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2 years ago

Help pls its attached as a file

Mathematics

1 answer:

ira [324]2 years ago

7 0

1. (6r + 7) + (13 + 7r)
1(6r + 7) + 1(13 + 7r)
6r + 7 + 13 + 7r
13r + 20

2. (13 - 3/2r) - (1 - r )
1(13 - 3/2r) - 1(1 - r )
13 - 3/2r - 1 + r
12 - (3/2r - 2/2r)
12 - 1/2r

3. (-8 - r) + (2r - 4)
1(-8 - r) + 1(2r - 4)
-8 - r + 2r - 4
-12 + r

4. (7r - 3/2) - (2/3 + 6r)
1(7r - 3/2) - 1(2/3 + 6r)
7r - 3/2 - 2/3 - 6r
r + (- 3/2 - 2/3)
r +(- 9 - 4 / 6)
r + -13/6
r - 13/6

Hope this helps!

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Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.

lesya [120]

Answer:

0.1994 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851$

P(sample of 20 boxers is more than 167 pounds)

$P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)$

$= 1 - P(z \leq 0.8438)$

Calculation the value from standard normal z table, we have,

$P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%$

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

3 0

3 years ago

Learning

andrezito [222]

Answer:

Step-by-step explanation:

6 0

3 years ago

Find the missing length <br> . Leave your answers in simplest radical form

KIM [24]

Use the Pythagorean theorem:

$x^2+7^2=14^2\\\\x^2+49=196\qquad\text{subtract 49 from both sides}\\\\x^2=147\to x=\sqrt{147}\\\\x=\sqrt{49\cdot3}\\\\x=\sqrt{49}\cdot\sqrt3\\\\\boxed{x=7\sqrt3\ mi}$

$x^2+9^2=(\sqrt{226})^2\\\\x^2+81=226\qquad\text{subtract 81 from both sides}\\\\x^2=145\to \boxed{x=\sqrt{145}\ km}$

5 0

3 years ago

Matching Equivalent Expressions

KiRa [710]

I hope this help

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8 0

3 years ago

What are the values of angles G and H?

Ulleksa [173]

Answer:

G = 78°

H = 98°

Step-by-step explanation:

Opposite angles in a cylic quadrilateral are supplementary. Therefore:

G + E = 180°

Substitute

8x + 1 + 11x + 8 = 180°

Add like terms

19x + 9 = 180°

19x + 9 - 9 = 180° - 9

19x = 171

19x/19 = 171/19

x = 9

G = 8x + 1

Plug in the value

G = 8(9) + 1 = 73°

H + F = 180° (opposite angles of a cyclic quadrilateral is equal?

Plug in the values

6y - 4 + 5y - 3 = 180°

Add like terms

11y - 7 = 180

11y = 180 + 7

11y = 187

y = 187/11

y = 17

H = 6y - 4

Plug in the value of y

H = 6(17) - 4

H = 98°

7 0

3 years ago