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2 years ago

11

PLEASE HELP SOLVE THE EQUATION USING ELIMINATION 3x+8y=-5 -2x+2y=18

1 answer:

Alexxx [7]2 years ago

6 0

By elimination:

X=-7 and y=2

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Question 21 is probably the hardest question on this homework

Snezhnost [94]

-15+5x= 20

5x= 35
x= 7

20= -15+5x

35= 5x

7 = x

7 0

3 years ago

Read 2 more answers

Lina spent 1/3 of her money. She now has $20 left. How much money did she have originally

PSYCHO15rus [73]

Answer:

60

Step-by-step explanation:

1/3 of 60 is 20

in other words

20 x 3 = 60

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2 years ago

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Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:

Vikentia [17]

Answer:

a

$\= x = 18.5$ , $\sigma = 5.15$

b

$15.505 < \mu < 21.495$

c

$14.93 < \mu < 22.069$

Step-by-step explanation:

From the question we are are told that

The sample data is 21, 14, 13, 24, 17, 22, 25, 12

The sample size is n = 8

Generally the ample mean is evaluated as

$\= x = \frac{\sum x }{n}$

$\= x = \frac{ 21 + 14 + 13 + 24 + 17 + 22+ 25 + 12 }{8}$

$\= x = 18.5$

Generally the standard deviation is mathematically evaluated as

$\sigma = \sqrt{\frac{\sum (x- \=x )^2}{n}}$

$\sigma = \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}$

$\sigma = 5.15$

considering part b

Given that the confidence level is 90% then the significance level is evaluated as

$\alpha = 100-90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.645$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.645 * \frac{5.15 }{\sqrt{8} }$

=> $E = 2.995$

The 90% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 2.995 < \mu < 18.5 + 2.995$

$15.505 < \mu < 21.495$

considering part c

Given that the confidence level is 95% then the significance level is evaluated as

$\alpha = 100-95$

$\alpha = 5\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.96 * \frac{5.15 }{\sqrt{8} }$

=> $E = 3.569$

The 95% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 3.569 < \mu < 18.5 + 3.569$

$14.93 < \mu < 22.069$

8 0

3 years ago

If a pool is 4 m wide and 16 m diagonally across, how long is the pool? (Pyth. Theorem)

alukav5142 [94]

Answer:

$L=4\sqrt{15}\ m$

or

$L=15.49\ m$

Step-by-step explanation:

we know that

Applying the Pythagorean Theorem in the rectangular pool

$d^2=L^2+W^2$

where

d is the diagonal of the pool

L is the length of the pool

W is the wide of the pool

we have

$d=16\ m\\W=4\ m$

substitute the given values

$16^2=L^2+4^2$

solve for L

$L^2=16^2-4^2\\L^2=240\\L=\sqrt{240}\ m$

simplify

$L=4\sqrt{15}\ m$

or

$L=15.49\ m$

4 0

2 years ago

A travel agency created a pamphlet for a city. They chose the high temperature for 10 random days during the year to calculate t

solong [7]

A. 68.4 degrees
Add all the temperatures together and divide by the number of days

6 0

2 years ago