PLEASE HELP! ONLY 4 MINUTES LEFT! WILL GIVE BRAINLIEST!

The spinner is divided into 8 equal parts. Find the theoretical probability

ella [17]

1/8 are you chances for all the colors

7 0

3 years ago

Referring to the figure, find the missing length.

vivado [14]

Answer:

5 units

Step-by-step explanation:

here;

perpendicular (p)= 3

base (b) = 4

hypotenuse (h) = c= ?

By Pythagorean relationship;

h²=p²+b²

or, h= √(p²+b²)

or, c= √(3²+9²)

or, c= √25

hence, c=5

5 0

3 years ago

If I take 1 hour to cook a batch of cookies and I have 15 ovens working 24 hours a day every day for 5 years, how long does it t

lawyer [7]

Answer:

Approximately 40,069 days

7 0

2 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

Help please number 4, 5, 6, and 7

FinnZ [79.3K]

Step-by-step explanation:

4. So, total time traveled is calculated by adding travel times in both directions.

From Detroit to Chicago, Charlie flew 300 miles at 150 miles per hour speed. That means that he traveled:

t1 = s / v1

t1 = 300 miles / 150 miles per hour

t1 = 2 hours

Now, let's do the same for the opposite direction. Since it's the same distance, he again flew 300 miles, but the speed this time was 100 miles per hour:

t2 = s / v2

t2 = 300 miles / 100 miles per hour

t2 = 3 hours

So, total time he traveled was t1 + t2 = 2 hours + 3 hours = 5 hours.

For the trip distance we need to add distances in both directions. We already said that the distance is the same, so the total trip distance is 300 miles + 300 miles = 600 miles.

Displacement shows how far are the ending and the starting point. Since Charlie started from Detroit, flew to Chicago, and then came back to Detroit, his ending point is the same as the starting, so the displacement is 0.

We can calculate average speed by dividing total distance with the total travel time. We already found that the total trip distance was 600 miles, and the total travel time was 5 hours. That means that:

v(average) = s(total) / t (total)

v(average) = 600 miles / 5 hours

v (average) = 120 miles per hour

As for the average velocity, we calculate it by dividing displacement by total travel time. Since the displacement was 0, average velocity will also be 0.

5. From the left graph we see that the object went from 0 to 2 meters (it moved 2 meters) in 1 second. That means that its speed was:

v1 = s1 / t1

v1 = 2 meters / 1 second

v1 = 2m/s

From there, it went from 2 to 4 meters (it moved 2 meters) in the next 4 seconds. That means that its speed, during that interval, was:

v2 = s2 / t2

v2 = 2 meters / 4 seconds

v2 = 0.5 m/s

So, our graph will show, for the first second, velocity of 2m/s, and for the next four seconds, the velocity of 0.5 m/s.

6. Again, when calculating speed, we use the equation:

v = s / t

In this case, the distance (s) is 325 miles and the time (t) is 5 hours. So, the speed will be:

v = 325 miles / 5 hours

v = 65 mph

7. Now, we have the same distance (s) of 325 miles ant the speed (v) of 70 mph. We want know how long will the drive take:

t = s / v

t = 325 miles / 70 mph

t = 4.64 hours.

6 0

3 years ago