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GalinKa [24]
2 years ago
14

B)When ABCD is drawn to scale, would

Mathematics
1 answer:
never [62]2 years ago
7 0

Hi there there's several ways this could be proven one way us to consider the allied angle theory where two angles formed between parallel lines are supplementary which in this case can be proven by

2(45)+90=180⁰ ✔

or 3(45)+45=180⁰✔

this would not be the case if it wasn't parallel

Consequently, you can also use the alternate angle theory where you essentially extend one of the lines and you'll see two equal alternate angles

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8 I guess because 50 divided by 6 is 8
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Two students, Raymond and Shauna, are trying to get city council members to vote for a new park along the river. 2/6 of the 24 c
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3 years ago
Write as a single power of 8.<br> 8 to the power of 8<br><br> ÷<br> 8 to the power of 2
OLEGan [10]

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7 0
3 years ago
Weights of cars passing over a bridge part 2<br>​
DaniilM [7]

Step-by-step explanation:

The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.

Given:

mean,  mu = 3550 lbs (hope I read the first five correctly, and it's not a six)

standard deviation, sigma = 870 lbs

weights are normally distributed, and assume large samples.

Probability to be estimated between W1=2800 and W2=4500 lbs.

Solution:

We calculate Z-scores for each of the limits in order to estimate probabilities from tables.

For W1 (lower limit),

Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069

From tables, P(Z<Z1) = 0.194325

For W2 (upper limit):

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From tables, P(Z<Z2) = 0.862573

Therefore probability that weight is between W1 and W2 is

P( W1 < W < W2 )

= P(Z1 < Z < Z2)

= P(Z<Z2) - P(Z<Z1)

= 0.862573 - 0.194325

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3 years ago
One student is selected at random from a group of 12 freshman, 16 sophomores,
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