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3 years ago

Simplify: -7x6 + 5x6

Mathematics

2 answers:

Ira Lisetskai [31]3 years ago

8 0

Answer:

-2x6^

Step-by-step explanation:

i think

Phantasy [73]3 years ago

4 0

Answer:

-2x12

Step-by-step explanation:

I don't know if this will help but at least I tried

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Alex Ar [27]

Answer:

The answer to the question provided is 11.

Step-by-step explanation:

$\: \: \: \: \: 5x + 9 = 6x - 2 \\ \frac{ - 5x \: = - 5x }{9 = 1x - 2} \\ \frac{ + 2 \: \: \: \: = + 2}{ \frac{11}{1} = \frac{1x}{1} } \\ 11 = x$

[I apologize if the answer to the question provided is inaccurate]

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2 years ago

Describe what is meant by a sampling unit. Explain why the sampling unit for verifying the existence of recorded sales differs f

xeze [42]

Answer:

Step-by-step explanation:

The sampling unit is the population item from which the auditor selects sample items. The major consideration in defining the sampling unit is making it consistent with the objectives of the audit tests. Thus, the definition of the population and the planned audit procedures usually dictate the appropriate sampling unit.

The sampling unit for verifying the existence of recorded sales would be the entries in the sales journal since this is the record the auditor wishes to validate.

The sampling unit for testing the possibility of omitted sales is the shipping document from which sales are recorded because the failure to bill a shipment is the exception condition of interest to the auditor.

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3 years ago

At Ralph's fruitstand three apples cost of $.90 you want to buy seven apples how much will they cost

Nata [24]

1 apple is $.30. $.30 times 7 = $2.10.

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3 years ago

A marketing research firm wishes to compare the prices charged by two supermarket chains—Miller’s and Albert’s. The research fir

AVprozaik [17]

Answer:

a) $F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

b) For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$ , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Step-by-step explanation:

Data given and notation

$n_1 = 10$ represent the sampe size for the Miller's Stores

$n_2 =10$ represent the sample size for the Albert's stores

$\bar X_1 =121.92$ represent the sample mean for Miller's store

$\bar X_2 =114.81$ represent the sample mean for Albert's store

$s_1 = 1.4$ represent the sample deviation for the Miller's store

$s_2 = 1.84$ represent the sample deviation for the Albert's stores

$s^2_2 = 12.25$ represent the sample variance for the utility stocks

$\alpha=0.05$ represent the significance level provided

Confidence =0.95 or 95%

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

$F=\frac{s^2_2}{s^2_1}$

Solution to the problem

System of hypothesis

We want to test if the variation for th two groups is the same, so the system of hypothesis are:

H0: $\sigma^2_1 = \sigma^2_2$

H1: $\sigma^2_1 \new \sigma^2_2$

a) Calculate the statistic

Now we can calculate the statistic like this:

$F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_1 -1 =10-1=9$ and for the denominator we have $n_2 -1 =10-1=9$ and the F statistic have 9 degrees of freedom for the numerator and 9 for the denominator. And the P value is given by:

P value

$p_v =2*P(F_{9,9}>1.727)=0.428$

And we can use the following excel code to find the p value:"=2*(1-F.DIST(1.727,9,9,TRUE))"

b) Critical value

For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$ , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Since our calcu

Conclusion

Since the $p_v > \alpha$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the two deviations are different at 5% of significance.