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4 years ago

I need some help on the question below please, and thank you.

Mathematics

1 answer:

White raven [17]4 years ago

6 0

It might be 31 if you add all the numbers together but I'm not positive.

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The sum of 1/9, 2/3, and 5/18is A. 8/30. B. 12/9. C. 19/18. D. 4/15.

mafiozo [28]

The least common multiple of 9, 3, and 18 is 18.

Convert 1/9 to 2/18.

Convert 2/3 to 12/18.

Then add the numerators:

2 + 12 + 5 = 19

So the answer is 19/18, or 1 1/18 if you prefer.

8 0

4 years ago

Read 2 more answers

Multiply(4×10^4)(3.63×10^-4) in scientifc notation

soldier1979 [14.2K]

Rewrite
(4)(3.63) * (10^4)(10^-4)
multiply
14.52 * (10^4)(10^-4)
multiplication of the coefficient is the addition of the exponent
so 4 -4 is 0

14.52

Hope this helps :)

5 0

4 years ago

What is the missing value? 85% of ____ is 136

spayn [35]

160 is the answer after solving.

7 0

4 years ago

Read 2 more answers

Calculating conditional probabilities - random permutations. About The letters (a, b, c, d, e, f, g) are put in a random order.

Evgesh-ka [11]

A="b is in the middle"

B="c is to the right of b"

C="The letter def occur together in that order"

a) b can be in 7 places, but only one is the middle. So, P(A)=1/7

b) X=i, "b is in the i-th position"

Y=j, "c is in the j-th position"

$P(B)=\displaystyle\sum_{i=1}^{6}(P(X=i)\displaystyle\sum_{j=i+1}^{7}P(Y=j))=\displaystyle\sum_{i=1}^{6}\frac{1}{7}(\displaystyle\sum_{j=i+1}^{7}\frac{1}{6})=\frac{1}{42}\displaystyle\sum_{i=1}^{6}(\displaystyle\sum_{j=i+1}^{7}1)=\frac{6+5+4+3+2+1}{42}=\frac{1}{2}$

P(B)=1/2

c) X=i, "d is in the i-th position"

Y=j, "e is in the j-th position"

Z=k, "f is in the i-th position"

$P(C)=\displaystyle\sum_{i=1}^{5}( P(X=i)P(Y=i+1)P(Z=i+2))=\displaystyle\sum_{i=1}^{5}(\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5})=\frac{1}{210}\displaystyle\sum_{i=1}^{5}(1)=\frac{1}{42}$

P(C)=1/42

P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420

$P(B\cap C)=\displaystyle\sum_{i=1}^{3} P(X=i)P(Y=i+1)P(Z=i+2)\displaystyle\sum_{j=i+3}^{6}P(V=j)P(W=j+1)=\displaystyle\sum_{i=1}^{3}\frac{1}{6}\frac{1}{7}\frac{1}{5}(\displaystyle\sum_{j=1+3}^{6}\frac{1}{4}\frac{1}{3})=1/420$

P(B∩A)=3*(1/7*1/6)=1/14

P(A|C)=P(A∩C)/P(C)=(1/420)/(1/42)=1/10

P(B|C)=P(B∩C)/P(C)=(1/420)/(1/42)=1/10

P(A|B)=P(B∩A)/P(B)=(1/14)/(1/2)=1/7

P(A∩B)=1/14

P(A)P(B)=(1/7)*(1/2)=1/14

A and B are independent

P(A∩C)=1/420

P(A)P(C)=(1/7)*(1/42)=1/294

A and C aren't independent

P(B∩C)=1/420

P(B)P(C)=(1/2)*(1/42)=1/84

B and C aren't independent

8 0

3 years ago

Evaluate. 21 ÷ 7 + 20 • 102 • 2 – 3

Novay_Z [31]

To avoid confusion, remember PEMDAS. This is the order of operation.

Parenthesis, Exponents, Multiply, Divide, Add, Subtract

21 ÷ 7 + 20 * 102 * 2 - 3

No parenthesis and no exponents, proceed to multiplication.

20 * 102 * 2 = 4,080

Then, perform division

21÷7 = 3

Then, do addition

3 + 4080 = 4083

Lastly, do subtraction

4,083 - 3 = 4,080

So, 21 ÷ 7 + 20 * 102 * 2 - 3 = 4,080

THEREFORE THE ANSWER IS 4,080

8 0

3 years ago