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4 years ago

Express 2100 as the product of its prime factors

Mathematics

1 answer:

aleksandrvk [35]4 years ago

8 0

2100
/     \
2     1050/       \
2      525/       \
3      175/       \
5      35/       \
5      7/       \
7      1

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julsineya [31]

× = .5 or 1/2 because .5×4=2

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3 years ago

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3 years ago

A fruit seller sold 50 kg of orange at the rate of rupees 80 per kg and gained rupees 800 calculate his purchasing rate.

Lorico [155]

Step-by-step explanation:

selling price of 1 kg oranges = rupees 80

so, selling price of 50 kg oranges :-

=》80 × 50

=》4000

now, profit = 800

so, Cost price (C.P) = selling price - profit

=》C.P = 4000 - 800

=》C.P = 3200

SO, purchasing rate per kg was

=》3200 ÷ 50 = 64

hence, purchasing rate = 64 rupees per kg

profit % =

$\frac{profit}{cost \: price} \times 100$

=》

$\frac{800}{3200} \times 100$

=》

$\frac{1}{4} \times 100$

=》

$25$

hence, profit % = 25%

3 0

3 years ago

7n + 7 = 42 help me please

Nata [24]

Answer:

Step-by-step explanation:

7n+7=42

7n=35

n=5

3 0

3 years ago

According to industry sources, online banking is expected to take off in the near future. The projected number of households (in

Airida [17]

Answer:

(a) The least-square regression line is: $y=4.662+2.709x$ .

(b) The number of households using online banking at the beginning of 2007 is 31.8.

Step-by-step explanation:

The general form of a least square regression line is:

$y=\alpha +\beta x$

Here,

y = dependent variable

x = independent variable

α = intercept

β = slope

(a)

The formula to compute intercept and slope is:

$\begin{aligned} \alpha &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} \\\beta &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} \end{aligned}$

The values of ∑X, ∑Y, ∑XY and ∑X² are computed in the table below.

Compute the value of intercept and slope as follows:

$\begin{aligned} \alpha &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} = \frac{ 68.6 \cdot 55 - 15 \cdot 218.9}{ 6 \cdot 55 - 15^2} \approx 4.662 \\ \\\beta &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} = \frac{ 6 \cdot 218.9 - 15 \cdot 68.6 }{ 6 \cdot 55 - \left( 15 \right)^2} \approx 2.709\end{aligned}$

The least-square regression line is:

$y=4.662+2.709x$

(b)

For the year 2007 the value of x is 10.

Compute the value of y for x = 10 as follows:

$y=4.662+2.709x$

$=4.662+(2.709\times10)\\=4.662+27.09\\=31.752\\\approx 31.8$

Thus, the number of households using online banking at the beginning of 2007 is 31.8.

5 0

3 years ago