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3 years ago

Solve the following: 2(x + 3) = x - 4 4(5x - 2)= 2(9x + 3)

Mathematics

2 answers:

pantera1 [17]3 years ago

4 0

Answer:

x = -10

x = 7

Step-by-step explanation:

2(x+3) = x - 4

2x + 6 = x - 4

2x -x = - 4 - 6

x = - 10

4(5x-2) = 2(9x+3)

20x - 8 = 18x +6

20x - 18x = 6 + 8

2x = 14

x = 7

have a nice day :D

vredina [299]3 years ago

4 0

2x(x + 3)= x - 4
2x + 6 = x - 4
2x - x - 4
X = -10

4(5x + -2) = 2(9x + 3)
20x - 8 = 18x + 6
20x -18x = 6 + 8
2x/2 = 14\2
X = 7

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Zachary earns $5 per hour, and he makes a $20 bonus.

stich3 [128]

Answer:

5x?

Step-by-step explanation:

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2 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

If each value of the data of the poplation function is doubled which of the following is not correct?

Anna35 [415]

Answer:

The variance is doubled; C

Step-by-step explanation:

Here, we want to select what will be wrong if every number in the dataset is doubled

A. This is correct

The mean is the average of the numbers which is calculated by dividing the sum of the numbers by the count of the numbers

By doubling the numbers, we have simply doubled the sum while the count is same

So what we have as the result of the division will be 2 times what we had initially making the mean doubled

B. The mode is doubled

Yes, this is correct

The mode is the highest in frequency

Doubling all numbers will not double their frequencies but their values. So the number that was the mode in the previous case will be the same but now doubled

D. The median is doubled

This is correct

The median deals with the middle number.

Since all numbers are doubled in the present case, the median to is also doubled along side as the numbers are now twice what they were evenly

C. This is wrong

The variance has to do with the spread of data

We are actually subtracting values

Since we are doubling across board, the difference between the numbers is the same as this, the variance cannot change

We can look at this from the angle that all others are measures of central tendency looking at having a central figure while the variance is a measure of spread.

3 0

2 years ago

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by: