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kykrilka [37]
1 year ago
5

Ellie has a delivery van which she is going to deliver boxes each box is a cuboid brain.ly 45cm by 40cm by 35cm the space in the

van is empty and by has maximum length 3.6m maximum width 1.6m maximum height of 2.1m Ellie wants to put as many boxes possible in the van she can put 3 boxes in the van in 1 minute how many minutes will it take ellie to put as many boxes in a van as possibe
Mathematics
1 answer:
jek_recluse [69]1 year ago
4 0

Answer:

It will take Ellie 64 minutes to put as many boxes as possible.

Step-by-step explanation:

Let us work with meters.

The dimensions of Ellie's boxes in meters are: 0.45m by 0.40m by 0.35 ( <em>to convert from centimeters to meters we just divide by 100, because 1m =100cm).</em> therefore the volume of each box is:

<em>V_{box}=0.45m*0.40m*0.35m=0.063m^3</em>

Now the dimensions of the empty van are 3.6m by 1.6m by 2.1 m, therefore its volume V_{van} is:

V_{van}=3.6m*1.6m*2.1m=12.096m^3.

So the amount of boxes that Ellie can put in the van is equal to the volume of the van V_{van} divided by the volume V_{box} of each box:

\frac{V_{van}}{V_{box}} =\frac{12.096}{0.063}=\boxed{192\:boxes }

So 192 boxes can be put into the van.

Now Ellie can put 3 boxes in the van in 1 minute, therefore the amount of time it will take her to put 192 boxes into the van will be:

\frac{192boxes}{3boxes/minute} =\boxed{64\:minutes}

So it takes Ellie 64 minutes to put as many boxes into the van as she can.

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A certain brand of electric bulbs has an average life of 300 hours with a standard deviation of 45. A random sample of 100 bulbs
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Answer:

P(\bar{x}

Step-by-step explanation:

Average life of electric bulbs = 300 hrs

standard deviation = 45

Total no of bulbs tested = 100

Probability that sample mean is less than 295 =?

Using Central Limit Theorem

P(\bar{x}

From table of normal distribution:

P(\bar{x}

3 0
1 year ago
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[Q71 Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters g = 71 inch and 02 = 6
viktelen [127]

Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

              (b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

Step-by-step explanation:

Given that,

                  Height (in inches) of a 25 year old man is a normal random variable with mean g=71 and variance o^{2} =6.25.

To find:  (a) What percentage of 25 year old men are 6 feet, 2 inches tall

               (b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.

Now,

(a) To calculate the percentage of men, we have to calculate the probability

P[Height of a 25 year old man is over 6 feet 2 inches]= P[X>74in]

                           P[X>74] = P[\frac{X-g}{o} > \frac{74-71}{2.5}]

                                         = P[Z > 1.2]

                                         = 1 - P[Z ≤ 1.2]

                                         = 1 - Ф (1.2)

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Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

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     P[X > 6ft 5in I X > 6ft] = P[X > 77 I X > 72]

                                          = \frac{P[X > 77]}{P[ X > 72]}

                                          = \frac{P[\frac{X - g}{o}>\frac{77-71}{2.5}]  }{P[\frac{X-g}{o} >\frac{72-71}{2.5}] }

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                                          =  \frac{1-P[Z\leq2.4] }{1-P[Z\leq0.4] }

                                          = \frac{1-0.9918}{1-0.6554}

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                                          = 0.024

Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

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Solve for x in the equation x^2+10x+12=36
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Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in
ASHA 777 [7]

Answer:

a) There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b) There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c) The expected number of Chrome users is 4.074.

d) The variance for the number of Chrome users is 3.2441.

The standard deviation for the number of Chrome users is 1.8011.

Step-by-step explanation:

For each Internet browser user, there are only two possible outcomes. Either they use Chrome, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

Google Chrome has a 20.37% share of the browser market. This means that p = 0.2037

20 Internet users are sampled, so n = 20.

a.Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

This is P(X = 8).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{20,8}.(0.2037)^{8}.(0.7963)^{12} = 0.0243

There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b.Compute the probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

Either there are less than 3 Chrome users, or there are three or more. The sum of the probabilities of these events is decimal 1. So:

P(X < 3) + P(X \geq 3) = 1

P(X \geq 3) = 1 - P(X < 3)

In which

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{20,0}.(0.2037)^{0}.(0.7963)^{20} = 0.0105

P(X = 1) = C_{20,1}.(0.2037)^{1}.(0.7963)^{19} = 0.0538

P(X = 2) = C_{20,2}.(0.2037)^{2}.(0.7963)^{18} = 0.1307

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0105 + 0.0538 + 0.1307 = 0.1950

P(X \geq 3) = 1 - P(X < 3) = 1 - 0.1950 = 0.8050

There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c.For the sample of 20 Internet browser users, compute the expected number of Chrome users

We have that, for a binomial experiment:

E(X) = np

So

E(X) = 20*0.2037 = 4.074

The expected number of Chrome users is 4.074.

d.For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

We have that, for a binomial experiment, the variance is

Var(X) = np(1-p)

So

Var(X) = 20*0.2037*(0.7963) = 3.2441

The variance for the number of Chrome users is 3.2441.

The standard deviation is the square root of the variance. So

\sqrt{Var(X)} = \sqrt{3.2441} = 1.8011

The standard deviation for the number of Chrome users is 1.8011.

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