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Olin [163]
3 years ago
5

Simplify 12p-8+6p+14

Mathematics
2 answers:
Klio2033 [76]3 years ago
5 0
18p+6 would be your answer
12p+6p=18p
-8+14=6
Tatiana [17]3 years ago
3 0
Combine the like terms: 12p & 6p ; 8 & 14

12p - 6p = 6p

8 + 14 = 22 ; -8 + 14 = 6

Problem becomes and your answer is ➡
18z + 6
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5 5/6+1 1/4+1 1/4, or 8 1/3 miles

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Jonathan wants to save up enough money so that he can buy a new sports equipment set that includes football,baseball,soccer ball
JulsSmile [24]

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Tyler sells 20 of his records (r) and has 74 left. Which equation best<br> represents this scenario?
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Equation: r-20=74

Step-by-step explanation:

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6 0
2 years ago
17. Your goal is to save at least $360.00 over the next 6 weeks. How much money must you save each week in order to meet that go
OLga [1]
<h2>Answer </h2>

D. 6x>_360; x>_60

<h2>Explanation</h2>

Let x be amount you will save each weeks

Since we know that you are saving over a period of 6 weeks, you will save 6x.

We also now that your goal is to save at least $360.00 over the period of 6 weeks, so saving more than $360.00 will be very desirable, but the goal is to save $360.00. We can rephrase this as: You need to save $360.00 or more; we can say the same using the inequality symbol \geq (greater on equal than)

Now we can combine our tow parts using the inequality symbol:

6x\geq 360

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4 0
3 years ago
The Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are
almond37 [142]

Answer:

For x = 0, P(x = 0) = 0.35

For x = 1, P(x = 1) = 0.54

For x = 2, P(x = 2) = 0.11

For x = 3, P(x = 3) = 0

Step-by-step explanation:

We are given that the Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected, one at a time, without replacement.

Let X = <u><em>the number that are not correctly calibrated.</em></u>

Number of altimeters that are correctly calibrated = 6

Number of altimeters that are not correctly calibrated = 2

Total number of altimeters = 6 + 2 = 8

(a) For x = 0: means there are 0 altimeters that are not correctly calibrated.

This means that all three selected altimeters are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = ^{8}C_3

The number of ways of selecting 3 altimeters from a total of 6 altimeters that are correctly calibrated = ^{6}C_3

So, the required probability = \frac{^{6}C_3}{^{8}C_3}  

                                              = \frac{20}{56}  = <u>0.35</u>

(b) For x = 1: means there is 1 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 1 is not correctly calibrated and 2 are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = ^{8}C_3

The number of ways of selecting 2 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = ^{6}C_2

The number of ways of selecting 1 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = ^{2}C_1

So, the required probability = \frac{^{6}C_2 \times ^{2}C_1 }{^{8}C_3}  

                                                = \frac{30}{56}  = <u>0.54</u>

(c) For x = 2: means there is 2 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 2 are not correctly calibrated and 1 is correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = ^{8}C_3

The number of ways of selecting 1 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = ^{6}C_1

The number of ways of selecting 2 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = ^{2}C_2

So, the required probability = \frac{^{6}C_1 \times ^{2}C_2 }{^{8}C_3}  

                                                = \frac{6}{56}  = <u>0.11</u>

(d) For x = 3: means there is 3 altimeter that is not correctly calibrated.

This case is not possible, so this probability is 0.

6 0
3 years ago
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