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tatyana61 [14]
4 years ago
15

Use log_2 7 is about 2.8074 and log_2 5 is about 2.3219 to approximate log_2 175?

Mathematics
2 answers:
Anton [14]4 years ago
5 0

Answer:

it would be log 1/2 5

Step-by-step explanation:

Stella [2.4K]4 years ago
4 0
Remembe
log_ax+log_ay=log_a(xy)


so
notice
7*5*5=175

so we want to multiply them
add

log_27+log_25+log_25=log_2175=
(2.8074)(2.3219)(2.3219)=15.135309933114
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Solve for x: 3x - 24 = 81<br> A-57/3<br> B-105/3<br> C-57<br> D-105<br> Helpppp please
expeople1 [14]

Solution,solve\:for\:x,\:3x-24=81\quad :\quad x=35

Steps: 3x-24=81

\mathrm{Add\:}24\mathrm{\:to\:both\:sides}, 3x-24+24=81+24

\mathrm{Simplify}, 3x=105

\mathrm{Divide\:both\:sides\:by\:}3, \frac{3x}{3}=\frac{105}{3}

Simplify, x=35

The correct answer is <u><em>B. 105/3</em></u>

<u><em>Hope this helps!!!</em></u>

<3 - austint1414

5 0
3 years ago
How does adding the log together automatically mean that it is a factorial?
andreyandreev [35.5K]

Answer: The answer is given below.

Step-by-step explanation:  We are given an equality involving logarithm and we are to show the implication of L.H.S. to R.H.S.

We will be using the following two properties of logarithm:

(i)~\log_ba=\dfrac{1}{\log_ab},\\\\\\(ii)~log_ab+\log_ac=\log_a(bc).

The proof is as follows:

L.H.S.\\\\\\=\dfrac{1}{\log_2N}+\dfrac{1}{\log_3N}+\dfrac{1}{\log_4N}+\cdots+\dfrac{1}{\log_{100}N}\\\\\\=\log_N2+\logN3+\log_N4+\cdots+\log_N100\\\\=\log_N\{2.3.4...100\}\\\\=\log_N\{1.2.3.4...100\}\\\\=\log_N{100!}\\\\=\dfrac{1}{\log_{100!}N}\\\\=R.H.S.

Hence proved.

6 0
3 years ago
12 to x and 4 to 117
Alex73 [517]
Write a proportion and solve for x ....

12 / x = 4 / 117

Cross multiply ...

12 * 117  =  1404

x  *  4   = 4 x

4 x = 1404

x =  351
7 0
4 years ago
Simple equation with division that equals 8
Nitella [24]

Answer:

64 divided by 8

Step-by-step explanation:

4 0
4 years ago
Read 2 more answers
A scientist claims that 9% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of
Lana71 [14]

The probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% is 0.006.

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

μρ=ρ

The standard deviation of this sampling distribution of sample proportion is:

σρ=\sqrt \frac{p(1-p)}{n}

The information provided is:

n = 533

p = 0.09

As the sample size is quite large, i.e. n = 533 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by a Normal distribution.

The mean and standard deviation are:

μρ=ρ=0.09

σρ=\sqrt \frac{p(1-p)}{n}=\sqrt\frac{0.09(1-0.09)}{533}=0.01239

Compute the probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% as follows:

p(p^-p > 0.03)=P(p^ > 0.12)                     \\=P(z > \frac{0.12-0.9}{0.012})                      \\= P(Z > 2.5)                     \\= 1-P(Z > 2.5)                    \\ = 1- 0.993                   \\  = 0.00621

Thus, the probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% is 0.006.

Learn more about PROBABILITY here

brainly.com/question/24756209

#SPJ4

4 0
2 years ago
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