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4vir4ik [10]
3 years ago
7

How to rewrite an expression using the GCF?

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
4 0
GCF is 3. Divide everything by 3 and get 1x + 4
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Which situation can be represented by this inequality?
sasho [114]

Answer:

C

Step-by-step explanation:

the greater than or less then symbol means at least and $425 means it has to be added in each month.

5 0
3 years ago
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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
Marcello is constructing the perpendicular bisector of AB. He has already constructed arcs as shown.
Andrews [41]
Did u get the answer please help me i am taking the test right now?!

8 0
3 years ago
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Find the coordinates of the reflection: B (8,-1) is reflected over the y-axis
Stolb23 [73]

Answer:

good for u

Step-by-step explanation:

3 0
3 years ago
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Which of the following is not one of the 8th roots of unity?
Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1.  Since z=1=1+0\cdot i, then

Rez=1,\\ \\Im z=0

and

|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.

This gives you \varphi=0.

Thus,

z=1\cdot(\cos 0+i\sin 0).

2. The 8th roots can be calculated using following formula:

\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.

Now

at k=0,  z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;

at k=1,  z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};

at k=2,  z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;

at k=3,  z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};

at k=4,  z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;

at k=5,  z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};

at k=6,  z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;

at k=7,  z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};

The 8th roots are

\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.

Option C is icncorrect.

5 0
3 years ago
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