Can some one help me i need help it is 6x-1/5=3/1 but its not hole numbers

Mathematics

2 answers:

baherus [9]2 years ago

8 0

X = 8/15
6x - 1/5 = 3/1
6x - 1/5 = 3
6x = 3 + 1/5
6x = 15 + 1 over 5
6x = 16/5
x = 16/30
Then, you simplify.
x = 8/15

VARVARA [1.3K]2 years ago

5 0

6x - 1/5 = 3/1...u want to get rid of the fractions...multiply everything by the common denominator which is 5

giving u :
30x - 1 = 15....add 1 to both sides
30x = 15 + 1
30x = 16...divide both sides by 30
x = 16/30 which reduces to 8/15

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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)

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