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Semenov [28]
2 years ago
13

Egbert is making trail mix out of 18 bags of nuts and 9 bags of dried fruit. He wants each new portion of trail mixto be identic

al, containing the same combination of bags of nuts and bagsof dried fruit, with no bags left over. What is the greatest number of portions of trail mix Egbert can make? ​
Mathematics
1 answer:
Over [174]2 years ago
8 0

Answer:

2 bags of deezznuts and 1 bag of fruit

Step-by-step explanation:

18 bags of nuts

9 bags of fruit

2 bags of nuts mixed with 1 bag of fruit, multiply that by nine and you have no remainder on any of the bags

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Elis [28]

Answer:

I think its BC and FD

4 0
2 years ago
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Solve the system by substitution <br><br> -x-y-z= -8<br><br> -4x+4y+5z= 7<br><br> 2x+2z= 4
Gennadij [26K]
- x - y - z = - 8 ------- (1)
- 4x + 4y + 5z = 7 -------- (2)
2x + 2z = 4 --------- (3)

From (3): x + z = 2
                      x = 2 - z --------- (4)

Substitute (4) into (1),
- x - y - z = - 8
- (2-z) - y - z = - 8
- 2 + z - y - z = -8
- 2 - y = - 8
2 + y = 8
Therefore, y = 6

Substitute y = 6 and (4) into (2),
- 4x + 4y + 5z = 7
- 4(2 - z) + 4(6) + 5z = 7
- 8 + 4z + 24 + 5z = 7
9z + 16 = 7
9z = - 9
Therefore, z = - 9

Substitute z = - 9 into (4),
x = 2 - z
x = 2 - (-9)
x = 2 + 9
Therefore x = 11

Solution:
x = 11
y = 6
z = -9
4 0
2 years ago
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Joseph squares a number and subtracts 8 from the answer. He then divides this result by 7. Finally he adds 12 and the result is
ivanzaharov [21]
Let the number be X, then we find
( {x}^{2} - 8) \div 7 + 12 = 20
( {x}^{2} - 8) \div 7 = 8
( {x}^{2} - 8) = 56
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x = \sqrt{64} \: \: \: \: \: x = - \sqrt{64}
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5 0
2 years ago
An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do
vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107

Then,

P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007

c) In this case, the distribution is B(1200,0.01)

P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366

5 0
2 years ago
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So the question tells to express the expression in your problem where N0 is N-naught and the symbol represent the lower case Greek letter lambda. So the best answer or expression would be that the lambda is the wavelength of the expression. I hope you are satisfied with my answer 
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