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Fittoniya [83]
2 years ago
12

What are the solutions to log subscript 6 baseline (x squared 8) = 1 log subscript 6 baseline (x)

Mathematics
1 answer:
n200080 [17]2 years ago
8 0

The solution of the logarithmic equation is given by the antilog. Then the value of x is 2 and 4.

<h3>What is a factorization?</h3>

It is the method to separate the polynomial into parts and the parts will be in multiplication. And the value of the polynomial at this point will be zero.

The logarithmic equation is given below.

\log _6(x^{2} +8)= 1+ \log _6x\\\\\log _6(x^{2} +8)= \log_66+ \log _6x\\\\\log _6(x^{2} +8)=  \log _66x\\

Take antilog, then we have

\begin{aligned} x^{2} +8 &= 6x\\\\x^2 -6x + 8 &= 0\\\\x^2 -4x - 2x +8 &=0\\\\(x-4)(x-2) &= 0\\\\x&= 2, 4\end{aligned}

Then the value of the x is 2 and 4.

More about the factorization link is given below.

brainly.com/question/6810544

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A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to mo
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This question is incomplete, the complete question;

A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to month. The results of a study of 193 fatal accidents were recorded. Is there enough evidence to reject the highway department executive's claim about the distribution of fatal accidents between each month? Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Fatal Accidents 11 13 22 11 15 29 15 9 13 15 22 18

Step 10 of 10 : State the conclusion of the hypothesis test at the 0.1 level of significance.

Answer: H₀ : The distribution of fatal accidents between each month is same

Step-by-step explanation:

H₀ : The distribution of fatal accidents between each month is same

H₁ : The distribution of fatal accidents between each month is different

Let the los be alpha 0.10

given data ;

       Observed    Expected  

Mon   Freq (Oi)   Freq Ei       (Oi-Ei)^2 /Ei

Jan      11    16.0833        1.606649

Feb      13    16.0833        0.591105

Mar      22    16.0833        2.176598

Apr       11    16.0833        1.606649

May     15    16.0833        0.072971

Jun     29    16.0833        10.373489

Jul     15    16.0833        0.072971

Aug       9    16.0833        3.119603

Sep      13    16.0833        0.591105

Oct      15    16.0833        0.072971

Nov      22    16.0833        2.176598

Dec      18    16.0833       0.228411

Total:   193     93                22.689119

Expected Freq ⇒ 193 / 12 = 16.08333

Test Statistic, x² : 22.6891

Num Categories: 12

Degrees of freedom: 12 - 1 = 11

Critical value X² : 24.725

P-Value: 0.0195

since Chi-square value < Chi-square critical value

and P-value > alpha 0.01 so we accept H₀

Therefore we conclude that the distribution of fatal accidents between each month is same

7 0
3 years ago
20/45 in simplest form
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20/45 would be 4/9 in simplest form. hope that helped
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Read 2 more answers
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