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AlladinOne [14]
2 years ago
15

Determine the total value of the merchandise using net realizable value. item quantity selling price commission doll 10 $7 $2 ho

rse 5 9 3 $115 $35 $25 $80
Mathematics
2 answers:
NemiM [27]2 years ago
5 0

Quantity of dolls and horses are 10 and 5 respectively .

Selling prices of dolls and horses are $7 and $ 9 respectively .

And the commission on them are $2 and $ 3 respectively .

So net price = 10*(7-2) + 5(9-3) = 10+5 + 5*6 = 50+30 = 80

So the total value of the merchandise using net realizable value is $ 80. Therefore out of the given options, correct option is the last option .

tankabanditka [31]2 years ago
4 0

Answer:

The correct option is 4.

Step-by-step explanation:

The given data table is

Item            quantity            selling price            commission

Doll                   10                       $7                       $2

Horse                 5                       $9                       $3

Total value of the merchandise using net realizable value is

Total value of the merchandise = Selling price -  Commission

The value of the merchandise using net realizable value for doll is

10(7-2)=10(5)\Rightarrow \$ 50

The value of the merchandise using net realizable value for horse is

6(9-3)=5(6)\Rightarrow \$ 30

Total value of the merchandise using net realizable value is

Total=\$ 50+\$ 30=\$ 80

Therefore the correct option is 4.

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Dividing by 150, we get ratio of Metal 1 thicknesses = 1:2:4.

Metal 2 thicknesses = 130lbs, 260lbs, 390lbs.

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<u>We can see the ratios of thicknesses of Metal 1 are 1:2:4 but it should be 1:2:3.</u>

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"Immediately after a ban on using hand-held cell phones while driving was implemented, compliance with the law was measured. A r
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    Alternate Hypothesis, H_A : p_1-p_2\neq 0  or  p_1\neq p_2

(b) We conclude that there is a statistical difference in these two proportions measured initially and then one year later.

Step-by-step explanation:

We are given that a random sample of 1,250 drivers found that 98.9% were in compliance. A year after the implementation, compliance was again measured to see if compliance was the same (or not) as previously measured.

A different random sample of 1,100 drivers found 96.9% compliance."

<em />

<em>Let </em>p_1<em> = proportion of drivers that were in compliance initially</em>

p_2<em> = proportion of drivers that were in compliance one year later</em>

(a) <u>Null Hypothesis</u>, H_0 : p_1-p_2=0  or  p_1= p_2      {means that there is not any statistical difference in these two proportions measured initially and then one year later}

<u>Alternate Hypothesis</u>, H_A : p_1-p_2\neq 0  or  p_1\neq p_2     {means that there is a statistical difference in these two proportions measured initially and then one year later}

The test statistics that will be used here is <u>Two-sample z proportion statistics</u>;

                     T.S.  = \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{ \frac{\hat p_1(1-\hat p_1)}{n_1} + \frac{\hat p_2(1-\hat p_2)}{n_2}} }  ~ N(0,1)

where, \hat p_1 = sample proportion of drivers in compliance initially = 98.9%

\hat p_2 = sample proportion of drivers in compliance one year later = 96.9%

n_1 = sample of drivers initially = 1,250

n_2 = sample of drivers one year later = 1,100

(b) So, <u><em>the test statistics</em></u>  =  \frac{(0.989-0.969)-(0)}{\sqrt{ \frac{0.989(1-0.989)}{1,250} + \frac{0.969(1-0.969)}{1,100}} }  

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<u>Now, P-value of the test statistics is given by;</u>

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Since in the question we are not given with the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical values between -1.96 and 1.96 for two-tailed test.

<em>Since our test statistics does not lies within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis.</em></u>

Therefore, we conclude that there is a statistical difference in these two proportions measured initially and then one year later.

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