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slega [8]
3 years ago
12

lara has violet and blue pansies in groups of 5. she has x groups of violets and y groups of pansies. which expression shows tha

t total number of pansies that lara has?
Mathematics
2 answers:
umka21 [38]3 years ago
7 0
5y+5x

y=number of groups of blue pansies
x=number of groups of violet pansies
Hope this helps :)
Bess [88]3 years ago
7 0
5Y+5X!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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Step-by-step explanation:

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The most common form of color blindness is an inability to distinguish red from green. However, this particular form of color bl
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Answer:

(a) The correct answer is P (CBM) = 0.79.

(b) The probability of selecting an American female who is not red-green color-blind is 0.996.

(c) The probability that neither are red-green color-blind is 0.9263.

(d) The probability that at least one of them is red-green color-blind is 0.0737.

Step-by-step explanation:

The variables CBM and CBW are denoted as the events that an American man or an American woman is colorblind, respectively.

It is provided that 79% of men and 0.4% of women are colorblind, i.e.

P (CBM) = 0.79

P (CBW) = 0.004

(a)

The probability of selecting an American male who is red-green color-blind is, 0.79.

Thus, the correct answer is P (CBM) = 0.79.

(b)

The probability of the complement of an event is the probability of that event not happening.

Then,

P(not CBW) = 1 - P(CBW)

                   = 1 - 0.004

                   = 0.996.

Thus, the probability of selecting an American female who is not red-green color-blind is 0.996.

(c)

The probability the woman is not colorblind is 0.996.

The probability that the man is  not color- blind is,

P(not CBM) = 1 - P(CBM)

                   = 1 - 0.004  

                   = 0.93.

The man and woman are selected independently.

Compute the probability that neither are red-green color-blind as follows:

P(\text{Neither is Colorblind}) = P(\text{not CBM}) \times  P(\text{not CBW})\\ = 0.93 \times  0.996 \\= 0.92628\\\approx 0.9263

Thus, the probability that neither are red-green color-blind is 0.9263.

(d)

It is provided that a one man and one woman are selected at random.

The event that “At least one is colorblind” is the complement of part (d) that “Neither is  Colorblind.”

Compute the probability that at least one of them is red-green color-blind as follows:

P (\text{At least one is Colorblind}) = 1 - P (\text{Neither is Colorblind})\\ = 1 - 0.9263 \\= 0.0737

Thus, the probability that at least one of them is red-green color-blind is 0.0737.

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2 years ago
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Answer:

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Step-by-step explanation:

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