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3 years ago

ANSWER INCLUDED: What is the solution of log3x + 4 4096 = 4?

Mathematics

1 answer:

allsm [11]3 years ago

6 0

Answer:

C $x=\frac{4}{3}$

Step-by-step explanation:

The given logarithmic equation is:

$\log_{3x+4}(4096)=4$

We rewrite in exponential form; to get;

$4096=(3x+4)^4$

We rewrite the LHS as a certain natural number exponent 4.

$8^4=(3x+4)^4$

The exponents are the same, hence the bases must also be the same.

$\implies 3x+4=8$

$\implies 3x=8-4$

$\implies 3x=4$

Divide both sides by 3;

$\implie x=\frac{4}{3}$

The correct answer is C

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Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

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a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05