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3 years ago

Find the angles of the triangle if they are proportional to the following 3 4 5

Mathematics

2 answers:

aleksklad [387]3 years ago

7 0

Answer: 90, and 1600 degrees.2500

Step-by-step explanation:

If the angles are proportional to 3, 4, 5,Then the angle measures are 3x, 4X and 5X

Step-by-step explanation:

Olin [163]3 years ago

6 0

Answer:

45 60 75

Step-by-step explanation:

Let's call the angles 3 x , 4 x and 5 x

They must add up to 180 o in a triangle, so:

3 x + 4 x + 5 x = 12 x = 180 → x = 180 / 12 = 15 So the angles are:

3 ⋅ 15 = 45 o 4 ⋅ 15 = 60 o 5 ⋅ 15 = 75 o Check answer:

45 + 60 + 75 = 180

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At maximum speed, an airplane travels 2,400 miles against the wind in 6 hours. Flying with the wind, the plane can travel the sa

Hatshy [7]

X - y = 2400/6 = 400 where x = speed of plane and y = speed of wind.

x + y = 2400/5 = 480 - flying with the wind

adding the 2 equations
2x = 880
x = 440

Speed of the plane with no wind = 440 mph

5 0

3 years ago

Read 2 more answers

What is the midpoint of the line segment with endpoints (3.2, 2.5) and (1.6, -4.5)

jok3333 [9.3K]

Answer:

(2.4, -1)

Step-by-step explanation:

Using midpoint formula plug in the info (x1+x2)/2, (y1+y2)/2

8 0

3 years ago

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0

3 years ago

Consider the continuous random variable x, which has a uniform distribution over the interval from 110 to 150. The probability t

Virty [35]

Answer:

The probability that x will take on a value between 120 and 125 is 0.14145

Step-by-step explanation:

For uniform distribution between a & b

Mean, xbar = (a + b)/2

Standard deviation, σ = √((b-a)²/12)

For 110 and 150,

Mean, xbar = (150 + 110)/2 = 130

Standard deviation, σ = √((150-110)²/12 = 11.55

To find the probability that x will take on a value between 120 and 125

We need to standardize 120 & 125

z = (x - xbar)/σ = (120 - 130)/11.55 = - 0.87

z = (x - xbar)/σ = (125 - 130)/11.55 = - 0.43

P(120 < x < 125) = P(-0.87 < x < -0.43)

We'll use data from the normal probability table for these probabilities

P(120 < x < 125) = P(-0.87 < x < -0.43) = P(z ≤ -0.43) - P(z ≤ -0.86) = 0.33360 - 0.19215 = 0.14145

Hope this Helps!!!

3 0

3 years ago

Find the slope between the two given points (-5,7) and (-5,11)

BartSMP [9]

Answer:

Slope (m) = infinity

Step-by-step explanation:

3 0

4 years ago