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3 years ago

Does (-5, -2) make the inequality -X- 2y 28 true?

Mathematics

1 answer:

yan [13]3 years ago

6 0

Answer:

Step-by-step explanation:

because if its -5 since it's in perenticies its multiplied negative five times negative two which equals negative 10 while x *2y which would be 28 equals x would be 14 and 2y would just be 2 so your answer is no

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For every blank squares, there are blank circles

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For every 3 squares there are 2 circles

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3 years ago

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Emily, Carl and Antony altogether has 121 stamps. How many stamps doeas each of them have, if it is known, that Emily has 40% mo

Leviafan [203]

Answer:

Step-by-step explanation:

1. First, put together the information we have. Total = 121. Emily has 40% more than Carl, and Carl has 60% more than Antony.

2. Next, set each person as a variable. Antony = x. Carl = 1.6x. Emily = 1.4 times 1.6x.

3. Next, form an equation using these variables.

x + 1.6x + (1.4 x 1.6x) = 121

x + 1.6x + 2.24x = 121

4.84x = 121

x = 25

4. Finally, plug in x to our previous variables in step #2 to find the number of stamps Emily and Carl have.

Antony: x = 25 stamps

Carl: 1.6x = 40 stamps

Emily: 1.4 times 1.6x = 56 stamps

By the way, is this for RSM? If so, I am working on that problem right now and I searched up the solution but couldn't find it, so I stumbled upon this. I hope this helped!

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3 years ago

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Answer:

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3 years ago

Classify the numbers as prime or composite. numbers:37, 65, 71, 82

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37:prime 65:composite 71:composite 82:composite

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3 years ago

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In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the

earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

A = video components need repair within 1 year

B = electronic components need repair within 1 year

C = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events A, B and C are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

$=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278$

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

$=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277$

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0

3 years ago