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Kamila [148]
3 years ago
14

Figure below shows two triangles EFG and KLM​

Mathematics
1 answer:
Vesna [10]3 years ago
3 0

Answer:

last option

Step-by-step explanation:

To prove that ΔEFG is also a right triangle, you must prove that KL = EF so that in ΔKLM c² = a² + b² which would make ΔEFG a right triangle.

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Can someone help me answer this?
andriy [413]

Answer:

m<S = 68 degrees.

m<D = 112 degrees

Step-by-step explanation:

Opposite angles of a parallelogram are equal so:

4x - 4 = 3x + 14      

4x - 3x = 14 + 4

x = 18.

So m < S = 4(18) - 4

= 72-4

= 68 degrees.

Angles on the same side of a parallelogram are supplementary, so

m < D = 180 - 68

= 112 degrees.

7 0
2 years ago
Solve the equation 4 (x-3) + 5=1
pishuonlain [190]
If your solving for x, x=2 
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3 years ago
4.
tiny-mole [99]

Answer:

7

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8 0
3 years ago
0.009 ia 1/10 of what number?
d1i1m1o1n [39]
0.009 x 10 = 0.09

it may not be correct, but here you go.
3 0
3 years ago
Read 2 more answers
There are big spenders among University of Alabama football season ticket holders. This data set Roll Tide!! shows the dollar am
Bas_tet [7]

Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:

n = 479, \pi = \frac{130}{479} = 0.2714

Hence the bounds of the interval are found as follows:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 - 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.2316

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 + 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.3112

The 95% confidence interval is (0.2316, 0.3112).

More can be learned about the z-distribution at brainly.com/question/25890103

7 0
2 years ago
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