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3 years ago

Please use method of substitution numbers 9;10;11;12

Mathematics

1 answer:

rodikova [14]3 years ago

5 0

Answer:

1. x = 1, y = 2.

12. x = 1, y = -3.

Step-by-step explanation:

I won't do all these for you, just the first and the last. The method is similar for all 12.

y = x + 1

x + y = 3

Substitute y = x + 1 into the second equation:

x + y = 3 so:

x + (x + 1) = 3

2x + 1 = 3

2x + 1 -1 = 3 - 1

2x = 2

x = 1.

Now substitute x = 1 into the first equation to get the value of y:

y = x + 1

y = 1 + 1

y = 2.

12.

4x = y + 7

3x + 4y + 9 = 0

Solve for y in the first equation:

4x = y + 7 Subtract 7 from both sides of the equation:

4x - 7 = y + 7 - 7

y = 4x - 7.

Now substitute y = 4x - 7 in the second equation:

3x + 4y + 9 = 0 so:

3x + 4(4x - 7) + 9 = 0

3x + 16x - 28 + 9 = 0

19x - 19 = 0

19x - 19 + 19 = 0 + 19

19x = 19

x = 19/19 = 1.

Now substitute x = 1 in the first equation:

4x = y + 7

4*1 = y + 7 Subtract 7 from both sides of the equation:

4 - 7 = y + 7 - 7

-3 = y.

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Step-by-step explanation:

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Find the area of a rectangle with a perimeter of 12 meters and a base of 4 meters

stepladder [879]

Considering the perimeter (P)
$p = 2 \times b + 2 \times h$
where b is the base and h is the height, then
$12 = 2 \times 4 \times 2 \times h \\ \\ h = \frac{12 - (2 \times 4)}{2} \\ \\ h = 2$
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3 years ago

Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind

Gemiola [76]

Answer:

(a)

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

(b)

$1\ girl = \{GBB, BBG, BGB\}$

$P(1\ girl) = 0.375$

ii.

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

$P(Atleast\ 2 \ girls) = 0.5$

iii.

$No\ girl = \{BBB\}$

$P(No\ girl) = 0.125$

Step-by-step explanation:

Given

$Children = 3$

$B = Boys$

$G = Girls$

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

And the number of elements are:

$n(S) = 8$

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

$1\ girl = \{GBB, BBG, BGB\}$

And the number of the list is;

$n(1\ girl) = 3$

The probability is calculated as;

$P(1\ girl) = \frac{n(1\ girl)}{n(S)}$

$P(1\ girl) = \frac{3}{8}$

$P(1\ girl) = 0.375$

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

And the number of the list is;

$n(Atleast\ 2 \ girls) = 4$

The probability is calculated as;

$P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}$

$P(Atleast\ 2 \ girls) = \frac{4}{8}$

$P(Atleast\ 2 \ girls) = 0.5$

Solving (biii) No girl

From the list of possible elements, we have:

$No\ girl = \{BBB\}$

And the number of the list is;

$n(No\ girl) = 1$

The probability is calculated as;

$P(No\ girl) = \frac{n(No\ girl)}{n(S)}$

$P(No\ girl) = \frac{1}{8}$

$P(No\ girl) = 0.125$

7 0

3 years ago

Derrick will need $39,500 in 10 years for college tuition. How much should his parents invest now at 9.5% annual interest, compo

erastova [34]

Answer:

His parents should invest $15,278.16 to reach this goal ⇒ 4th answer

Step-by-step explanation:

* Lets explain how to solve the problem

- Derrick will need $39,500 in 10 years for college tuition

∴ The future amount is $39,500

∴ The time for investment is 10 years

- P is the money his parents invest now at 9.5% annual interest,

compounded daily

∴ The rate is 9.5% per year compounded daily

- The formula of the compounded interest is:

$A=P(1+\frac{r}{n})^{nt}$ , where

# A is the future value of money

# P is the value of investment

# r is the rate of interest in decimal

# t is the time of investment

# n is the period of the time

∵ A = $39,500

∵ t = 10

∵ r = 9.5/100 = 0.095

∵ n = 365 ⇒ compounded daily

- Lets use the formula above to find P

∴ $39500=P(1+\frac{0.095}{365})^{365*10}$

∴ $39500=p(2.58539)$

- Divide both sides by 2.58539

∴ P = $15278.16

∴ His parents should invest $15,278.16 to reach this goal

7 0

3 years ago