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sashaice [31]
3 years ago
6

A log raft is moving downstream at a speed of 3 km/h. To insure that there are no river jams a supervisor travels along the raft

on a motor boat speed of which in still water is 15 km/h. What is the length of the log raft, if it takes the supervisor 16 minutes to travel from one end of the raft to another and back?
Mathematics
1 answer:
skad [1K]3 years ago
8 0

Answer:

2000 meters

Step-by-step explanation:

When the motorboat is moving downstream, its speed relative to the ground is 18 km/h, so its speed relative to the raft is 15 km/h.

When the motorboat is moving upstream, its speed relative to the ground is 12 km/h, so its speed relative to the raft is again 15 km/h.

Converted to m/min, the relative speed is:

15 km/h × (1000 m/km) × (1 h / 60 min) = 250 m/min

It takes the motorboat 16 minutes to travel to the front of the raft and back.  Since the speed is the same in both directions, the motorboat takes 8 minutes to travel the length of the raft.

So the length of the raft is 250×8 = 2000 meters.

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the size of the second application is 3.45 MB less than he first application. What is the size of the second application
mr_godi [17]

The size of the second application given the size of the first application and the expression ( x - 3.45 mb) for the size of the second application is 293.55 MB.

<h3>Equation</h3>

Let

  • Size of the first application = x

  • Size of the second application= x - 3.45 mb

For instance,

if the size of the first application is 297 MB

Size of the second application= x - 3.45 mb

= 297 MB - 3.45 MB

= 293.55 MB

Therefore, the size of the second application given the size of the first application and the expression for the size of the second application is 293.55 MB

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2 years ago
I arrive at a bus stop at a time that is normally distributed with mean 08:00 and SD 2 minutes. My bus arrives at the stop at an
Nimfa-mama [501]

Answer:

0.0485 = 4.85% probability that you miss the bus.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

When two normal distributions are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this question:

We have to find the distribution for the difference in times between when you arrive and when the bus arrives.

You arrive at 8, so we consider the mean 0. The bus arrives at 8:05, 5 minutes later, so we consider mean 5. This means that the mean is:

\mu = 0 - 5 = -5

The standard deviation of your arrival time is of 2 minutes, while for the bus it is 3. So

\sigma = \sqrt{2^2 + 3^2} = \sqrt{13}

The bus remains at the stop for 1 minute and then leaves. What is the chance that I miss the bus?

You will miss the bus if the difference is larger than 1. So this probability is 1 subtracted by the pvalue of Z when X = 1.

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

Z = \frac{1 - (-5)}{\sqrt{13}}

Z = \frac{6}{\sqrt{13}}

Z = 1.66

Z = 1.66 has a pvalue of 0.9515

1 - 0.9515 = 0.0485

0.0485 = 4.85% probability that you miss the bus.

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Match the inequality to the graph of its solution.<br> a. n/4 ≤ -1<br> b. -10 ≥ -100<br> c. 5x ≥ 20
postnew [5]
<span>a. n/4 ≤ -1

Multiply both sides by 4 => n ≤ - 4, which is all the real numbers less or equal than - 4.

That in the real number line is all the numbers to the left of - 4 (including  -4)

The matching graph is the B.


b. -10n ≥ -100

Divide both sides by - 10 => n ≤ 10

That is all the real numbers less or equal than 10.

In the real number line it is all the numbers to the left of 10, including 10.

So, the matching graph is the A.


c. 5x ≥ 20

Divide both sides by 5 => x ≥ 4

That is all the real numbers greater or equal to 4.

In the real number line it is all the numbers to the right of 4, including 4.

The matching graph is C.</span>
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Answer:

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