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2 years ago

Can someone tell me how to solve this?

Mathematics

1 answer:

liberstina [14]2 years ago

4 0

These two angles are equivalent because they're corresponding angles created by two parallel lines.

Since they're equivalent, we can set them equal to each other.

23x - 5 = 21x + 5

Now solve for x

23x - 5 = 21x + 5

Add 5 to both sides

23x = 21x + 10

Subtract 21x from both sides

2x = 10

Divide both sides by 2

x = 5

Now, plug in 5 as "x" in one of the equations

Angle 1 = (23*5) - 5

= 115 - 5

= 110 degrees

Since they're equivalent, Angle 2 should also be 110 degrees, but let's check in case.

Angle 2 = (21*5) + 5

= 105 + 5

= 110 degrees

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3/4 divided by 6 and 2/3

Talja [164]

It would be 0.125 for the first answer

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2 years ago

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For parallelogram LMNO , if m

natali 33 [55]

I'm just going to assume you're given the volume and LMNO are sides of the parallelogram

Area = LMNO

M=?

M= $\frac{area}{LNO}$

5 0

2 years ago

Write and solve an equation to find your final score. Use x to represent your final score.

attashe74 [19]

Let your final bowling score be x. It is 14 pins less than your friends final score. Your friends final bowling score is 105. Then x is 14 pins less than 105.
1 way: This means that 105 is greater than x by 14. Mathematically,
105-x=14,
-x=14-105,
-x=-91,
x=91.
2 way: When you add 14 to x, you'll get 105.
x+14=105,
x=105-14,
x=91.
Answer: Your final score is 91.

8 0

1 year ago

Compare -4 ____ -12 9 ___ -35

omeli [17]

-4 < -12 and 9 > -35

Hope this helps!

5 0

2 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

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)

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

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

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

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

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

8 0

2 years ago