Help please! A. B. C. D.

a bird is flying at a true bearing of 75 degree's at a speed of 40 miles per hour. Write the velocity of the bird as a vector in

Allisa [31]

Answer: V = (10.4 mph, 38.6 mph)

Step-by-step explanation:

The velocity is written as (vx, vy)

where vx is the component of the velocity in the x-axis and vy is the component of the velocity in the y-axis.

In usual notation, the angles are measured counterclockwise from the positive x-axis.

We know that the angle is 75°, this means that the velocity in the x-axis will be equal to the total velocity of the bird projected in the x-axis (suppose a triangle rectangle, where the velocity is the hypotenuse, the x component is a cathetus and the y component is other cathetus)

vx = 40mph*cos(75°) = 10.4 mph

vy = 40mph*sin(75°) = 38.6mph

Then the vector of velocity is V = (10.4 mph, 38.6 mph)

5 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

1. What is the volume, in cubic centimeters, of the sphere shown below? Use 3.14 for Pi

prohojiy [21]

Answer:

113.04 cm³

Step-by-step explanation:

diameter = 6cm

so radius = 3cm

Volume of sphere = $\frac{4}3}$ $\pi r^3$

= 4 * 3.14 * 3 * 3 *3 / 3

= 4 * 28.26

= 113.04 cm³

6 0

2 years ago

Which values are solutions to the inequality? 5r≤6r−8 Select each correct answer. r = 7 r = 8 r = 9 r = 10

Ratling [72]

Answer:

r = 8,9,10

Step-by-step explanation:

The given inequality is :

5r≤6r−8 ...(1)

We need to find the value of r.

Subtracting 5r to both sides of the inequality .

5r-5r≤6r-5r−8

0≤r−8

r ≥ 8

Hence, the values of r are 8,9,10.

8 0

2 years ago

Read 2 more answers

Ms. Wall will roll a single number cubes. What is the probability that she will roll an even number?

Arturiano [62]

Well single number, meaning from 1 - 9. She has a higher chance of getting an odd number for sure, as there are more odd numbers on the dice :) So, the ratio is 4:5 and fraction: 4/5. %44.444 is your answer :)

Thank you,
Darian D.

4 0

3 years ago