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3 years ago

1/5(x)( 5/1)=6/3(5/1) What is the value of the x?

Mathematics

2 answers:

VladimirAG [237]3 years ago

8 0

Answer:

The value of x would be 10. I hope this helped :)

GalinKa [24]3 years ago

4 0

Answer:

Step-by-step explanation:

on the left side 1/5 and 5/1 time each other is 1. Which means the x is already alone. On the right 6/3 times 5/1 is 10. So x = 10.

You might be interested in

Find the minimum sample size necessary to be 99% confident that the population mean is within 3 units of the sample mean given t

SVEN [57.7K]

Answer:

The minimum sample size required is 207.

Step-by-step explanation:

The (1 - α) % confidence interval for population mean μ is:

$CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}$

The margin of error of this confidence interval is:

$MOE=z_{\alpha /2}\frac{\sigma}{\sqrt{n}}$

Given:

$MOE=3\\\sigma=29\\z_{\alpha /2}=z_{0.01/2}=z_{0.05}=2.576$

*Use a z-table for the critical value.

Compute the value of n as follows:

$MOE=z_{\alpha /2}\frac{\sigma}{\sqrt{n}}\\3=2.576\times \frac{29}{\sqrt{n}} \\n=[\frac{2.576\times29}{3} ]^{2}\\=206.69\\\approx207$

Thus, the minimum sample size required is 207.

3 0

3 years ago

Part of the proceeds from a garage sale was $455 worth of $5 bills and $20 bills. If there were 6 more $5 bills than 20$ bills f

Feliz [49]

Answer:

$\large \boxed{\text{17 \$20 bills and 23 \$5 bills}}$

Step-by-step explanation:

Let x = the number of $20 bills.

Then x + 6 = the number of $5 bills and

20x + 5(x + 6) = 455

$\begin{array}{rcl}20x + 5(x + 6) & = & 455\\20x + 5x + 30 & = & 455\\25x + 30 & = & 455\\25x & = & 425\\x & = &\dfrac{425}{25}\\\\x & = &\mathbf{17}\\\end{array}$

x = 17 = number of $20 bills

x + 6 = 17 + 6 = 23 = number of $5 bills

$\text{There were }\large \boxed{\textbf{17 \$20 bills and 23 \$5 bills}}$

Check:

$\begin{array}{rcl}20\times17+ 5(17 + 6) & = & 455\\340 + 5\times23& = & 455\\340 + 115 & = & 455\\455 & = & 455\\\end{array}$

OK.

7 0

3 years ago

-5 - (x - 6) = 7x - 23

drek231 [11]

The answer is x = 3. I hope this helped

3 0

4 years ago

Read 2 more answers

Factor completely and then place the factors in the proper location on the grid a^2-a-20

Gnoma [55]

Answer:

$(x-5)(x+4)$

Step-by-step explanation:

We have the following quadratic equation

$a^2-a-20$

The factored expression will have the following form

$(x+b)(x+c)$

Where b and c are two real numbers which when added together result in -1, and multiplying them results in -20

$b*c=-20\\b+c=-1$

You can verify that the numbers that meet this condition are:

-5 and 4

Then the factors are:

$(x-5)(x+4)$

3 0

3 years ago

#5 of the way to solve it?

erma4kov [3.2K]

Put 12/100 and see how many times 100 can go into 300 and the number you multiplied multiply that number by 12.

5 0

4 years ago