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jarptica [38.1K]
3 years ago
10

Find 30% decrease from 95 68.40 63.20 66.50

Mathematics
1 answer:
Kay [80]3 years ago
8 0

Answer:

66.50

Step-by-step explanation:

90 decreased by 30% is 66.50!

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Is finding 9/10-1/2 the same as finding 9/10-1/4-1/4?explain.
zavuch27 [327]
Yes it is the same. 1/4 +1/4 = 1/2. so 9/10-1/2=9/10-1/2
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Please help, pic attached!
RUDIKE [14]

Answer:

(y − 9) (y − 2)

Step-by-step explanation:

Factor  y^2 − 11y + 18 using the AC method.

(y − 9) (y − 2)

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Maria just got a raise at work. She was making $9 per hour, but now she makes 150% of that amount.
Stolb23 [73]

Answer:

13.50

Step-by-step explanation:

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100. 9

5 0
3 years ago
Read 2 more answers
The most common form of color blindness is an inability to distinguish red from green. However, this particular form of color bl
Alecsey [184]

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.

6 0
3 years ago
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