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2 years ago

Anyone wanna give me stuff in rocket league pc hehehehehe

Mathematics

2 answers:

adell [148]2 years ago

7 0

Answer:

not really hehehehehe

Step-by-step explanation:

Juliette [100K]2 years ago

3 0

Maybe. What’s your name?

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This circle is centered at the origin, and the length of its radius is 8. What is

Anton [14]

Answer: The equation of a circle with center (h,k) and radius r is given by (x−h)2+(y−k)2=r2 . For a circle centered at the origin, this becomes the more familiar equation x2+y2=r2

Step-by-step explanation:

5 0

3 years ago

Radio tower had a special on rechargeable batteries. It sold AA for $1 and AAA for $0.75. It sold 42 batteries on a single day a

PIT_PIT [208]

AA = $1
AAA= $0.75

AA + AAA = 42

$1AA + $0.75AAA= $37

AA + AAA = 42
AA + AAA-AAA= 42- AAA
AA = 42- AAA

$1(42- AAA) + $0.75AAA= $37
$42 - AAA +0.75AAA = $37
$42 -0.25AAA= $37
$42-$42 -0.25AAA= $37 -$42
-0.25AAA= -5
-0.25AAA/-0.25 = -5/-0.25
AAA= 20

AA + AAA= 42
AA + 20 = 42
AA +20 -20 = 42-20
AA= 22

Check
$1AA + $0.75AAA= $37
$1(22)+ $0.75(20)= $37
$22 + $15 =$37
$37 = $37

5 0

3 years ago

If x-1/x=5 find x^3+1/x^3=?

ipn [44]

$\dfrac{x-1}{x}=5\ \ \ \ /x\neq0\\\\\dfrac{x-1}{x}=\dfrac{5}{1}\ \ \ \ |\text{cross multiply}\\\\5x=x-1\ \ \ \ |-x\\\\5x=1\ \ \ \ |:5\\\\x=0.2\\\\\dfrac{x^3+1}{x^3}=\dfrac{x^3}{x^3}+\dfrac{1}{x^3}=1+\dfrac{1}{x^3}\to1+\dfrac{1}{0.2^3}=1+\dfrac{1}{0.008}\\\\=1+\dfrac{1}{\frac{8}{1,000}}=1+\dfrac{1,000}{8}=1+125=126\\\\\text{Answer:}\ \dfrac{x^3+1}{x^3}=126$

5 0

3 years ago

Read 2 more answers

Given: Quadrilateral PQRS is a rectangle. Prove: PR = QS

lord [1]

Answer:

Given: Quadrilateral P QR S is a rectangle.

To prove :PR= Q S

Construction : Join PR and Q S.

Proof: In Rectangle PQRS, and

→ taking two triangles PSR and Δ QRS

1. PS = Q R

2. ∠ PS R = ∠ Q RS [Each being 90°]

3. S R is common.

→ ΔP SR ≅ Δ Q RS → [Side-Angle-Side Congruency]

∴ PR =Q S [ corresponding part of congruent triangles ]

Hence proved.

6 0

3 years ago

Memory module consists of 9 chips. The device is designed with redundancy so that it works even if one of its chips is defective

soldier1979 [14.2K]

Answer:

a) $P[C]=p^n$

b) $P[M]=p^{8n}(9-8p^n)$

c) n=62

d) n=138

Step-by-step explanation:

Note: "Each chip contains n transistors"

a) A chip needs all n transistor working to function correctly. If p is the probability that a transistor is working ok, then:

$P[C]=p^n$

b) The memory module works with when even one of the chips is defective. It means it works either if 8 chips or 9 chips are ok. The probability of the chips failing is independent of each other.

We can calculate this as a binomial distribution problem, with n=9 and k≥8:

$P[M]=P[C_9]+P[C_8]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)\\\\P[M]=p^{8n}(p^{n}+9(1-p^n))\\\\P[M]=p^{8n}(9-8p^n)$

$P[M]=(0.999)^{8n}(9-8(0.999)^n)=0.9$

This equation was solved graphically and the result is that the maximum number of chips to have a reliability of the memory module equal or bigger than 0.9 is 62 transistors per chip. See picture attached.

d) If the memoty module tolerates 2 defective chips:

$P[M]=P[C_9]+P[C_8]+P[C_7]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1+\binom{9}{7}P[C]^7(1-P[C])^2\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])+36P[C]^7(1-P[C])^2\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)+36p^{7n}(1-p^n)^2$

We again calculate numerically and graphically and determine that the maximum number of transistor per chip in this conditions is n=138. See graph attached.

6 0

3 years ago