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2 years ago

If ce = 7x+4, find the value of x

Mathematics

2 answers:

Vanyuwa [196]2 years ago

6 0

Answer:

x=ce-4/7

Step-by-step explanation:

ce = 7x +4

ce + (-4) = (7x +4) + (-4)

ce - 4 = 7 X + 4 - 4

x= ce-4/7

Kryger [21]2 years ago

4 0

Answer:

x = $\frac{y}{7} - \frac{4}{7}$

Step-by-step explanation:

Since it will be easier, just set "ce" as one term. Let ce = y

You are solving for the variable, x. Note the equal sign, what you do to one side, you do to the other.

Do the opposite of PEMDAS.

First, subtract 4 from both sides:

y = 7x + 4

y (-4) = 7x + 4 (-4)

y - 4 = 7x

Next, Divide 7 from both sides:

(y - 4)/7 = (7x)/7

(y - 4)/7 = x

x = (y/7) - (4/7)

Your value of x is $\frac{y}{7} - \frac{4}{7}$

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In a function, y varies inversely with x. The constant of variation is 4. Which table could represent the function?

Nimfa-mama [501]

The equation will be y=4/x. u can make the table of variables by inserting the values of x in the equation. for suppose here, if x=1 then y=4 , x=2 then y=2 and so on.

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1 year ago

Read 2 more answers

(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX

pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable X follows a Normal distribution with mean μ = 10 and variance σ² = 36.

(a)

Compute the value of P (X > 5) as follows:

$P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z$

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

$P(4$

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

$P(X$

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

$P(X$

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

$P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z$

Thus, the value of P (X > 16) is 0.1587.

**Use a z-table for the probabilities.

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2 years ago

If r(x) = 3x – 1 and s(x) = 2x + 1, which expression is equivalent to (StartFraction r Over s EndFraction) (6)?

Airida [17]

Answer:

$\dfrac{r}{s}(6)=\dfrac{3(6)-1}{2(6)+1}$

Step-by-step explanation:

We know that for any two function f(x) and g(x) ,

$\dfrac{f}{g}(x)=\dfrac{f(x)}{g(x)}$

Given functions : $r(x)=3x-1$ and $s(x)=2x+1$

Then, $\dfrac{r}{s}(x)=\dfrac{r(x)}{s(x)}$

$\Rightarrow\ \dfrac{r}{s}(x)=\dfrac{3x-1}{2x+1}$

At x= 6 , we get

$\dfrac{r}{s}(6)=\dfrac{3(6)-1}{2(6)+1}$

The , The expression is equivalent to $\dfrac{r}{s}(6)=\dfrac{3(6)-1}{2(6)+1}$

When we further simplify it , we get $\dfrac{r}{s}(6)=\dfrac{18-1}{12+1}=\dfrac{17}{13}$

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2 years ago

David can proofread15 reports in an hour while Armando can proofread 37reports in an hour. If they have 400 reports to proofread

ycow [4]

Together, it would take them 6 hours

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1 year ago

Read 2 more answers

Mrs.Steﬀen’s third grade class has 30 students in it. The students are divided into three groups(numbered 1, 2,and 3),each havin

qaws [65]

Answer:

a. $\\ 10! = 3628800$ ;

b. $\\ 10!*10!*10! = 47784725839872000000 = 4.7784725839872*10^{19}$

Step-by-step explanation:

We need here to apply the Multiplication Principle or the Fundamental Principle of Counting for each answer. Answer b needs an extra reasoning for being completed.

The Multiplication Principle states that if there are n ways of doing something and m ways of doing another thing, then there are n x m ways of doing both (Rule of product (2020), in Wikipedia).

<h3>In how many ways can ten students line up? </h3>

There are ten students. When one is selected, there is no other way to select it again. So, no repetition is allowed.

Then, in the beginning, there are 10 possibilities for 10 students; when one is selected, there are nine possibilities left. When another is selected, eight possibilities are left to form the file, and so on.

Thus, we need to multiply the possibilities after each selection: that is why the Multiplication Principle is important here.

This could be expressed mathematically using n!:

$\\ n! = n * (n-1)! * (n-2)! *...* 2*1.$

For instance, $\\ 5! = 5 * (5-1)! * (5-2)! *...*2*1 = 5 * 4 * 3 * 2 * 1 = 120.$

So, for the case in question, the ten students can line up in:

$\\ 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800$ ways to line up in a single file.

<h3>Second Question</h3>

For this question, we need to consider the former reasoning with extra consideration in mind.

The members of Group 1 can occupy only the following places in forming the file:

$\\ G1 = \{ 1, 4, 7, 10, 13, 16, 19, 22, 25, 28\}^{th}$ places.

The members of Group 2 only:

$\\ G2 = \{ 2, 5, 8, 11, 14, 17, 20, 23, 26, 29\}^{th}$ places.

And the members of Group 3, the following only ones:

$\\ G3 = \{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30\}^{th}$ places.

Well, having into account these possible places for each member of G1, G2 and G3, there are: 10! ways for lining up members of G1; 10! ways for lining up members of G2 and, also, 10! ways for lining up members of G3.

After using the Multiplication Principle, we have, thus:

$\\ 10! * 10! * 10! = 47784725839872000000 = 4.7784725839872 *10^{19}$ ways the students can line up to come in from recess.

3 0

1 year ago