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adoni [48]
2 years ago
14

What additional information could you use to show that ΔSTU ≅ ΔVTU using SAS? Check all that apply.

Mathematics
1 answer:
GalinKa [24]2 years ago
4 0

Options

A. UV = 14 ft and m∠TUV = 45°

B. TU = 26 ft

C. m∠STU = 37° and m∠VTU = 37°

D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°

E. m∠UST = 98° and m ∠TUV = 45°

Answer:

A. UV = 14 ft and m∠TUV = 45°

D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°

Step-by-step explanation:

Given

See attachment for triangle

Required

What proves that: ΔSTU ≅ ΔVTU using SAS

To prove their similarity, we must check the corresponding sides and angles of both triangles

First:

\angle UST must equal \angle UVT

So:

\angle UST = \angle UVT = 98

Next:

UV must equal US.

So:

UV = US = 14

Also:

ST must equal VT

So:

ST = VT = 20

Lastly

\angle TUV must equal \angle TUS

So:

\angle TUV = \angle TUS = 45

Hence: Options A and D are correct

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Jose must plant 380 trees. In the past 7 days Jose planted 266 trees. If he continues at this rate, how many more days will it t
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A store ships cans by weight. A small box can hold 3 to 5 pounds. A medium box can hold 5 to 8 pounds. A large box can hold 8 to
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The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar
Llana [10]

Answer:

a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.

b) The probability that a random movie is longer than 2.3 hours is 0.0918.

c) The length of movie that is shorter than 94% of the movies is 1.4 hours

Step-by-step explanation:

In the above question, we would solve it using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation

a) A random movie is between 1.8 and 2.0 hours

z = (x-μ)/σ,

x1 = 1.8,

x2 = 2.0

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z1 = (1.8 - 1.9)/0.3

z1 = -1/0.3

z1 = -0.33333

Using the z score table

P(z1 = -0.33) = 0.3707

z2 = (2.0 - 1.9)/0.3

z1 = 1/0.3

z1 = 0.33333

p(z2 = 0.33) = 0.6293

= P(- 0.33 ≤ z ≤ 0.33)

= 0.6293 - 0.3707

= 0.2586

The probability that a random movie is between 1.8 and 2.0 hours = 0.2586

b) A movie is longer than 2.3 hours

z = (x-μ)/σ,

x1 = 2.3

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z = (2.3 - 1.9)/0.3

z = 4/0.3

z = 1.33333

P(z = 1.33) = 0.90824

P(x>2.3) = = 1 - 0.90824

= 0.091759

≈ 0.0918

The probability that a random movie is longer than 2.3 hours is 0.0918.

3) The length of movie that is shorter than 94% of the movies.

z = (x-μ)/σ

Probability (z ) = 94% = 0.94

Movie that is shorter than 0.94

= P(1 - 0.94) = P(0.06)

Finding the P (x< 0.06) = -1.555

≈ -1.56

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

-1.56 = (x - 1.9)/ 0.3

Cross multiply

-1.56 × 0.3 = x - 1.9

- 0.468 + 1.9 = x

= 1.432 hours

≈ 1.4 hours

Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours

5 0
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