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3 years ago

50 points if you answer correctly

Mathematics

2 answers:

Andrei [34K]3 years ago

4 0

Answer:

1. D. 15

2. A. 3

3. A. c= 6g+5

4. A. All real numbers

5. D

rewona [7]3 years ago

3 0

Answer:

$f(3) = {3}^{2} + 5(3) - 9 \\ f(3) = 9 + 15 - 9 \\ f(3) = 15$

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P= $1,000, t= 6, r= 11%, compounded annually (k = 1)

kobusy [5.1K]

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3 years ago

4(3x - 2) - x = 3x - 4x

Oduvanchick [21]

Answer:

2/3

Step-by-step explanation:

Step 1:

4 ( 3x - 2 ) - x = 3x - 4x

Step 2:

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4 0

3 years ago

Read 2 more answers

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.

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

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3 years ago

Solve the inequality 3 - 5(2x + 1) 10.

inn [45]

3 - 50(2x + 1)
3 - 100x - 50
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3 years ago