1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
postnew [5]
2 years ago
8

Two bags with 3 oranges in each bag is the same amount as 3 bags with 2 oranges in each bag. Which property of multiplication do

es this represent?
Mathematics
1 answer:
kodGreya [7K]2 years ago
7 0

Answer:

Commutative Property

Step-by-step explanation:

The property of multiplication that is being demonstrated is known as Commutative Property. This property basically shows that when multiplying two numbers the actual order in which they are multiplied does not matter and does not affect the result at all. For example, the in this scenario to get the total number of oranges we have to multiply the number of bags by the number of oranges in each bag, but whatever way we do this they equal the same

2 bags * 3 oranges per bag = 6 oranges

3 bags * 2 oranges per bag = 6 oranges

Therefore,

2 * 3 = 3 * 2

You might be interested in
What is the probability of choosing a king from a deck of cards? Type your answer as a fraction.
Vikki [24]

Answer:

4/52 or 1/13

Step-by-step explanation:

there are only 4 kings; diamond, Heart, Spades and Clubs.

4 0
3 years ago
Read 2 more answers
A square field has a side of 42 m. Find the cost of fencing its<br>boundary at Rs. 8 per metré.​
Rufina [12.5K]
I have to say something bc I am sleepy
3 0
3 years ago
An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating co
levacccp [35]

Answer:

1. The 95% confidence interval would be given by (24.8190;27.8010)  

2. 4.7048 \leq \sigma^2 \leq 16.1961

3. t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871    

t_{crit}=1.753

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

4. t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306

\chi^2 =24.9958

Since our calculated value is less than the critical value we don't hav enough evidence to reject the null hypothesis at the significance level provided.

Step-by-step explanation:

Previous concepts

\bar X=26.31 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=2.8 represent the sample standard deviation

n=16 represent the sample size

Part 1

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

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

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=16-1=15

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that t_{\alpha/2}=2.13

Now we have everything in order to replace into formula (1):

26.31-2.13\frac{2.8}{\sqrt{16}}=24.819    

26.31+2.13\frac{2.8}{\sqrt{16}}=27.801

So on this case the 95% confidence interval would be given by (24.8190;27.8010)  

Part 2

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=16-1=15

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,15)" "=CHISQ.INV(0.95,15)". so for this case the critical values are:

\chi^2_{\alpha/2}=24.996

\chi^2_{1- \alpha/2}=7.261

And replacing into the formula for the interval we got:

\frac{(15)(2.8)^2}{24.996} \leq \sigma^2 \leq \frac{(15)(2.8)^2}{7.261}

4.7048 \leq \sigma^2 \leq 16.1961

Part 3

We need to conduct a hypothesis in order to determine if actual mean operating cost is at most 25 cents per mile , the system of hypothesis would be:    

Null hypothesis:\mu \leq 25      

Alternative hypothesis:\mu > 25      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

Calculate the statistic      

We can replace in formula (1) the info given like this:      

t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871    

Critical value  

On this case we need a critical value on th t distribution with 15 degrees of freedom that accumulates 0.05 of th area on the right and 0.95 of the area on the left. We can calculate this value with the following excel code:"=T.INV(0.95,15)" and we got t_{crit}=1.753

Conclusion      

Since our calculated valu it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

Part 4

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 2.3, so the system of hypothesis are:

H0: \sigma \leq 2.3

H1: \sigma >2.3

In order to check the hypothesis we need to calculate the statistic given by the following formula:

t=(n-1) [\frac{s}{\sigma_o}]^2

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 15 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,15)". And our critical value would be \chi^2 =24.9958

Since our calculated value is less than the critical value we FAIL to reject the null hypothesis.

7 0
3 years ago
It costs 15 dollars to send 3 packages through a certain shipping company consider the number of packages pero dollar?
docker41 [41]

solution:

= $15 for 3 packages

= 5 packages per dollar (15/3)


5 0
3 years ago
What is this number in standard form? plzzzzzzzzz i need this
Nookie1986 [14]
The answer is 82.046
3 0
3 years ago
Other questions:
  • the area of a parallelogram is 96 cm squared. The height of the parallelogram is 4 cm less than the length of the base. What is
    11·2 answers
  • 1.ADD (5x4 + 3x3 -x2 +16x)+(2x3 -3x2 -10x4 +x -1)
    9·1 answer
  • Which are the solutions of x = -7x - 8?
    5·1 answer
  • A baseball player's batting average is found by dividing the number of hits by the number of at-bats and rounding the result to
    9·1 answer
  • Given a square with two vertices of one side located at (-5, -3) and (-5, 12), in square units what is its area?
    10·1 answer
  • A trial balance has total debits of $36,000 and total credits of $48,500. Which one of the following errors would create this im
    11·1 answer
  • A soccer team is selling doughnuts to raise money for new uniforms. They are selling the doughnuts for $8 per dozen. They must p
    9·1 answer
  • HELP WITH GEOMETRY TEST QUESTION PLEASE
    8·1 answer
  • after one side of a square was enlarged by 3 inches and the other side was enlarged by 3 inches, the area of the shape was doubl
    7·1 answer
  • What is Seventeen less than three times a number
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!