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elena-14-01-66 [18.8K]

3 years ago

14

Tom's Game Shop ordered 18 computer games. Two-thirds of the games were baseball games. How many of the game were not baseball g

ames?

2 answers:

Kaylis [27]3 years ago

6 0

Answer:

Your answer would be 12

juin [17]3 years ago

5 0

2/3=?/18

18÷3=6

3×6=18

2×6=12

?=12

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Simplify: 7.2(2.8r – 4.5s) + 9.34r – 1.25s

Free_Kalibri [48]

Answer:

29.5r - 33.65s

Step-by-step explanation:

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3 years ago

In the diagram below of circle O, OB and OC are radii, and chords AB, BC,

Vikki [24]

Answer:

It would be number 2. measure of angle BAC is equal to half the measure of angle BOC.

Step-by-step explanation:

This is because of inscribed angles.

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3 years ago

Solve the equation. Show all steps in the solution<br>2\3x + 1 = 5<br>

Troyanec [42]

Answer: $x=6$

<u>Subtract 1 from both sides</u>

<u></u> $2/3x+1-1=5-1$ <u></u>

<u></u> $2/3x=4$ <u></u>

<u></u>

<u>Multiply both sides by 3/2</u>

<u></u> $(3/2)*(2/3x)=(3/2)*4\\x=6$ <u></u>

3 0

3 years ago

Read 2 more answers

Consider a small population consisting of the 100 students enrolled in an introductory statistics course. Students in the class

Ray Of Light [21]

Answer:

5 ; 0.728 ; 0.00002

Step-by-step explanation:

According to the Central limit theorem, for a normal distribution, the population mean equals the sample mean., Hence the population mean, μ = sample mean = 5

The standard deviation of sampling distribution :

Standard deviation / sqrt(n)

σ / sqrt(n)

σ = 3 ; n = 17

3 / sqrt(17)

3 / 4.1231056

= 0.7276

= 0.728

3.)

P(x < 2) :

Using :

Z = (x - mean) ÷ s/sqrt(n)

P(x < 2) = (2 - 5) ÷ 3/sqrt(17)

P(x < 2) = -3 / 0.7276068

P(x < 2) = - 4.1231060

Z < - 4.123 = 0.00002 (Z probability calculator).

7 0

3 years ago

15.30 find the inverse laplace transform of: 1. (a) f1(s) = 6s 2 8s 3 s(s 2 2s 5) 2. (b) f2(s) = s 2 5s 6 (s 1) 2 (s 4) 3. (c) f

EleoNora [17]

The solution of the inverse Laplace transforms is mathematically given as

$f_{1}(t)=e^{-t}\sin (2 t)$

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$

<h3>What is the inverse Laplace transform?</h3>

1)

Generally, the equation for the function is mathematically given as

$F_{1}(s)=\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}$

By Applying the Partial fractions method

$\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}=\frac{A}{s}+\frac{B s+C}{s^{2}+2 s+5}$

$6 s^{2}+8 s+3=A\left(s^{2}+2 s+5\right)+(B s+C) s$

$\begin{aligned}&3=5 A \\&A=\frac{3}{5}\end{aligned}$

Considers s^2 coefficient

$\begin{aligned}&6=A+B \\&B=6 \cdot A \\&B=\frac{27}{5}\end{aligned}$

Consider s coeffici ent

$\begin{aligned}&8=2 A+C \\&C=8-2 A \\&C=\frac{34}{5}\end{aligned}$

Putting these values into the previous equation

$&F_{1}(s)=\frac{3}{5 s}+\frac{27 s+34}{5\left(s^{2}+2 s+5\right)} \\\\&F_{1}(s)=\frac{3}{5 s}+\frac{27(s+1)}{5\left((s+1)^{2}+4\right)}+\frac{7 \times 2}{10\left((s+1)^{2}+4\right)}$

By taking Inverse Laplace Transforms

$f_{1}(t)=\frac{3}{5}+\frac{27}{5} e^{-t} \cos (2t) + \frac{7}{10}\\\\$

$f_{1}(t)=e^{-t}\sin (2 t)$

For B

$F_{2}(s)=\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}$

By Applying Partial fractions method

$\begin{aligned}&\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\\\&s^{2}+5 s+6=A(s+1)(s+4)+B(s+4)+C(s+1)^{2}\end{aligned}$

at s=-1

$1-5+6=3 B \\\\B=\frac{2}{3}$

at s=-4

$&16-20+6=9 C \\\\&9 C=2 \\\\&C=\frac{2}{9}$

at s^2 coefficient

1=A+C

A=1-C

A=7/9

inputting Variables into the Previous Equation

$\begin{aligned}&F_{2}(s)=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\&F_{2}(s)=\frac{7}{9(s+1)}+\frac{2}{3(s+1)^{2}}+\frac{2}{9(s+4)}\end{aligned}$

By taking Inverse Laplace Transforms

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

For C

$F_{3}(s)=\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}$

Using the strategy of Partial Fractions

$\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}=\frac{A}{s+1}+\frac{B s+C}{s^{2}+4 s+8}$

$10=A\left(s^{2}+4 s+8\right)+(B s+C)(s+1)$

S=-1

10=(1-4+8) A

A=10/5

A=2

Consider constants

10=8 A+C

C=10-8 A

C=10-16

C=-6

Considers s^2 coefficient

0=A+B

B=-A

B=-2

inputting Variables into the Previous Equation

$&F_{3}(s)=\frac{2}{s+1}+\frac{-2 s-6}{\left((s+2)^{2}+4\right)} \\\\&F_{3}(s)=\frac{2}{s+1}-\frac{2(s+2)}{\left((s+2)^{2}+4\right)}-\frac{2}{\left((s+2)^{2}+4\right)}$

Inverse Laplace Transforms

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$

Read more about Laplace Transforms

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3 0

2 years ago