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densk [106]
3 years ago
6

Please help ASAP!!!! You have a bag of change. Your bag consist of 5 Quarters, 20 Pennies, 10 Dimes and 5 Nickels. If you take o

ne out to give to a friend, what are than chances that you would pull a Nickle then a Penny without replacing them. Placing your answer in
simplest fraction form.
Mathematics
1 answer:
butalik [34]3 years ago
5 0

Answer:

5/78 chance to get a Nickel and then a penny without replacing

Step-by-step explanation:

The bag contains a total of 40 coins which show that the probability of getting a Nickel is 5/40 and the probability of getting a penny after taking out a nickel is 20/39. In order to find the Probability of two events together you must multiply the probabilities so

(5/40) * (20/39) = 100/1560 = 5/78

You might be interested in
If point p partitions line BA,what would be the part to whole ratio
IRINA_888 [86]

Answer:

4 / 5

Step-by-step explanation:

From the image attached, Line BA is divided into a total of 5 equal proportions (or intervals). Point P is placed at the 4th proportion of the divided line, also the distance between point P and point A is 1 propotion. The ratio in which point P divides the line BA is given as:

Ratio = BP : PA = 4 : 1

Ratio = 4 : 1

The part to whole ratio If point p partitions line BA = BP / BA = 4 / 5

Part to whole ratio = 4/5

3 0
3 years ago
The price of an item yesterday was $135 . Today, the price rose to $216 . Find the percentage increase.
Softa [21]

Answer:

60% increase

Step-by-step explanation:

Work out the difference increase between the two numbers you are comparing.Then you increase the new number by the original number. Next you divide the increase by the original number and multiply the answer by 100. Finally divide the increase number and the original number then multiply by 100. That will give you your answer.

5 0
3 years ago
Read 2 more answers
What is the GCF of 26 48 62
Bingel [31]
What factors to they have in common?

First, they're all even, so 2

if we divide all of them by 2, we get:
13,24,31

now, 13,31 are primes: so that means that those numbers don't have any more common factors!

So the greatest Common Factor of those 3 numbers is 2.
8 0
3 years ago
Can you plz help? Plzzzzz
katen-ka-za [31]
First question:b 
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3 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 "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" 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 <em>random variable</em> <em>normally distributed</em> (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 <em>normal distribution</em> 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:

  1. We will use the concept of the <em>standard normal distribution</em>, 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 <em>z-score</em>.
  2. A z-score is a kind of <em>standardized value</em> which tells us the <em>distance of a raw score from the mean in standard deviation units</em>. The formula for it is: \\ z = \frac{x - \mu}{\sigma}. Where <em>x</em> is the value for the raw score (in this case x = 5000 miles).
  3. The values for probabilities for the standard normal distribution are tabulated in the <em>standard normal table</em> (available in Statistics books and on the Internet). We will use the <em>cumulative standard normal table</em> (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 <em>below</em> 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)

\\ 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 "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" 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 <em>12 trucks</em>, and this is a problem of <em>the sampling distribution of the means</em>.

In this case, we have samples from a <em>normally distributed data</em>, 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

\\ P(z

\\ 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
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