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lana66690 [7]
3 years ago
9

claudia is making a gelatin desseert for a party.She plans on making 12 servings for every 5 people.If each pan claudia uses to

make the dessert holds 8 servings, what is the minimum number of these pans that shpe needs in order to make enough to feed 10 people?
Mathematics
1 answer:
prisoha [69]3 years ago
5 0
First we need to find the total amount of servings needed.

(x2) 12 servings/5 people (x2)
        24 servings/10 people

Now we can find the amount of pans needed

24 servings/8 servings (1 pan)
=3 pans

She needs at least 3 pans
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Answer:

3. The missing angle is 56°

4. x = 7

Step-by-step explanation:

3.

We know sum of 3 angles in a triangle is 180°.

Looking at the top triangle, we can figure out the third angle. Let third angle be x:

85 + 35 + x = 180

120 + x = 180

x = 180 - 120

x = 60

<u>The angle "x" and the angle that is missing from the "bottom" triangle in the figure, are vertical angles, and hence, are EQUAL.</u>

So the bottom triangle now has 2 angles, 60 and 64 (given). Let the third angle be y(the one with a question mark). So we can write:

60 + 64 + y = 180

124 + y = 180

y = 180 - 124

y= 56

This is the missing angle.

4.

10x - 5 AND 8x + 9 are vertical angles. They ARE EQUAL.

Thus we can write the equation:

(10x-5) =  (8x+9)

10x-8x=9+5

2x=14

x=14/2

x=7

So x = 7

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Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random
lesya692 [45]
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The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
P(A|B)= \frac{P(A \cap B)}{P(B)}

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is \frac{1}{3}

Part A:
To find <span>the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}

</span><span>P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

</span><span>Thus P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}

Therefore, P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}


Part B:
</span>To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime<span>, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
</span><span>
P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}

</span><span>P(D)= \frac{1}{3}</span><span>

Therefore, P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}</span>
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