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STatiana [176]
3 years ago
15

use the pythagorean theorem to prove whether or not each set of numbers represent the side length of a right triangle in your fi

nal answer, include your proof A. 6,12,15 B. 5,,12, and 13
Mathematics
1 answer:
Natalija [7]3 years ago
4 0

Answer:

a. no, it is not a right triangle

b.  yes, it is a right triangle

Step-by-step explanation:

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(0,-9) could be a solution
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3 years ago
To divide the sector into two congruent sectors we can use the _________________ construction. A) angle bisector B) copy and ang
12345 [234]

Answer:

Option A) angle bisector

Step-by-step explanation:

Angle Bisector:

  • An angle bisector is a line that divides an angle into two equal parts.
  • The angle bisector divide the angle in two equal parts.
  • An angle bisector is equidistant from the sides of the angle when measured along a segment perpendicular to the sides of the angle.
  • It cuts the angle into half.
  • Thus, a sector can be divided into two equal sectors by drawing an angle bisector.

To divide the sector into two congruent sectors we can use the angle bisector construction.

Thus, the correct answer is

Option A) angle bisector

4 0
3 years ago
Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we
Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N(\mu = 7 lbs , \sigma^{2}  = 0.875^{2}  lbs)

The standard normal z distribution is given by;

              Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, X bar = sample mean weight

             n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{7.5-7}{\frac{0.875}{\sqrt{4} } }  ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0
4 years ago
A pet shop has rabbits and gerbils for sale. They have 20 animals altogether. They sell two rabbits. Then one of the gerbils has
Triss [41]
20 animals - 2 rabbits = 18 animals
18 animals + 2 gerbils = 20 animals.
20 animals / 4 = 5 animals
There are 5 rabbits and 15 gerbils now.
At the start 2 rabbits were sold so add that on.
5 + 2 = 7
There were 7 rabbits.
There weren't any twin gerbils then.
15 - 2 = 13.
There were 13 gerbils and 7 rabbits to start with.

Hope This Helps You!
Good Luck Studying :)
4 0
3 years ago
One batch of cookies requires the following ingredients: 2 13 cups of flour 34 of a cup of chocolate chips 25 of a cup of choppe
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Answer:

B,AD

Step-by-step explanation:

BECAUSE I SAY IT

8 0
3 years ago
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