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Dovator [93]
3 years ago
9

Find the measures of the interior angles of the triangle

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
6 0

Answer:

A = 75°

B = 90°

C = 15°

Step-by-step explanation:

The diagram tells us that C = 15°.

The diagram places a square in the angle of B, showing you the angle is a 90° right angle.

You can solve for A because all angles of a triangle must equal 180°. 180° - 90° (B) = 90°. 90° - 15° (C) = 75°, the measure of A.

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Answer:

About <em>4 hours.</em>

Step-by-step explanation:

To find the answer we will <em>add both percentages</em> of <u><em>Watching TV</em></u> and <u><em>Homework</em></u>.  8% + 13% = <em>21%</em>.  Now we will <em>divide 100 by 21</em> and we get<em> 4 hours a day.</em>

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Please help with these! or at least some of them.​
Fiesta28 [93]

Answer:

<h2>1) 9/2 = 5/4</h2><h2>5) False</h2><h2>6) $66</h2><h2>7) x=7</h2><h2>8) x=69</h2>

Step-by-step explanation:

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3 years ago
Which number line shows the solution to the inequality -2(3x-1)&lt;8
Zigmanuir [339]

Answer:

x  > -1

Step-by-step explanation:

Given

-2(3x - 1) < 8

Required

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3 years ago
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

8 0
3 years ago
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