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rusak2 [61]
3 years ago
6

Alex is working to simplify

Mathematics
2 answers:
r-ruslan [8.4K]3 years ago
7 0
What’s the problem.

Explanation:
stich3 [128]3 years ago
4 0

Answer:

is this all?

Step-by-step explanation:

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Sveta_85 [38]

Answer:

check online for more information

6 0
2 years ago
What is the surface area of the cylinder
Hitman42 [59]

Answer:

In terms of pi: 164.5pi cm squared

Substituting pi for 3.14: 516.53 cm squared

Step-by-step explanation:

3.5^2 = 12.25pi

12.25*2 = 24.5pi

7pi*20 = 140pi

140 + 24.5 = 164.5pi

164.5*3.14 = 516.53

5 0
2 years ago
Maria had $100 at the beginning of summer vacation. She plans to mow lawns and charge $20 per lawn. Which graph could be used to
Illusion [34]

Answer:

Line graph

Step-by-step explanation:

7 0
3 years ago
How do yuh evaluate the expression <br> Log 4096<br> 4
Furkat [3]
log_{4}(4096)

4096=4^{6}

let's replace it

log_{4}(4^6)

Them

log_4(4^6)=6log_{4}(4)

log_4(4)=1

\boxed{\boxed{log_4(4096)=6}}
6 0
3 years ago
Read 2 more answers
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
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