Applying the concepts. Need help finding the value.

Mathematics

1 answer:

luda_lava [24]3 years ago

5 0

we can use formula

$log_ab=\frac{ln(b)}{ln(a)}$

$log_23*log_34*log_45*log_56*log_6 7 *log_7 8=\frac{ln(3)}{ln(2)}*\frac{ln(4)}{ln(3)}*\frac{ln(5)}{ln(4)}*\frac{ln(6)}{ln(5)}*\frac{ln(7)}{ln(6)}*\frac{ln(8)}{ln(7)}$

now, we can cancel terms

and then we can simplify it

$log_23*log_34*log_45*log_56*log_6 7 *log_7 8=\frac{ln(8)}{ln(2)}$

$log_23*log_34*log_45*log_56*log_6 7 *log_7 8=\frac{ln(2^3)}{ln(2)}$

$log_23*log_34*log_45*log_56*log_6 7 *log_7 8=\frac{3ln(2)}{ln(2)}$

now, we can simplify it

and we get

$log_23*log_34*log_45*log_56*log_6 7 *log_7 8=3$ ...........Answer

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A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye

Katena32 [7]

Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we select appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{x}=n\mu$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{x}=\sqrt{n}\sigma$

The information provided is:

μ = $970

σ = $129

n = 102

Since the sample size is quite large, i.e. n = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

$\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)$