4 years ago

Which pair show equivalent expressions

1 answer:

zysi [14]4 years ago

6 0

Second one if you distribute the first expression you get the second equation so the expressions are equivalent

You might be interested in

Dmitry_Shevchenko [17]

$\displaystyle\\|\Omega|=\binom{15}{3}=\dfrac{15!}{3!12!}=\dfrac{13\cdot14\cdot15}{2\cdot3}=455$

$\displaystyle\\|A|=\binom{10}{3}=\dfrac{10!}{3!7!}=\dfrac{8\cdot9\cdot10}{2\cdot3}=120\\\\P(A)=\dfrac{120}{455}=\dfrac{24}{91}\approx26.4\%$

$\displaystyle\\|A|=\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdot5}{2}=10\\\\P(A)=\dfrac{10}{455}=\dfrac{2}{91}\approx2.2\%$

$\displaystyle\\|A|=\binom{10}{2}\cdot5=\dfrac{10!}{2!8!}\cdot5=\dfrac{9\cdot10}{2}\cdot5=225\\\\P(A)=\dfrac{225}{455}=\dfrac{45}{91}\approx49.5\%$

$A$ - at least one is spoilt

$A'$ - none is spoilt

$P(A)=1-P(A')$

We calculated $P(A')$ in a).

Therefore

$P(A)=1-\dfrac{24}{91}=\dfrac{67}{91}\approx73.6\%$

7 0

2 years ago

Basile [38]

Answer:

sorry but I really don't know the answer.

Step-by-step explanation:

8 0

3 years ago

MrRissso [65]

The answer is 65. :)

4 0

3 years ago

Tamiku [17]

I believe the answer you're looking for is $9.88

3 0

3 years ago

horrorfan [7]

Yes.

-- Take 10/50.

-- Divide numerator and denominator by 2.5 ,
(Or multiply them by 0.4 .)

-- You now have 4/20 .

5 0

3 years ago