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1 year ago

prove that ( sin theta cos theta = cot theta ) is not a trigonometric identity by producing a counterexample

Mathematics

1 answer:

user100 [1]1 year ago

8 0

Answer:

To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.

Assume θ with a value and substitute with it.

Let θ = 45°

So, L.H.S = sin θ cos θ = sin 45° cos 45° = (1/√2) * (1/√2) = 1/2

R.H.S = cot θ = cot 45 = 1

So, L.H.S ≠ R.H.S

So, sin θ cos θ = cot θ is not a trigonometric identity.

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J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0

2 years ago

There are 30 businesses with 4 executives each for the new office building. Each business needs one office for each of its

Zanzabum

Answer:

Step-by-step explanation:

B is the right one because the question says that every 4 executives can go into 1 office

8 0

2 years ago

A company purchased a delivery van for $28,000 with a salvage value of $3,000 on September 1, Year 1. It has an estimated useful

Tems11 [23]

Answer:

Depreciation till December 31, Year 1 will be equal to 1,250$

Step-by-step explanation:

Purchasing Cost = 28,000$

Salvage Value = 3,000$

Total Depreciation:

Total Depreciation over 5 years (60 Months) = Purchasing Cost - Salvage Value

Total Depreciation over 5 years (60 Months) = 28,000 - 3,000

Total Depreciation over 5 years (60 Months) = 25,000$

Monthly Depreciation:

Using the unity method we have monthly depreciation by dividing the total depreciation by the total no. of months as below:

Total Depreciation over a single month =25,000/60

Total Depreciation over a single month = 416.67$ (Monthly Depreciation)

Depreciation till December 31, Year 1

As from September 1, Year 1 to December 31, Year 1, its been 3 months therefore total depreciation will be = 3 * Monthly Depreciation

Depreciation till December 31, Year 1 = 3 * 416.67

Depreciation till December 31, Year 1 = 1,250$

4 0

2 years ago

How much profit was generated by the sales of gold label and black label combined?

blsea [12.9K]

The profit genereted by the sales of gold label and black label combined is the sum of the profits generated by the sales of gold label and of the profits generated by the sales of black label.

1. From the table

$\begin{array}{ccc}&\text{Number of bottles sold}&\text{Average profit (per bottle)}\\\text{Gold label}&1,000&\$2.75\end{array}$