \ a \ polygon \ with \ n \ sides= \dfrac{n \times (n - 3)}{2}

Step-by-step explanation:

A diagonal is defined in geometry as a line connecting to two non adjacent  vertices.

Therefore, the minimum number of sides a polygon must have in order to have a diagonal n - 3 sides as the 3 comes from the originating vertex and the other two adjacent vertices

Given that the polygon has n sides, the number of diagonals that can be drawn from each of those n sides gives the total number of diagonals as follows;

Total possible diagonals = n × (n - 3)

However, half of the diagonals drawn within the polygon are the same diagonals drawn in reverse. Therefore, the total number of distinct diagonals that can be drawn in a polygon is given as follows;

The \ total \ number \ of \ distinct \ diagonals \ in \ a \ polygon \ with \ n \ sides= \dfrac{n \times (n - 3)}{2}

6 0
3 years ago
Suppose you take the small prism and stack it on top of the larger prism. What will be the volume of the composite figure?
Degger [83]
Find the volume of the first figure, then find the volume of the second figure, then add them together.
V=lwh
V(1)=(15)(12)(6)
= 1,080in^3
V(2)=(12)(6)(6)
= 432in^3
--------------------------------------------
1,080+432= 1,512 in^3
--------------------------------------------
Your answer should be 1,512
3 0
3 years ago
The lengths of the sides of a triangle are consecutive integers and the largest angle is twice the smallest angle. Find the meas
eduard

Answer:

  41°

Step-by-step explanation:

I could not think of an easy way to solve this, apart from having a graphing calculator do it. In the end, I found I could solve it analytically using a combination of the law of sines and the law of cosines.

Let x represent the length of the shortest side, and θ the smallest angle. Then the <em>law of sines</em> tells you ...

  sin(θ)/x = sin(2θ)/(x+2)

Cross-multiplying and using the trig identity for sin(2θ), we have ...

  (x +2)sin(θ) = 2x·sin(θ)cos(θ)

Dividing out sin(θ), we see that ...

  cos(θ) = (x+2)/(2x)

___

The law of cosines for the shortest side and smallest angle tells you ...

  x^2 = (x+1)^2 + (x+2)^2 - 2(x+1)(x+2)·cos(θ)

Substituting the above expression for cos(θ), this can be rewritten as ...

  0 = (x^2 +2x +1) +(x^2 +4x +4) -x^2 -(x+1)(x+2)^2/x

  0 = x^2 +6x +5 -(x+1)(x+2)^2/x . . . . . . collect terms outside the fraction

  0 = x(x+5)(x+1) -(x+1)(x+2)^2 . . . . . . . . factor and multiply by x

We know that x=-1 is not a solution, so we can divide by that factor:

  0 = x^2 +5x -(x^2 +4x +4) . . . . . multiply it all out

  0 = x -4 . . . . . . . . . . . . . . . . . . . . collect terms

  4 = x

so, cos(θ) = (4+2)/(2·4) = 6/8 = 3/4

and the angle of interest is ...

  θ = arccos(3/4) ≈ 41.40962° ≈ 41°

_____

The attachment shows a triangle-solver's result using the consecutive integers for side lengths. It confirms the answer we have here.

7 0
3 years ago
Creating and solving inequality ​
Tom [10]

Answer:

I decided to take a picture since it was easier for me that way. Hope you don't mind. Oh and if my number line isn't clear, the answer is the first number line

5 0
3 years ago
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