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ki77a [65]
3 years ago
8

How do i know if this linear? on both

Mathematics
2 answers:
Gre4nikov [31]3 years ago
4 0
This isn't linear if it wasn't linear, the last y wouldn't be 4
Serggg [28]3 years ago
3 0
Neither is linear , when a number is over your variable x it will not be linear , if the ten were beside it , it would be linear since that would be the slope but in this case it is above , and the second one wouldn’t either since neither x or y remains constant the entire time. Hope this helped !
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Please answer ASAP! An explanation is a MUST REQUIREMENT in order to receive points and the Brainliest answer. Thank you.
alexdok [17]
Group and factor/undistribute

(x^2y^3-2y^3)+(-2x^2+4)
(y^3)(x^2-2)+(-2)(x^2-2)
see the (x^2-2) is common term so undistribute
(x^2-2)(y^2-2)

last one
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Dr. Watson combines 400 mL of detergent 800 ml of alcohol and 1500 mL of water how many liters of solution does he have
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Dr.Watson has 2.7 liters of solution. 
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Simplify (square root)2/^3(square root)2
Whitepunk [10]

Answer: Personally I would do option "B" 2 1/3 because it sounds right.

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3 years ago
The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article "Testing P
zysi [14]

Answer:

Step-by-step explanation:

Hello!

Given the variable

X: breakdown voltage of an insulating liquid under certain conditions.

a)

Graphic in attachment.

Box: The median is closer to the third quartile than the first quartile showing skewness to the left.

The black dot in the middle of the box represents the sample mean (X[bar]) as you can see it is also moved to the right side of the distribution, affected by the presence of an outlier.

Whiskers:

The upper whisker is smaller than the lower one and there is a value that can be considered an outlier.

Altogether you can say that the distribution of the data set is skewed to the left.

b)

To estimate the population mean you need the population to be at least normal. The box plot shows that the distribution is not symmetrical but, considering that the sample size is n= 48 you can apply the Central Limit Theorem and approximate the sampling distribution to normal.

X[bar]≈N(μ;σ²)

As the distribution is approximate, using the sample standard deviation (since we don't know the population value) for the distribution standard deviation is also valid ⇒ Z= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

∑X= 2626 ∑X²= 144950 n= 48

X[bar]= ∑X/n= 2626/48= 54.71

S²= 1/n[∑X²- (∑X)²/n]= 1/47[144950-(2626)²/48]= 27.36

S= 5.23

Z_{1-\alpha /2}= Z_{0.975}= 1.96

54.71 ± 1.96 * \frac{5.23}{\sqrt{48} }

[53.22; 56.18]kV

c)

95% CI

amplitude a=2 kV  i.e. semi amplitude d= 1 kV

The semi amplitude or margin of error for the CI can be calculated as:

d= Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

From this formula you can clear the sample size

\frac{d}{Z_{1-\alpha /2}} = \frac{S}{\sqrt{n} }

(\frac{d}{Z_{1-\alpha /2}} )*\sqrt{n} = S

\sqrt{n} = S* (\frac{Z_{1-\alpha /2}}{d} )

n = (S* (\frac{Z_{1-\alpha /2}}{d} ))^2

n= (5.23*(\frac{1.96}{1} ))^2= 105.08= 106

3 0
3 years ago
How can we find the volume of the fish tank
tamaranim1 [39]

Answer:

times the height, width and base

Step-by-step explanation:

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