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2 years ago

Rosen 15 how many solutions are there to the equation x1 x2 x3 x4 x5

1 answer:

4 0

The number of answers always corresponds to the highest number of exponents. 

Therefore, if x⁵ is the highest exponent, then there are 5 possible answers.

However, it is important to note that not all of these have real answers for each one. In some cases, an x⁵ equation will have a number of real answers and a number of imaginary answers. Nevertheless, they will always add to 5.

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xxMikexx [17]

Answer:

Step-by-step explanation:jol

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2 years ago

Solve for x for #5 and 6

Alex777 [14]

Answer:

In 5:

UM and MS are equal since the diagonals US and VT have perpendicularly bisected each other.

So,15=6x-3

15+3=6x

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2 years ago

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It’s B!!!! Hope this helped;)

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2 years ago

A collection of 30 gems, all of which are identical in appearance, are supposedto be genuine diamonds, but actually contain 8 wo

levacccp [35]

Answer:

Step-by-step explanation:

From the given information:

There are 30 collections of gems, of which 8 are worthless;

Thus, the number of the genuine diamonds = 30 - 8 = 22.

Let X = random variable;

X consider the value as 0 (for 2 worthless stone selection),

X = 1200(1 worthless stone & 1 genuine stone)

X = 2400 (2 genuine stones selected)

However, the numbers of ways of selecting and chosen Gems can be estimated as:

$(^n_r) = (^{30}_2) \\ \\ \implies \dfrac{30!}{2!(30-2)!} \\ \\ \implies \dfrac{30!}{2!(28)!} \\ \\ \implies \dfrac{30*29*28!}{2!(28)!} \\ \\ \implies \dfrac{30*29}{2*1} \\ \\ \implies 435$

Thus;

$Pr (X = 0) = \dfrac{(^8_2)}{435}$

$Pr (X = 0) = \dfrac{\dfrac{8!}{2!(8-2)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8!}{2!(6)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8*7*6!}{2!(6)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8*7}{2*1}}{435} \\ \\ Pr (X = 0) = 0.0644$

$P(X =1200) = \dfrac{(^{8}_{1})(^{22}_{1})}{435}$

$P(X =1200) = \dfrac{ \dfrac{8!}{1!(8-1)!}) ( \dfrac{22!}{1!(22-1)!}) }{435}$

$P(X =1200) = \dfrac{ (8) ( 22) }{435}$

$P(X =1200) =0.4046$

$Pr (X = 2400) = \dfrac{(^{22}_2)}{435}$

$Pr (X = 2400) = \dfrac{\dfrac{22!}{2!(22-2)!}}{435} \\ \\ Pr (X = 2400) = \dfrac{\dfrac{22!}{2!(20)!}}{435} \\ \\ Pr (X = 2400) = \dfrac{\dfrac{22*21*20!}{2!(20)!}}{435} \\ \\ Pr (X =2400) = \dfrac{\dfrac{22*21}{2*1}}{435} \\ \\ Pr (X = 2400) = 0.5310$

To find E(X):

E(X) = (0 × 0.0644) + (1200 × 0.4046) + (2400 × 0.5310)

E(X) = 0 + 485.52 + 1274.4

E(X) = 1759.92

8 0

2 years ago

What is the missing number in this pattern? 0, 2, 6, ----, 30, 62

mario62 [17]

Answer:

Step-by-step explanation:

so the pattern is they add 2 then they add 4 then they add 8 then the add 16 basically they are double the number being added each number so the missing number is 14

5 0

2 years ago

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