The width of a laptop is 11.25 inches. The width is 0.75 times the length. What is the length of the laptop?

ollegr [7]

Hello!

For a:

How I would do this, is I would first say if all 46 animals (heads) were chickens, how many legs would there be? Each chicken has 2 legs, so 46 * 2 = 92. The total amount of legs is 96 as stated in the question, so if all of the animals were chickens, the farmer would be 4 legs short.

Now to add rabbits into the equation. Rabbits have 4 legs, and chickens have 2. You want to find the difference between the two, because as you add rabbits to the animals the farmer has, then you have to take away chickens at the same time. 4-2 = 2, so for each rabbit you replace, you add 2 legs.

Since the farmer is 4 legs short with all chickens, then you just divide that 4 by the 2 legs you add by replacing a chicken with a rabbit.

4 / 2 = 2 rabbits

So that means there are 2 rabbits. Since there are 46 heads in total, if 2 are rabbits, that means there are 44 chickens.

So there are 44 chickens and 2 rabbits.

b)

You can follow the same steps: I'm assuming all are child tickets for now:

3.05 * 100 = $305

And now you find how much money short you are.

498.6 - 305 = 193.6

Next, you find the difference in the ticket costs.

5.25 - 3.05 = 2.20

And you divide to find the number of adult tickets.

193.6 / 2.2 = 88

Since 100 tickets were sold, and 88 adult tickets were sold, that means 12 child tickets were sold.

7 0

3 years ago

Write the letter of the correct answer on your answer sheet. If your answer is not found among

s344n2d4d5 [400]

Answer:

Hello dear asker, you have to put the picture of the graph, then I would be happy to help

4 0

3 years ago

Michael bought 0.44 pound of slice turkey. what is the value of the digit in the hundredths place?

jarptica [38.1K]

Answer:

4

Step-by-step explanation:

You have 0.44

0 is in the ones place

4 is in the tenths place (0.4)

And the other 4 is in the hundredths place (0.04)

8 0

3 years ago

Read 2 more answers

2. The time between engine failures for a 2-1/2-ton truck used by the military is

OLEGan [10]

Answer:

A truck "will be able to travel a total distance of over 5000 miles without an engine failure" with a probability of 0.89435 or about 89.435%.

For a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

Step-by-step explanation:

We have here a random variable normally distributed (the time between engine failures). According to this, most values are around the mean of the distribution and less are far from it considering both extremes of the distribution.

The normal distribution is defined by two parameters: the population mean and the population standard deviation, and we have each of them:

$\\ \mu = 6000$ miles.

$\\ \sigma = 800$ miles.

To find the probabilities asked in the question, we need to follow the next concepts and steps:

We will use the concept of the standard normal distribution, which has a mean = 0, and a standard deviation = 1. Why? With this distribution, we can easily find the probabilities of any normally distributed data, after obtaining the corresponding z-score.
A z-score is a kind of standardized value which tells us the distance of a raw score from the mean in standard deviation units. The formula for it is: $\\ z = \frac{x - \mu}{\sigma}$ . Where x is the value for the raw score (in this case x = 5000 miles).
The values for probabilities for the standard normal distribution are tabulated in the standard normal table (available in Statistics books and on the Internet). We will use the cumulative standard normal table (see below).

With this information, we can solve the first part of the question.

The chance that a truck will be able to travel a total distance of over 5000 miles without an engine failure

We can "translate" the former mathematically as:

$\\ P(x>5000)$ miles.

The z-score for x = 5000 miles is:

$\\ z = \frac{5000 - 6000}{800}$

$\\ z = \frac{-1000}{800}$

$\\ z = -1.25$

This value of z is negative, and it tells us that the raw score is 1.25 standard deviations below the population mean. Most standard normal tables are made using positive values for z. However, since the normal distribution is symmetrical, we can use the following formula to overcome this:

$\\ P(z$

So

$\\ P(z$

Consulting a standard normal table available on the Internet, we have

$\\ P(z$

Then

$\\ P(z1.25)$

However, this value is for P(z<-1.25), and we need to find the probability P(z>-1.25) = P(x>5000) (Remember that we standardized x to z, but the probabilities are the same).

In this way, we have

$\\ P(z>-1.25) = 1 - P(z$

That is, the complement of P(z<-1.25) is P(z>-1.25) = P(x>5000). Thus:

$\\ P(z>-1.25) = 1 - 0.10565$

$\\ P(z>-1.25) = 0.89435$

In words, a truck "will be able to travel a total distance of over 5000 miles without an engine failure" with a probability of 0.89435 or about 89.435%.

We can see the former probability in the graph below.

The chance that a fleet of a dozen trucks will have an average time-between-failures of 5000 miles or more

We are asked here for a sample of 12 trucks, and this is a problem of the sampling distribution of the means.

In this case, we have samples from a normally distributed data, then, the sample means are also normally distributed. Mathematically:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

In words, the samples means are normally distributed with the same mean of the population mean $\\ \mu$ , but with a standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ .

We have also a standardized variable that follows a standard normal distribution (mean = 0, standard deviation = 1), and we use it to find the probability in question. That is

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z \sim N(0, 1)$

Then

The "average time-between-failures of 5000" is $\\ \overline{x} = 5000$ . In other words, this is the mean of the sample of the 12 trucks.

Thus

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{5000 - 6000}{\frac{800}{\sqrt{12}}}$

$\\ z = \frac{-1000}{\frac{800}{\sqrt{12}}}$

$\\ z = \frac{-1000}{230.940148}$

$\\ z = -4.330126$

This value is so low for z, that it tells us that P(z>-4.33) is almost 1, in other words it is almost certain that for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is almost 1.

$\\ P(z$