Ask question

Ask question

All categories

3 years ago

10

using the numbers 3, 5, and 8, can you write nine proper fractions and nine improper fractions? You may use each number only onc

e within a fraction.

1 answer:

Bond [772]3 years ago

6 0

Answer:

First, we can write a fraction as a/b

Where a is the numerator, and b is the denominator.

A proper fraction is a fraction where the numerator is smaller than the denominator.

Using only the given numbers (only once per fraction), some examples of proper fractions are:

3/5

3/8

5/8

3/85

3/58

5/83

5/38

8/35

8/53

You can see that in all of them the denominator is larger than the numerator.

The improper fractions are those where the numerator is equal or larger than the denominator.

The 9 examples using the given numbers are:

5/3

8/5

8/3

35/8

38/5

53/8

58/3

83/5

85/3

You might be interested in

Seclate all the numbers that are solutions to the inequalitt w>1.

natima [27]

Answer:

The numbers should be A. 5 and C. 0.

Hope I Helped

6 0

2 years ago

Read 2 more answers

There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

8 0

3 years ago

Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 t

musickatia [10]

Answer:

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Step-by-step explanation:

Assuming this complete question: Which statement gives a valid interpretation of the interval?

The store manager is 90% confident that the average amount spent by the 36 sampled customers is between S38.01 and $44.29.

There is a 90% chance that the mean amount spent by all customers is between S38.01 and $44.29.

There is a 90% chance that a randomly selected customer will spend between S38.01 and $44.29.

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

8 0

3 years ago

What is the value of x to the nearest tenth? <br> <br><br> A. 9.7 B. 8.1 C. 5.9 D. 7.9

8090 [49]

Use the Pythagorean theorem. The equation for this right triangle would be a^2+b^2=c^2. Since you have one leg and the hypotenuse, you can plug these into the equation. It would be 7^2+b^2=12^2. Do the exponents. You will get 49+b^2=144. You need to get b by itself so subtract each side by 49. You get the equation b^2=95. Square root each side to get b=9.746. This rounds to 9.7. So the answer is A. Hope this helps! ;)

7 0

3 years ago

Read 2 more answers

jen and sarah go to lunch at the Green Grill.Their meals total $28.00 .The tax is 6%.What is the total cost including tax?

nordsb [41]

6%=0.06

28(0.06)
= 1.68 ---> tax

28+1.68
= 29.68

Jen and Sarah spent 29.68 dollars including tax

I hope this answer helped you and have a wonderful day!

5 0

3 years ago

Read 2 more answers