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3 years ago

What Value of m will make this equation true? 3 (m - 5) + 8 = 7 - 1/2 (4m - 11)

Mathematics

1 answer:

KatRina [158]3 years ago

6 0

Answer:

m = 7.1

Step-by-step explanation:

3 (m -5) + 8 = 7 -1/2 (4m - 11)

So, what I did first was distribute the values in the parenthesis to get,

3m - 15 + 8 = 7 -2m -5.5

Now that we have the parenthesis taken care of we can do the simpler math,

3m - 23 = 7 -2m - 5.5

I just added the 15 and 8, so now I move the -5.5 to the opposite side by adding.

3m - 23 = 7 -2m -5.5

3m - 28.5 = 7 - 2m

Here we can do the same with the -2m.

3m - 28.5 = 7 - 2m

5m - 28.5 = 7

To get rid of the -28.5 I added it to the 7 getting an answer of,

5m - 28.5 = 7

5m = 35.5

Finally, divide 5m and 35.5 both by 5.

5m/5 = m

35.5/5 = 7.1

Answer: m=7.1

Sorry it's really long, but I hope this helps! Have a great day!

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A manufacturer produces piston rings for an automobile engine. It is known that ring diameter is normally distributed with σ=0.0

oksian1 [2.3K]

<h2>Answer with explanation:</h2>

The confidence interval for population mean is given by :-

$\overline{x}\pm z^* SE$ (1)

, where $\overline{x}$ = sample mean

z* = critical value.

SE = standard error

and $SE=\dfrac{\sigma}{\sqrt{n}}$ , $\sigma$ = population standard deviation.

n= sample size.

As per given , we have

$\overline{x}=74.021$

$\sigma=0.001$

n= 15

It is known that ring diameter is normally distributed.

$SE=\dfrac{0.001}{\sqrt{15}}=0.000258198889747\approx0.000258199$

By z-table ,

The critical value for 95% confidence = z*= 1.96

A 99% two-sided confidence interval on the true mean piston diameter :

$74.021\pm (2.576) (0.000258199)$ (using (1))

$74.021\pm 0.000665120624$

$74.021\pm 0.000665120624\\\\=(74.021- 0.000665120624,\ 74.021+ 0.000665120624)\\\\=(74.0203348794,\ 74.0216651206)\approx(74.020,\ 74.022)$ [Rounded to three decimal places]

∴ A 99% two-sided confidence interval on the true mean piston diameter = (74.020, 74.022)

By z-table ,

The critical value for 95% confidence = z*= 1.96

A 95% lower confidence bound on the true mean piston diameter:

$74.021- (1.96) (0.000258199)$ (using (1))

$74.021- 0.00050607004=74.02049393\approx74.020$ [Rounded to three decimal places]

∴ A 95% lower confidence bound on the true mean piston diameter= 74.020

7 0

3 years ago

Expand the polynomial and simplify: (4xy− 1 8 y^2)^2

Vinvika [58]

Answer: $4y^{2}(2x-9y)^{2}$

Step-by-step explanation:

We have the following polynomial:

$(4xy-18y^{2})^{2}$

This is a polynomial of the form $(a-b)^{2}=a^{2}-2ab+b^{2}$ . Following this rule to expand it, we have:

$(4xy)^{2}-2(4xy)(18y^{2})+(18y^{2})^{2}$

$16x^{2}y^{2}-144xy^{3}+324y^{4}$

Applying common factor $4y^{2}$ :

$4y^{2}(4x^{2}-36xy+81y^{2})$

Note the polynomial inside the parenthesis is a perfect square trinomial, which can be factored to $(2x-9y)^{2}$ . Hence, the final simplification is: