Find the product of (n-3)(n2-2n+4)

9. An article was sold at a profit of 12%. Had it been sold for Nu 80 more, there would be a profit of

Kamila [148]

Answer:

The cost price of the article was Nu 1,000

The original selling price was Nu 1,120

The second selling price was Nu 1,200

Step-by-step explanation:

System of Equations

Let's call:

x = Original selling price of the article

y = Cost of the article

The first condition implies that the difference between the selling price and the cost price (the profit) is 12%. Recall the profit % is always calculated with respect to the cost, that is, the profit at the first condition is:

0.12y = Original profit

Setting up the equation, we have:

$x-y=0.12y$

The second condition states the selling price is 80 Nu more than before, that is:

x+80= Second selling price of the article

y = Cost of the article (it doesn't change)

The profit of this condition is 20%, thus:

0.2y = Second profit

This produces a second equation:

$x+80-y=0.2y$

Let's put both equations together to form a system of equations:

$x-y=0.12y$

$x+80-y=0.2y$

Operating on both equations:

$x=1.12y$

$x=1.2y-80$

Equating both equations, we have an equation with only y's:

$1.12y=1.2y-80$

Subtracting 1.2y on each side:

$1.12y-1.2y=-80$

Operating:

$-0.08y=-80$

Solving:

$\displaystyle y=\frac{-80}{-0.08}=1,000$

a)

The cost price of the article was Nu 1,000

b) The selling price of the article will be calculated in both cases:

Original selling price: x=1.12y= 1.12*1000=1,120

The original selling price was Nu 1,120

The second selling price was x+80=1,200

The second selling price was Nu 1,200

8 0

3 years ago

Directions for questions 4 & 5: We selected a random sample of 100 StatCrunchU students, 67 females and 33 males, and analyz

GaryK [48]

Answer:

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

The lower limit on the confidence interval is -$2164.21.

The upper limit on the confidence interval is -$299.13.

Step-by-step explanation:

The sample data is:

Gender Mean Std. dev. n

Female 2577.75 1916.29 67

Male 3809.42 2379.47 33

We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The sample 1, of size n1=67 has a mean of 2577.75 and a standard deviation of 1916.29.

The sample 2, of size n2=33 has a mean of 3809.42 and a standard deviation of 2379.47.

The difference between sample means is Md=-1231.67.

$M_d=M_1-M_2=2577.75-3809.42=-1231.67$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1916.29^2}{67}+\dfrac{2379.47^2}{33}}\\\\\\s_{M_d}=\sqrt{54808.468+171572.045}=\sqrt{226380.513}=475.795$