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kogti [31]
2 years ago
12

Select the correct answer from the drop-down menu.

Mathematics
2 answers:
Genrish500 [490]2 years ago
6 0

ANSWER: control group

EXPLANATION: valid for edmentum/plato

Fittoniya [83]2 years ago
3 0

Answer:

control group

Step-by-step explanation:

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Jorge wrote the following key for a scale drawing of his garden: 1 cm 2 ft Express this scale as inches to feet.
Andru [333]
We know that

the scale (cm/feet)-----------> 1/2

1 in-------------> 2.54 cm
x in---------------> 1 cm
X=1/2.54------> 0.3937 in

so
 the new scale (in/feet)------------> 0.3937 in/ 2 feet-------> 0.1969 in/feet

the answer is 
0.1969 in/feet

7 0
3 years ago
The Original Price of an item is $150.00. It is on sale for 30% off. You also have a store coupon for 10% off of the sale price.
kumpel [21]
30% + 10% = 40% off
150 * 0.40 = $60 off
150 - 60 = $90
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3 years ago
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M&M plain candies come in various colors. According to the M&M/Mars Department of Consumer Affairs, the distribution of
lyudmila [28]

Answer:

A) 0.12. Yes. Choosing a green and blue M&M is possible

B) 0.43. Yes. Choosing a yellow and red M&M is possible

C) 0.78

Step-by-step explanation:

First of all, the summation of the distribution of all colours is;

Σ(all colors ) = 22% + 20% + 23% + 10% + 6% + 6% + 13% = 100%, or 1.

Thus;

a) P(green candy or blue candy) is;

P(GREEN ∪ BLUE) = P(G) + P(BL)

P(GREEN ∪ BLUE) = 6%+6%

P(GREEN ∪ BLUE) = 12% or 0.12

Now, due to the fact that we have to choose ONE candy and only ONE candy at random, then they are mutually exclusive: Yes. Choosing a green and blue M&M is possible

b)P(yellow candy or red candy is;

P(YELLOW ∪ RED) = P(Y) + P(R)

P(YELLOW ∪ RED) = 20% + 23% = 43% or 0.43

Yes. Choosing a yellow and red M&M is possible

c) P(NOT PURPLE)

the probability of having a purple is;

P(PURPLE) = 22% or 0.22

So, the Probability of NOT having a PURPLE is 1 - 0.22 = 0.78

4 0
3 years ago
Why 5/8 is bigger than 7/12 ?
Andrew [12]
Because the larger the denominator, the smaller the value (especially if the numerators are the same).
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