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Fantom [35]
2 years ago
11

A triangle has sides with lengths of 51 miles,68 miles and 85 miles. Is it a right triangle.?

Mathematics
1 answer:
WITCHER [35]2 years ago
7 0

Answer: YES

Step-by-step explanation: It is

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The weights of professional wrestlers are approximately normally distributed with a mean of 220 pounds and a standard deviation
nikdorinn [45]

Answer:

1.6%

Step-by-step explanation:

The cumulative distribution function (CDF) of a random variable X is denoted by F(x), and is defined as

F(x) = P(X ≤ x). where x is the largest possible value of X that is less than or equal to x

z = (x-μ)/σ,

where:

x is the raw score = 205

μ is the population mean, = 220 pounds

σ is the population standard deviation = 7 pounds

205 -220/7

z = -15/7

z = -2.1428571429

Using the normal cdf function on your graphing calculator,the cumulative distribution is

normalcdf( -2.1428571429, 100)

= 0.01606229

In percent form = 0.01606229 × 100

= 1.6%

5 0
3 years ago
(40 points to best answer) (help ASAP)
aleksandrvk [35]

Answer:

x=3.5

Step-by-step explanation:

We can use proportions to solve this problem.  Put the side of one triangle over the same side of the other triangle.

6           4.2

------ = ----------

5           x

Using cross products

6x = 5*4.2

6x =21

Divide by 6

6x/6 = 21/6

x=3.5

5 0
3 years ago
I need the answers if you know them you’ll get 80 points
katrin2010 [14]
For number 1, it says based on the first two clues, so you would still include the even numbers ranging from 70-81. Then for number 2 we would use clue C because it gives us the most information about our number.
4 0
3 years ago
Read 2 more answers
The triangles shown above are similar. State the pairs of corresponding angles the images are not drawn to scale
melamori03 [73]
We know that
∠X≈∠B
∠Y≈∠C
∠Z≈∠A

XZ≈BA
XZ=12
BA=9

XY≈BC
XY=14
BC=10.5

ZY≈AC
ZY=16
AC=12

the scale factor=12/16-----> 0.75

therefore

the answer is the option D
7 0
4 years ago
Can u help me with my homework
I am Lyosha [343]
Ur dad Ur dad Ur dad
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3 years ago
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