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At the local grocery store, a 16-ounce bottle of apple juice costs $3.20. What is the cost of the apple juice per ounce? Show wo

oee [108]

The answer is $0.20

When you say 3.20 ÷ 16, it gives you 0.2, but you need to add a zero at the end to make it understandable, so that is how you end up with $0.20

4 0

3 years ago

PLSS PLS PLS HELP ME I NEED THE ANSWER question: solve for this solution using substitution.

Ira Lisetskai [31]

Answer:point form= (-10,16)

Equation form : x=-10,y=16

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Please help

nalin [4]

Answer:

12870ways

Step-by-step explanation:

Combination has to do with selection

Total members in a tennis club = 15

number of men = 8

number of women = 7

Number of team consisting of women will be expressed as 15C7

15C7 = 15!/(15-7)!7!

15C7 = 15!/8!7!

15C7 = 15*14*13*12*11*10*9*8!/8!7!

15C7 = 15*14*13*12*11*10*9/7 * 6 * 5 * 4 * 3 * 2

15C7 = 15*14*13*12*11/56

15C7 = 6,435 ways

Number of team consisting of men will be expressed as 15C8

15C8 = 15!/8!7!

15C8 = 15*14*13*12*11*10*9*8!/8!7!

15C8 = 15*14*13*12*11*10*9/7 * 6 * 5 * 4 * 3 * 2

15C8 = 6,435 ways

Adding both

Total ways = 6,435 ways + 6,435 ways

Total ways = 12870ways

Hence the required number of ways is 12870ways

4 0

2 years ago

A square pyramid has a height of 12 feet. Each side to of base is 7 feet long. use the following formula to calculate the pyrami

MissTica

Let's find the area of the base first:
7×7=49
The S stands for the sides of the base. Since you are multiplying them by each other, you are basically squaring them.

Let's now multiply the area of the base by the height.
49×12=588

Now, let's multiply by 1/3.
588×1/3 or 588÷3
196

So, the volume is 196 ft³.

4 0

3 years ago

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.

$\\ P(z$

The complement of P(z<-4.33) is:

$\\ P(z>-4.33) = 1 - P(z$ or practically 1.

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

7 0

3 years ago