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Irina-Kira [14]
3 years ago
6

PLZ HELP IF WRONG ILL REPORT

Mathematics
2 answers:
dimulka [17.4K]3 years ago
7 0
I HONESTLY WOULD THINK THIS COULD BE BIAS ONLY BECAUSE DIFFERENT OPINION ARE INCLUDED BUT AT THE SAME TIME BESTBUY NEVER SAID ANYTHING ABIUT THEM HATING BESTBUY
kvv77 [185]3 years ago
6 0

Answer:

The store calls and asks every 4th person who bought a television within the past year whether he or she likes it.

<u>It is biased because opinion was shared..</u>

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Polimerlerin su ile parçalanarak monomerlerine ayrılmasına ne denir?<br>​
kolbaska11 [484]

Answer:

tarlak iefdiebdid didbdkd

5 0
3 years ago
Help please! 16 points
emmainna [20.7K]

Answer:

3

Step-by-step explanation:

divide the top number by the bottom to always find the constant of proportionality :D

6 0
3 years ago
Read 2 more answers
PLEASE HURRY IM TIMED!!!
andriy [413]

<u>Answer-</u>

<em>A. Brandon’s sound intensity level is 1/4th as compared to Ahmad’s.</em>

<u>Solution-</u>

Given that, loudness measured in dB is

L=10\log \frac{I}{I_0}

Where,

I   = Sound intensity,

I₀ = 10⁻¹² and is the least intense sound a human ear can hear

Given in the question,

I₁ = Intensity at Brandon's = 10⁻¹⁰

I₂ = Intensity at Ahmad's  = 10⁻⁴

Then,

L_1=10\log \frac{I_1}{I_0}=10\log \frac{10^{-10}}{10^{-12}}=10\log \frac{1}{10^{-2}}=10\log 10^{2}=2\times10\log 10=20

L_2=10\log \frac{I_2}{I_0}=10\log \frac{10^{-4}}{10^{-12}}=10\log \frac{1}{10^{-8}}=10\log 10^{8}=8\times10\log 10=80

\therefore \frac{L_1}{L_2} =\frac{20}{80} =\frac{1}{4}

5 0
3 years ago
Write each expression as an equivalent expression with a single logarithm. Assume xx, yy, and zz are positive real
Alekssandra [29.7K]

Answer:

\ln(\frac{xx + yy}{zz})^{\frac{1}{2}}

Step-by-step explanation:

Given:

1/2(ln(xx + yy) − ln(zz))

Now,

From the properties of log function,

1)  n × ln(x) = ln(xⁿ)

and,

2)  ln(A) - ln(B) = \ln\frac{A}{B}

applying the properties in the given equation

we get the above equation as:

\frac{1}{2}(\ln\frac{xx + yy}{zz})        

( using the property 2 we get (ln(xx + yy) − ln(zz) = \ln\frac{xx + yy}{zz}  

or

⇒ \ln(\frac{xx + yy}{zz})^{\frac{1}{2}}    ( using the property 1 i.e n × ln(x) = ln(xⁿ) )

expression as an equivalent expression with a single logarithm is \ln(\frac{xx + yy}{zz})^{\frac{1}{2}}

3 0
3 years ago
HELP PLEASE!!! LOL PLEASE!
Tpy6a [65]

Answer:

thats is the answer write it yourself and learn math

3 0
3 years ago
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