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3 years ago

What does algebra mean in math

2 answers:

8 0

Algebra is a question that uses any letter of the alphabet to replace a missing box, example:
4y equals 20 because 4 x 5 is 20

Ghella [55]3 years ago

5 0

Algebra is solving for the unknown so
if you have 10 apples and you give 5 to your friend, how many apples would you have so that is also 10-5=x solve for x

or it could be more complicated like (x-2)/6+4x-3=17

You might be interested in

Can someone help me with this THX

Sergio [31]

Answer:

4 x 4 = 16 56 divided by 16 is 3.5 I'm pretty sure the answer is 3.5

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Use logarithmic properties and the fact that ln(2) ≈ 0.69 and ln(3) ≈ 1.10 to approximate the value of each of the following log

bija089 [108]

Answer:

ln6 = 1.79

Step-by-step explanation:

The logarithm here has base e so it is a natural logarithm.

$log_ex=lnx$

$lnx=y\\\Rightarrow e^y=x$

According to the product rule of logarithms

$ln(a\times b)=ln(a)+ln(b)$

$6=3\times 2$

$ln6=ln(3\times 2)\\\Rightarrow ln6=ln3+ln2\\\Rightarrow ln6=1.10+0.69\\\Rightarrow ln6=1.79$

∴ ln6 = 1.79

3 0

4 years ago

the slide at the playground has a height of 6 feet.The base of the slide measured on the ground is 8 feet.what is the length of

NemiM [27]

The formula for finding the length is
${a}^{2} + {b}^{2} = {c}^{2}$
All we have to do is plug in the numbers.

${6}^{2} + {8}^{2} = {c}^{2}$
36 + 64 = 100

Since 100 =
${c}^{2}$
we have to find the square root.

$\sqrt{100} = 10$

The length of the slide is 10 feet.

6 0

3 years ago

Find the value of x. If necessary, write your answer in simplest radical form. x=?

Sever21 [200]

Pythagoras’ theorem

a^2 + b^2 = c^2

Where a and b are two sides on a triangle and c is the hypotenuse.

Rearrange to find one to the sides that isn’t the hypotenuse.

a^2 = c^2 - b^2
a^2 = 25^2 - 10^2
a^2 = 625 - 100
a^2 = 525

(square root)

a = √525

= 5√21

5 0

2 years ago

Consider the given data. x 0 2 4 6 9 11 12 15 17 19 y 5 6 7 6 9 8 8 10 12 12 Use the least-squares regression to fit a straight

levacccp [35]

Answer:

See below

Step-by-step explanation:

By using the table 1 attached (See Table 1 attached)

We can perform all the calculations to express both, y as a function of x or x as a function of y.

Let's make first the line relating y as a function of x.

y as a function of x

(y=response variable, x=explanatory variable)

$\bf y=m_{yx}x+b_{yx}$

where

$\bf m_{yx}$ is the slope of the line

$\bf b_{yx}$ is the y-intercept

In this case we use these formulas:

$\bf m_{yx}=\frac{(\sum y)(\sum x)^2-(\sum x)(\sum xy)}{n\sum x^2-(\sum x)^2}$

$\bf b_{yx}=\frac{n\sum xy-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}$

n = 10 is the number of observations taken (pairs x,y)

Note: Be careful not to confuse

$\bf \sum x^2$ with $\bf (\sum x)^2$

Performing our calculations we get:

$\bf m_{yx}=\frac{(83)(95)^2-(95)(923)}{10*1277-(95)^2}=176.6061$

$\bf b_{yx}=\frac{10*923-(95)(83)}{10(1277)-(95)^2}=0.3591$

So the equation of the line that relates y as a function of x is

In order to compute the standard error $\bf S_{yx}$ , we must use Table 2 (See Table 2 attached) and use the definition

$\bf s_{yx}=\sqrt{\frac{(y-y_{est})^2}{n}}$

and we have that standard error when y is a function of x is

$\bf s_{yx}=\sqrt{\frac{39515985}{10}}=1987.8628$

Now, to find the line that relates x as a function of y, we simply switch the roles of x and y in the formulas.

So now we have:

x as a function of y

(x=response variable, y=explanatory variable)

$\bf x=m_{xy}y+b_{xy}$

where

$\bf m_{xy}$ is the slope of the line

$\bf b_{xy}$ is the x-intercept

In this case we use these formulas:

$\bf m_{xy}=\frac{(\sum x)(\sum y)^2-(\sum y)(\sum xy)}{n\sum y^2-(\sum y)^2}$

$\bf b_{xy}=\frac{n\sum xy-(\sum x)(\sum y)}{n(\sum y^2)-(\sum y)^2}$

n = 10 is the number of observations taken (pairs x,y)

Note: Be careful not to confuse

$\bf \sum y^2$ with $\bf (\sum y)^2$

Remark: If you wanted to draw this line in the classical style (the independent variable on the horizontal axis), you would have to swap the axis X and Y)

Computing our values, we get

$\bf m_{xy}=\frac{(95)(83)^2-(83)(923)}{10*743-(83)^2}=1068.1072$

$\bf b_{xy}=\frac{10*923-(95)(83)}{10(743)-(83)^2}=2.4861$

and the line that relates x as a function of y is

To find the standard error $\bf S_{xy}$ we use Table 3 (See Table 3 attached) and the formula

$\bf s_{xy}=\sqrt{\frac{(x-x_{est})^2}{n}}$

and we have that standard error when y is a function of x is

$\bf s_{xy}=\sqrt{\frac{846507757}{10}}=9200.5856$

In both cases the correlation coefficient r is the same and it can be computed with the formula:

$\bf r=\frac{\sum xy}{\sqrt{(\sum x^2)(\sum y^2)}}$

Remark: This formula for r is only true if we assume the correlation is linear. The formula does not hold for other kind of correlations like parabolic, exponential,..., etc.

Computing the correlation coefficient :

$\bf r=\frac{923}{\sqrt{(1277)(743)}}=0.9478$

5 0

4 years ago