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7 days ago

Rearrange the following steps in the correct order to locate the last occurrence of the smallest element in a finite list of int

egers, where the integers in the list are not necessarily distinct.
Rank the options below.
if min > ai then
procedure last smallest a1,a2,....,an: integers)
return location
min = a1 and location = 1
min ai and location = i
for i = 2 to n

Mathematics

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Clarissa's income puts her in the bottom tax bracket (10%) last year. During the same year, she earned $250 in dividends and $75

Leona [12618]

Answer:

75 dollars in coupons.

250 dollars in dividends.

She achieved a profit of 600 - 425 = 175 from her stock investment.

Her complete income sums up to 250 + 75 + 175 = 500.

If all this is taxed at 10%, her tax will be 500 *.1 = 50.

8 0

1 month ago

mridul jogs around a rectangular park of 100m by 80m at the rate of 9km per hour. how much time will he take in jogging 6 rounds

tester [12383]

14 minutes. Step-by-step explanation: First, find the perimeter which can be calculated as P=100+80+100+80 = 360 meters, and then multiply it by 6 for the total distance = 2160 meters. To convert meters to kilometers, note that 1000 m = 1 km, so you would find how many kilometers are in 2160 m by cross-multiplying to get 2.16 km. Lastly, you need to calculate the time required represented by T = D/S = 2.16 km / 9 km/h, which simplifies down and gives you the time.

3 0

1 month ago

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

Svet_ta [12734]

Response:

a) The average is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% likelihood that the average time of visitors falls within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step clarification:

To figure this out, we must comprehend the normal probability distribution alongside the central limit theorem.

Normal Probability Distribution

Issues involving normal distributions can be resolved using the z-score formula.

For a data set with mean $\mu$ and standard deviation $\sigma$ , the z-score related to a measure X is defined as:

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

The z-score illustrates how many standard deviations the measure is positioned from the mean. Once the z-score is calculated, we refer to the z-score table to find the p-value matching this z-score. This p-value reflects the likelihood that the measurement is less than X, which means it indicates the percentile of X. To determine the chance that the measure exceeds X, we subtract the p-value from 1.

Central Limit Theorem

The Central Limit Theorem states that for a normally distributed random variable X, defined by mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of sample means of size n can be approximated as a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For non-normal distributions, this theorem applies if n is at least 30.

In this case, the distribution has a mean of 10 minutes and a standard deviation of 2 minutes.

This indicates that $\mu = 10, \sigma = 2$

Assuming 64 visitors independently access the site.

This implies that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and variance of the average time of the visitors.

Employing the Central Limit Theorem, the mean is 10 and the variance is (0.25)^2 = 0.0625.

b. The chance that the mean visit duration is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, thus between 9.75 and 10.25 seconds, which corresponds to the p-value of z when X = 10.25 minus the p-value of z when X = 9.75.

X = 10.25

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

Through the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{10.25 - 10}{0.25}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

X = 9.75

$Z = \frac{X - \mu}{s}$

$Z = \frac{9.75 - 10}{0.25}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

0.8413 - 0.1587 = 0.6826.

0.6826 = 68.26% chance that the average visitor time is within 15 seconds of 10 minutes.

c. The value surpassed by the average visitor time with a probability of 0.01.

$Z = \frac{X - \mu}{s}$ This corresponds to the 99th percentile, denoting X when Z equals a p-value of 0.99, hence X when Z = 2.327.

$2.327 = \frac{X - 10}{0.25}$

$X - 10 = 2.327*0.25$

$X = 10.58$

Thus, the result is 10.58 minutes.

6 0

1 month ago

You are in an airplane 5.7 miles above the ground. What is the measure of BD⌢

AnnZ [12381]

Result:

6.1°; 425.86 m.

Step-by-step breakdown:

The information provided states that the airplane is at an altitude of 5.7 miles above ground level, while the "radius of Earth is about 4000 miles." Thus,

θ = 2 × cos^-1 (a/ (a + b)), where a = 4000 miles, and b = 5.7 miles.

θ = 2 × cos^-1 (4000/ (4000 + 5.7)) = 6.1°.

To calculate the distance in meters:

Change in distance = 6.1° /360° × 2π × 4000 miles = 425.86 meters.

Consequently, BD⌢ measures at 6.1° and the distance corresponding to this section of Earth is 425.86 meters.

6 0

2 months ago