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scoundrel [369]

3 years ago

6

abc pens sells pens in boxes of 12 their competitor xyz sells pens in boxes of 144 an office building is considering purchasing

either 1000 boxes from abc pens or 100 boxes from xyz how many pens are in 1000 boxes of abc pens

2 answers:

AfilCa [17]3 years ago

8 0

Abc- 12 pens in each box
1,000 boxes are needed of ABC pens
1000x12=12,000
Answer is 12,000 pens are in 1000 Abc boxes of pens.

MAVERICK [17]3 years ago

6 0

Answer:

Step-by-step explanation:

Abc- 12 pens in each box

1,000 boxes are needed of ABC pens

1000x12=12,000

Answer is 12,000 pens are in 1000 Abc boxes of pens

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Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of

frez [133]

Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

3) $\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

4) $df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 21 37 58 116

Not Married 59 63 42 164

Total 80 100 100 280

Part 1

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is $\alpha=0.05$

Part 2

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part 3

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{80*116}{280}=33.143$

$E_{2} =\frac{100*116}{280}=41.429$

$E_{3} =\frac{100*116}{280}=41.429$

$E_{4} =\frac{80*164}{280}=46.857$

$E_{5} =\frac{100*164}{280}=58.571$

$E_{6} =\frac{100*164}{280}=58.571$

And the expected values are given by:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 33.143 41.429 41.429 116

Not Married 46.857 58.571 58.571 164

Total 80 100 100 280

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

Part 4

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

7 0

4 years ago

To which płace would you round to estimate 6,904-4,111 explain

Rainbow [258]

To estimate you would round the numbers, so it would be 7,000 - 4,000 = 3,000.

8 0

3 years ago

Find the soultion(s) to the system of equations. Select all that apply

neonofarm [45]

Answer:

(0,-3)

(3,0)

Step-by-step explanation:

The solutions to the system of equations are where the two graphs cross

The first is at x=0 and y=-3

The second is at x=3 and y=0

3 0

3 years ago

Sooooooooo<br> technically i've neevr lost a fight with a tiger

Aloiza [94]

Answer:

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Step-by-step explanation:

3 0

3 years ago

A food company sells salmon to various customers. The mean weight of the salmon is 37 lb with a standard deviation of 2 lbs. The

brilliants [131]

Answer:

The standard deviation for the mean weigth of Salmon is 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount order stores.

Step-by-step explanation:

The mean sample of the sum of n random variables is

$\overline{X} = \frac{X_1+X_2+...+X_n}{n}$

If $X_1, ..., X_n$ are indentically distributed and independent, like in the situation of the problem, then the variance of $X_1 + .... + X_n$ will be the sum of the variances, in other words, it will be n times the variance of $X_1$ .

However if we multiply this mean by 1/n (in other words, divide by n), then we have to divide the variance by 1/n², thus $\overkine{X} = \frac{V(X_1)}{n}$ and as a result, the standard deviation of $\overline{X}$ is the standard deviation of $X_1$ divided by $\sqrt{n}$ .

Since the standard deviation of the weigth of a Salmon is 2 lbs, then the standard deviations for the mean weigth will be:

Restaurants: We have boxes with 9 salmon each, so it will be $\frac{2}{\sqrt{9}} = \frac{2}{3}$
Grocery stores: Each carton has 49 salmon, thus the standard deviation is $\frac{2}{\sqrt{49}} = \frac{2}{7}$
Discount outlet stores: Each pallet has 64 salmon, as a result, the standard deviation is $\frac{2}{\sqrt{64}} = \frac{1}{4}$

We conclude that de standard deivation of the mean weigth of salmon of the types of shipment given is: 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount outlet stores.

7 0

3 years ago