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Montano1993 [528]
3 years ago
6

A kiddie pool holds miniature plastic ducks in different colors. Sophia randomly picks up a duck and then replaces it. She repea

ts this process 100 times. The table shows the results of her experiment.
Duck Color Times Picked
red 30
green 20
yellow 40
blue 10

What is the relative frequency of choosing a red duck? What color of duck is there likely to be the most of in the pool?

A.
The relative frequency is 0.3. It is likely that most ducks are blue.
B.
The relative frequency is 0.2. It is likely that most ducks are yellow.
C.
The relative frequency is 0.3. It is likely that most ducks are yellow.
D.
The relative frequency is 0.2. It is likely that most ducks are green.
Mathematics
2 answers:
Elis [28]3 years ago
8 0

Answer:

C.

The relative frequency is 0.3. It is likely that most ducks are yellow.

Step-by-step explanation:

aliya0001 [1]3 years ago
4 0

Answer:

A is the answer

Step-by-step explanation:

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Prove that the triangle EDF is isosceles. Give reasons for your answer.
Gekata [30.6K]

Answer:

\triangle EDF is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

\text{External }\angle DFG = 90 +\dfrac{y}{2}

Let us try to find out \angle E and \angle DFE. After that we will compare them.

<u>Finding </u>\angle DFE<u>:</u>

Side EG is a straight line so \angle GFE = 180

\angle GFE is sum of internal \angle DFE and external \angle DFG

\angle GFE = 180 = \angle DFE  + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)

<u>Finding </u>\angle E<u>:</u>

<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external \angle DFG = \angle E + \angle EDF

\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2}  -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)

Comparing equations (1) and (2):

It can be clearly seen that:

\angle DFE = \angle E =90-\dfrac{y}{2}

The two angles of \triangle EDF are equal hence \triangle EDF is isosceles.

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Answer:

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that's it

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