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Pavel [41]
3 years ago
10

In the haden company, the standard material cost for the silk used in making a dress is $27.00 based on three square feet of sil

k at a cost of $9.00 per square foot. the production of 1,000 dresses resulted in the use of 3,400 square feet of silk at a cost of $9.20 per square foot. the materials quantity variance is
Mathematics
1 answer:
ad-work [718]3 years ago
5 0
To solve for the materials quantity variance, we have the formula below:
Materials Quantity Variance = (Actual Quantity Used * Standard Rate) - (Standard Quantity Allowed * Standard Rate)

The solution is shown below:
Materials Quantity Variance = ($9.20 - $9.00)*3400
Materials Quantity Variance = $680

The answer is $680.

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The sector of a circle with a 12 in. radius has a central angle measure of 60°. What is the exact area of the sector in terms of
IRINA_888 [86]
The sector of the circle formed by its two radius of length 12 inches has an area of 24 pi. This is obtained using the formula, Area is equal to the product of pi, radius squared and central angle all over 360.
4 0
3 years ago
Christina has a stamp collection she is very proud of. She buys a frame to display her favorite stamps. The frame is 6 inches wi
vladimir1956 [14]

Answer:

6/7 inches

Step-by-step explanation:

If 7 stamps fit perfectly, we can find the width of each stamp by dividing the width of the frame by 7:

= 6/7

So, the width of each stamp is 6/7 inches

8 0
3 years ago
Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa
xxMikexx [17]

Answer:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

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} >19.221)=0.000067

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

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      32   18    50

Medium                                 68     7    75

Large                                     89    11    100

_____________________________________

Total                                     189    36   225

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

H0: independence between heath insurance coverage and size of the company

H1:  NO independence between heath insurance coverage and size of the company

The statistic to check the hypothesis is given by:

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

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{50*189}{225}=42

E_{2} =\frac{50*36}{225}=8

E_{3} =\frac{75*189}{225}=63

E_{4} =\frac{75*36}{225}=12

E_{5} =\frac{100*189}{225}=84

E_{6} =\frac{100*36}{225}=16

And the expected values are given by:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      42    8    50

Medium                                 63     12    75

Large                                     84    16    100

_____________________________________

Total                                     189    36   225

And now we can calculate the statistic:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

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} >19.221)=0.000067

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

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

3 0
3 years ago
P=13 z=4
hram777 [196]

Answer:

you have more outfits than your friend

Step-by-step explanation:

you:

5*13=65outfits

friend:

4*9=36outfits

3 0
3 years ago
An individual needs a daily supplement of at least 464 units of vitamin C and 252 of vitamin E and agrees to obtain this supplem
hichkok12 [17]

Answer:

z (min)  = 705

x₁  = 10

x₂  = 9

Step-by-step explanation:

Let´s call  x₁  quantity of food I ( in ou )  and x₂ quantity of food II ( in ou)

          units of vit. C   units of vit.E   Cholesterol by ou

x₁                 32                      9                         48

x₂                16                      18                          25

Objective function z

z  =   48*x₁   +  25*x₂        To minimize

Subject to:

1.-Total units of vit. C at least  464

32*x₁   +  16*x₂     ≥   464

2.- Total units of vit. E at least  252

9*x₁  + 18*x₂   ≥  252

3.- Quantity of ou per day

x₁   +   x₂   ≤  35

General constraints    x₁  ≥ 0    x₂   ≥ 0

Using the on-line simplex method solver (AtoZmaths) and after three iterations the solution is:

z (min)  = 705

x₁  = 10

x₂  = 9

5 0
3 years ago
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