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torisob [31]
2 years ago
5

2)(40pts) A class contains 18 girls and 14 boys. For all parts of this question, each boy and girl are distinguishable from one

another. Answer the following questions:a)In how many ways can a committee of one boy and one girl be chosen
Mathematics
1 answer:
V125BC [204]2 years ago
4 0

Answer:

A committee of one boy and one girl can be chosen in 252 different ways.

Step-by-step explanation:

Given that a class contains 18 girls and 14 boys, and for all parts of this question, each boy and girl are distinguishable from one another, to determine in how many ways can a committee of one boy and one girl be chosen, the following calculation has to be done:

18 x 14 = X

252 = X

Therefore, a committee of one boy and one girl can be chosen in 252 different ways.

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Ms. Jan brought in cookies of her class. She gave out half of them in the morning. At lunch, she gave out 12 more. She then had
Svetach [21]

Answer:

Jan brought 44 cookies

Step-by-step explanation:

To solve this problem, we need to work backwards. First, we have 10 cookies left at the end of the day. At lunch, she gave away 12 cookies. This means that Ms. Jan had 22 cookies at lunch since  

10 + 12 = 22

In the morning, Ms. Jan gave out half of her cookies. This means that she started with twice as many cookies as she had at the beginning of lunch.  

22  x 2 = 44

Ms. Jan brought 44 cookies.

7 0
3 years ago
Someone please help with this question please​
natima [27]

Step-by-step explanation:

Since Angle DAE = Angle BCE, lines AD amd BC are parallel (by Z-angles).

This means that Angle ADE = Angle CBE (by Z-angles).

We have 2 congruent angles and 1 congruent side (AD = BC, given).

By ASA congruence, triangles AED and CEB are congruent.

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2 years ago
Read 2 more answers
Step by step show answer for problem 3-3×6+2
storchak [24]
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the answer would be     -13

4 0
3 years ago
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SOMEONE PLZ HELP MEEE
Step2247 [10]

Answer:

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Step-by-step explanation:

6 0
3 years ago
Find the equation of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit.
leonid [27]

<u>Answer-</u>

The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are

12x-5y+14=0 \\ 12x-5y-14=0

<u>Solution-</u>

Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)

The applying the distance formula,

\Rightarrow \left | \frac{12h-5k-1}{\sqrt{12^2+5^2}} \right |=1

\Rightarrow \left | \frac{12h-5k-1}{\sqrt{169}} \right |=1

\Rightarrow \left | \frac{12h-5k-1}{13} \right |=1

\Rightarrow 12h-5k-1=\pm 13

\Rightarrow 12h-5k=\pm 14

\Rightarrow 12h-5k=14,\ 12h-5k=-14

\Rightarrow 12h-5k-14=0,\ 12h-5k+14=0

\Rightarrow 12x-5y-14=0,\ 12x-5y+14=0

Two equations are formed because one will be upper from the the given line and other will be below it.

4 0
2 years ago
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