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Ainat [17]
3 years ago
11

3(5 − 2 x) = −2(6 – 3 x) − 10 x

Mathematics
1 answer:
kobusy [5.1K]3 years ago
5 0

Answer:

15-6x= -12-4x

15-2x= -12

-2x= -27

x= -13.5

Step-by-step explanation:

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Write 0.65% as a decimal
MAXImum [283]
The answer would be: 0.0065
4 0
3 years ago
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A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 17 randomly selected pens yield
Leno4ka [110]

Answer:

0.762 = 76.2% probability that this shipment is accepted

Step-by-step explanation:

For each pen, there are only two possible outcomes. Either it is defective, or it is not. The probability of a pen being defective is independent from other pens. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

17 randomly selected pens

This means that n = 17

(a) Find the probability that this shipment is accepted if 10% of the total shipment is defective. (Use 3 decimal places.)

This is P(X \leq 2) when p = 0.1. So

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{17,0}.(0.1)^{0}.(0.9)^{17} = 0.167

P(X = 1) = C_{17,1}.(0.1)^{1}.(0.9)^{16} = 0.315

P(X = 2) = C_{17,2}.(0.1)^{2}.(0.9)^{15} = 0.280

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.167 + 0.315 + 0.280 = 0.762

0.762 = 76.2% probability that this shipment is accepted

8 0
4 years ago
How many different ways can you make 82 cents using current u.s. currency
andrew11 [14]
 <span>You can probably just work it out. 

You need non-negative integer solutions to p+5n+10d+25q = 82. 

If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80. 


So this is the same as n + 2d + 5q ≤ 16 

So now you simply have to "crank out" the cases. 

Case q=0 [ n + 2d ≤ 16 ] 

Case (q=0,d=0) → n = 0 through 16 [17 possibilities] 
Case (q=0,d=1) → n = 0 through 14 [15 possibilities] 
... 
Case (q=0,d=7) → n = 0 through 2 [3 possibilities] 
Case (q=0,d=8) → n = 0 [1 possibility] 

Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81 

Case q=1 [ n + 2d ≤ 11 ] 
Case (q=1,d=0) → n = 0 through 11 [12] 
Case (q=1,d=1) → n = 0 through 9 [10] 
... 
Case (q=1,d=5) → n = 0 through 1 [2] 

Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42 

Case q=2 [ n + 2 ≤ 6 ] 
Case (q=2,d=0) → n = 0 through 6 [7] 
Case (q=2,d=1) → n = 0 through 4 [5] 
Case (q=2,d=2) → n = 0 through 2 [3] 
Case (q=2,d=3) → n = 0 [1] 

Total from case q=2: 1 + 3 + 5 + 7 = 16 

Case q=3 [ n + 2d ≤ 1 ] 
Here d must be 0, so there is only the case: 
Case (q=3,d=0) → n = 0 through 1 [2] 

So the case q=3 only has 2. 

Grand total: 2 + 16 + 42 + 81 = 141 </span>
3 0
3 years ago
Guys I need help, this is so important for my grade
Dmitrij [34]

Answer:

For Part C) use numbers 1 2 3 4.

Step-by-step explanation:

8 0
3 years ago
What is the answer to this x-7;x=23
blagie [28]

Answer:

16

Step-by-step explanation:

if x=23, 23-7=16

4 0
2 years ago
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