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elena-14-01-66 [18.8K]
1 year ago
8

In ΔKLM, k = 890 inches, ∠M=143° and ∠K=34°. Find the length of m, to the nearest 10th of an inch.

Mathematics
1 answer:
Vesna [10]1 year ago
4 0

Answer: 957.8

Step-by-step explanation:

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What’s the answer? Because am having a hard time with this math problem.
Studentka2010 [4]

9514 1404 393

Answer:

  34.5 square meters

Step-by-step explanation:

We assume you want to find the area of the shaded region. (The actual question is not visible here.)

The area of the triangle (including the rectangle) is given by the formula ...

  A = 1/2bh

The figure shows the base of the triangle is 11 m, and the height is 1+5+3 = 9 m. So, the triangle area is ...

  A = (1/2)(11 m)(9 m) = 49.5 m^2

The rectangle area is the product of its length and width:

  A = LW

The figure shows the rectangle is 5 m high and 3 m wide, so its area is ...

  A = (5 m)(3 m) = 15 m^2

The shaded area is the difference between the triangle area and the rectangle area:

  shaded area = 49.5 m^2 - 15 m^2 = 34.5 m^2

The shaded region has an area of 34.5 square meters.

7 0
1 year ago
Consider a set of 7500 scores on a national test whose score is known to be distributed normally with a mean of 510 and a standa
german
\mathbb P(X>600)=\mathbb P\left(\dfrac{X-510}{85}>\dfrac{600-510}{85}\right)=\mathbb P(Z>1.059)\approx0.145

So approximately 14.5% of the scores are higher than 600. This means in a sample of 7500, one could expect to see 0.145\times7500\approx10.86 scores above 600.
5 0
2 years ago
Grogg typed the following $1000$ expressions into his calculator, one by one: \[\sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}, \dots, \
sergey [27]

Answer:

pano po yan hinde ko po alam

Step-by-step explanation:

paturo po

5 0
2 years ago
The thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters. Determine the foll
Mrrafil [7]

Answer

given,

thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters.

X = U[0.95,1.05]           0.95≤ x ≤ 1.05

the cumulative distribution function of flange

F(x) = P{X≤ x}=\dfrac{x-0.95}{1.05-0.95}

                     =\dfrac{x-0.95}{0.1}

b) P(X>1.02)= 1 - P(X≤1.02)

                   = 1- \dfrac{1.02-0.95}{0.1}

                   = 0.3

c) The thickness greater than 0.96 exceeded by 90% of the flanges.

d) mean = \dfrac{0.95+1.05}{2}

              = 1

   variance = \dfrac{(1.05-0.95)^2}{12}

                  = 0.000833

4 0
2 years ago
The geometric average of -12%, 20%, and 25% is _________.
Grace [21]
<span>20.28% is the answer</span>
8 0
2 years ago
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