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IceJOKER [234]
3 years ago
10

One angle of a triangle has a measure of 3 times as large as the smallest. The measurement of the 3rd angle is 40 degrees more t

han that of the smallest..what is the measurement of the largest angle?
Mathematics
1 answer:
MrMuchimi3 years ago
3 0

The largest of the three angles is 84°.

Hope this helps! :)

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7) What is the slope-intercept form of this equation? Show all your work. 6x + 2y = 12​
mariarad [96]

Answer:

3

Step-by-step explanation:

6x+2y=12

2y=6x-12

y=6x/2- 12/2

y=3x-6

7 0
2 years ago
Need help with school
Scorpion4ik [409]

Answer:

x = 7.2m

Step-by-step explanation:

Because it is a right triangle, you can use the Pythagorean theorem:

4^2+6^2 = x^2

16+36 = x^2

52 = x^2

sqrt52 = x

7.2 = x

5 0
2 years ago
Read 2 more answers
The film starts at 6.45 pm The film last 102 minutes what time dose the film finish
Brrunno [24]

Answer:

102 minutes=1 hour and 42 minutes

6:45+1 hour and 42 minutes=8:27 p.m.

hence, the answer is 8:27 pm

hope this helped.

8 0
1 year ago
If the point (x,square root 3/2) is on the unit circle, what. is x?
galina1969 [7]

Answer:

x=\pm\frac{1}{2}

Step-by-step explanation:

On a unit circle, (x,y)=(\cos\theta,\sin\theta), so \sin\theta=\frac{\sqrt{3}}{2} in this case. If you look at the attached circle, the only time that the y-coordinate is \frac{\sqrt{3}}{2} is when x=-\frac{1}{2} and x=\frac{1}{2}, which correspond to angles of \theta=\frac{2\pi}{3} and \theta=\frac{\pi}{3} respectively

3 0
2 years ago
The time needed to complete a final examination in a particular college course is normallydistributed with a mean of 80 minutes
Artist 52 [7]

Answer:

a) 0.023

b) 0.286

c) 10 students will be unable to complete the exam inthe allotted time.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 80 minutes

Standard Deviation, σ = 10 minutes

We are given that the distribution of time to complete an exam is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(completing the exam in one hour or less)

P(x < 60)

P( x < 60) = P( z < \displaystyle\frac{60 - 80}{10}) = P(z < -2)

Calculation the value from standard normal z table, we have,  

P(x < 60) =0.023= 2.3\%

b) P(complete the exam in more than 60 minutes but less than 75 minutes)

P(60 \leq x \leq 75) = P(\displaystyle\frac{60 - 80}{10} \leq z \leq \displaystyle\frac{75-80}{10}) = P(-2 \leq z \leq -0.5)\\\\= P(z \leq -0.5) - P(z < -2)\\= 0.309- 0.023 = 0.286= 28.6\%

c) P(completing the exam in more than 90 minutes)

P(x > 90)

P( x > 90) = P( z > \displaystyle\frac{90 -80}{10}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,  

P(x > 90) = 1 - 0.8413 = 0.1587 = 15.87\%

15.87% of children of class will require more than 90 minutes to complete the test.

Number of children =

\dfrac{15.87}{100}\times 60 = 9.52\approx 10

Approximately, 10 students of class will require more than 90 minutes to complete the test.

4 0
3 years ago
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