Isolate the absolute value expression
The point with the greatest distance to the origin is given by:
B. (-3, 3).
<h3>What is the distance between two points?</h3>
Suppose that we have two points,
and
. The distance between them is given by:

The origin is given by point (0,0), hence the distance of a point (x,y) to the origin is given by:
D = sqrt(x² + y²).
Hence the distances for each point given in the problem are:
- A. Distance = sqrt((-4)² + (-1)²) = sqrt(17).
- B. Distance = sqrt((-3)² + (3)²) = sqrt(18).
- C. Distance = sqrt((4)² + 0²) = sqrt(16).
- D. Distance = sqrt((2)² + 3²) = sqrt(13).
Hence option B has the greatest distance.
More can be learned about the distance between two points at brainly.com/question/18345417
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Step-by-step explanation:
(5.6) × 1/8

Here the unit fraction is 1/8
multiply 5.6 times 1
so its 5.6. then divide the decimal by 8
we know that 8 times 7 is 56
so 56 divide by 8 is 7
we have decimal point 1 number to the left
so we move decimal point 2 place left
Hence 5.6 divide by 8 is 0.7

Answer: 0.7
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):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
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
= 0.668248
= 0.67 (to the hundredth)
Answer:
you could buy 8 t shirst
Step-by-step explanation: