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3 years ago

In need of help , can someone please help ?

Mathematics

2 answers:

Ira Lisetskai [31]3 years ago

6 0

He final number is 5 times the original number

emmainna [20.7K]3 years ago

5 0

the answer is C.

I multiply 5x30 then I got 150 then I divided by 5 and I got 30 then I subtract 30-6 and got 24

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Tìm một đoạn nghiệm của phương trình 3/x+1 +4/(x+1)^2+5/(x+1)^3+6/(x+1)^4=10

SVETLANKA909090 [29]

Answer:

btfrt34598fc

Step-by-step explanation:

6 0

3 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.

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

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3 years ago

Orlando says that he can find the length of the diagonal, c, of the rectangular prism by

Zanzabum

Answer:

Step-by-step explanation:

6 0

3 years ago

14+12n=11n-37 what is N

zaharov [31]

Combine like terms and then simplify

8 0

3 years ago

Read 2 more answers

When using Cramer's Rule to solve a system of equations, if the determinant of the coefficient matrix equals zero and neither nu

o-na [289]

The answer would be A. When using Cramer's Rule to solve a system of equations, if the determinant of the coefficient matrix equals zero and neither numerator determinant is zero, then the system has infinite solutions. It would be hard finding this answer when we use the Cramer's Rule so instead we use the Gauss Elimination. Considering the equations:

x + y = 3 and 2x + 2y = 6
Determinant of the equations are 
| 1 1 | 
| 2 2 | = 0

the numerator determinants would be
| 3 1 | . .| 1 3 | 
| 6 2 | = | 2 6 | = 0.
Executing Gauss Elimination, any two numbers, whose sum is 3, would satisfy the given system. For instance (3, 0), (2, 1) and (4, -1). Therefore, it would have infinitely many solutions.

3 0

3 years ago