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monitta
3 years ago
15

1. 5h-9+-16+6h

Mathematics
1 answer:
seraphim [82]3 years ago
6 0

Answer:

1) 11h - 25

2) 8

3) C

Step-by-step explanation:

1) 5h-9+-16+6h

+ and - together makes -

5h - 9 - 16 + 6h

Arrange the like terms together

5h + 6h - 9 - 16

= 11h - (9+16)

= 11h - 25


2) 4x+4=9x-36

Subtract 4x from both sides

4x+4-4x = 9x-36-4x

Cancl out 4x and -4x from the left side

4 = 9x - 4x -36

=> 4 = 5x -36

Add 36 to both sides

4+36 = 5x -36 + 36

Cancel out -36 and +36 from the right side

=> 40 = 5x

Flip both the sides of the equation

5x = 40

Dividing both sides by 5

\frac{5x}{5} = \frac{40}{5}

Cancel out the 5's on the top and bottom of the left side

x = 8

3) 7x+5 = 4x+5+3x

Arrange the like terms together on the right side

7x+5 = 4x+3x+5

=> 7x+5 = 7x+5

Subtract 5 from both sides

7x+5-5 = 7x+5-5

Cancel out +5 and -5 on both the sides

7x = 7x

Divide both sides by 7

\frac{7x}{7} = \frac{7x}{7}

Cancel out the 7's from the top and bottom of both sides

x = x

which can be true for infinite value of x like 1=1, 2=2, 3=3,.............

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Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44

In point a)  

\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}

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value= (1- \frac{1}{k^2}) \times 100 \% =75\%

In point b)

\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma  \\\\= |(x-\mu)|\leq 2.5 \times 4.44  \\\\  = |(x-\mu)|\leq 1.11 \\\\ =  (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)

In point c)

\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\

In point d)

using standard normal variate

x=20.17\\\\  z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17

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