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Ulleksa [173]
3 years ago
15

Does anyone know this

Mathematics
1 answer:
Rudik [331]3 years ago
6 0

Answer:

its the first one. you're right.

A reflection over y-axis.

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PLEASE HELP ME IN MATH IM CONFUSED
Butoxors [25]

Answer:

the second choice is correct

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = \frac{3}{4} x - 9 is in this form

with slope m = \frac{3}{4}

• Parallel lines have equal slopes, hence

y = \frac{3}{4} x + c ← is the partial equation of the parallel line

To find c substitute (- 8, - 18) into the partial equation

- 18 = - 6 + c ⇒ c = - 18 + 6 = - 12

y = \frac{3}{4} x - 12 ← equation of parallel line



4 0
3 years ago
The exterior angles of an octagon are 42°, 55°, 39°, 20°, 62°, 45°, and 47°. What is the measure of the eighth exterior angle? S
marysya [2.9K]
The sum of exterior angles of a polygon = 360<span>°

Thus, the eighth exterior angle is equal to
360</span>° - (42° + 55° + 39° + 20° + 62° + 45° + 47°) = 50<span>°</span> 
6 0
4 years ago
Read 2 more answers
) the top and bottom margins of a poster are 4 cm and the side margins are each 5 cm. if the area of printed material on the pos
grin007 [14]
If x represents the width of the poster (including borders), the area of the finished poster can be written as
.. a = x*(390/(x -10) +8)
.. = 8x +390 +3900/(x -10)

Then the derivative with respect to x is
.. da/dx = 8 -3900/(x -10)^2
This is zero at the minimum area, where
.. x = √(3900/8) +10 ≈ 32.08 . . . . cm
The height is then
.. 390/(x -10) +8 = 8 +2√78 ≈ 25.66 . . . . cm

The poster with the smallest area is 32.08 cm wide by 25.66 cm tall.

_____
In these "border" problems, the smallest area will have the same overall dimension ratio that the borders have. Here, the poster is 10/8 = 1.25 times as wide as it is high.
8 0
3 years ago
If RS = 17, ST = x + 6, and RT = 3x - 5, What is ST.
tigry1 [53]

Answer:

ST = 20

Step-by-step explanation:

Line segment RS and segment ST must add up to line segment RT

Step 1: Define

RS = 17

ST = x + 6

RT = 3x - 5

Step 2: Set up equation

mRS + mST = mRT

17 + x + 6 = 3x - 5

Step 3: Solve for <em>x</em>

x + 23 = 3x - 5

23 = 2x - 5

28 = 2x

x = 14

Step 3: Find mST

ST = x + 6

ST = 14 + 6

ST = 20

3 0
4 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3
In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that \mu = 8.3, \sigma = 3.3.

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

s = \frac{3.3}{\sqrt{37}} = 0.5425

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

This probability is the pvalue of Z when X = 8.65. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{8.65 - 8.3}{0.5425}

Z = 0.65

Z = 0.65 has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is subtracted by the pvalue of Z when X = 7.43

Z = \frac{X - \mu}{\sigma}

Z = \frac{7.43 - 8.3}{0.5425}

Z = -1.60

Z = -1.60 has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is the pvalue of Z when X = 8.65 subtracted by the pvalue of Z when X = 7.43.

So:

From a), we have that for X = 8.65, we have Z = 0.65, that has a pvalue of 0.7422.

From b), we have that for X = 7.43, we have Z = -1.60, that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0
3 years ago
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