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Sidana [21]
3 years ago
8

Which of the following expressions represent the distance between - 3 and 5 on the number line?

Mathematics
1 answer:
garri49 [273]3 years ago
7 0

Answer:

ahah woavwoqbwbaiwviabwoa owbwvuqbwiabw iaba i

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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
3 years ago
P( x) =x^3-6x^2-5x-14 What is the remainder R when the polynomial p(x) is divided by (x-7)
Igoryamba

Answer:

The remainder is zero

Step-by-step explanation:

To find the remainder we will use the long division

\frac{x^{3}-6x^{2}-5x-14}{x-7}=x^{2}\frac{x^{2}-5x-14 }{x-7}⇒(1)

\frac{x^{2}-5x-14}{x-7}=x\frac{2x-14}{x-7}⇒(2)

\frac{2x-14}{x-7}=2⇒(3)

From (1) , (2) and (3)

The quotient of the long division is x^{2}+x+2 and no remainder

So the remainder is zero

* If you want to check your answer Multiply the quotient by the divisor

(x^{2}+x+2)(x-7)=x^{3}-6x^{2}-5x-14

8 0
3 years ago
Find the tangent plane to the given surface of f(x,y)=6- 6/5 x-y at the point (5, -1, 1). Make sure tat your final answer for th
HACTEHA [7]

Answer:

Required equation of tangent plane is z=\frac{6}{5}(x-5y-11).

Step-by-step explanation:

Given surface function is,

f(x,y)=6-\frac{6}{5}(x-y)

To find tangent plane at the point (5,-1,1).

We know equation of tangent plane at the point $(x_0,y_0,z_0)[/tex] is,

z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\hfill (1)

So that,

f(x_0,y_0)=6-\frac{6}{5}(5+1)=-\frac{6}{5}

f_x=-\frac{6}{5}y\implies f_x(5,-1,1)=\frac{6}{5}

f_y=-\frac{6}{5}x\implies f_y(5,-1,1)=-6

Substitute all these values in (1) we get,

z=\frac{6}{5}(x-5)-6(y+1)-\frac{6}{5}

\therefore z=\frac{6}{5}(x-5y-11)

Which is the required euation of tangent plane.

4 0
3 years ago
What is (5a2 − 3a + 8) + (−4a2 − 1) + (15a + 11) expressed in its simplest form?
murzikaleks [220]

The Answer is:

(5a2-3a+8)+(-4a2-1)+(15a+11)

3 0
3 years ago
Read 2 more answers
35 points!! How much does a fish weigh if its tail weighs 4 kilograms, it's head weighs as much as it's tail and half its body,
NeX [460]

Let t, h, b represent the weighs of tail, head, and body, respectively.

t = 4 . . . . given

h = t + b/2 . . . . the head weighs as much as the tail and half the body

b/2 = h + t . . . . half the body weighs as much as the head and tail

_____

Substituting for b/2 in the second equation using the expression in the third equation, we have

... h = t + (h + t)

Subtracting h from both sides gives

... 0 = 2t . . . . . . in contradiction to the initial statement about tail weight.

Conclusion: there's no solution to the problem given here.

4 0
3 years ago
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