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3 years ago

Solve the inequality 3(x-1)<-3(2-2x)

Mathematics

2 answers:

borishaifa [10]3 years ago

8 0

i believe the answer would be x > 1?

sattari [20]3 years ago

5 0

Answer:

x > 1

Step-by-step explanation:

3(x-1)<-3(2-2x)

=> 3x +3 < 6x

=> 1<x

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Mia is three years older than twice her sister brooke's age. The sum of their ages is less than 30. what is the greatest age Bro

Pachacha [2.7K]

Answer: 9

Step-by-step explanation: You can start by setting this up as an equation, where X is Brooke’s age.

X + 2(x)+ 3 = 30

The 2(X) + 3 signifies the part of the problems identifying that Mia is twice her sisters age plus 3 years.

Combine x’s

3x + 3 = 30

Subtract 3 from both sides to isolate the x term

3x= 27

Divide both sides by 3 to reduce to a single x

X= 9 , which is brooke’s age

7 0

3 years ago

Can someone plzz help me on thiss!

aev [14]

Answer:

Step-by-step explanation:

because it goes up constantly by 12

5 0

3 years ago

Please answer QUICKLY!!!

sergiy2304 [10]

Answer:

the range is {7,5,1}

Step-by-step explanation:

To find the range, just plug in the domain into the function.

f(-1) = -2(-1) + 5 = 7

f (0) = -2(0) + 5 = 5

f(2) = -2(2) + 5 = 1

8 0

1 year 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)

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