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3 years ago

Solving Exponential Equations (lacking a common base)(0.52)^9=4

Mathematics

2 answers:

Vladimir79 [104]3 years ago

7 0

Answer:

q= -2.584

Step-by-step explanation:

(0.52)^q = 4.

We exponent are written in compound forms, having a different base, we solve such taking the logs of both sides of the equation.

(0.52)^q = 4

We take the log of both sides!

q * log 0.52 = log 4.

q * -0.284 = 0.602

Divide through by -0.233

q = 0.602/ - 0.233

q = -2.584

matrenka [14]3 years ago

3 0

Answer:

$q\approx-2.12$

Explanation:

The exponent of 0.52 is not 9 but q. Thus, you need so solve the equattion to find the value of the exponent, q.

To solve exponential equations with different bases, you must use logarithms, which is the inverse function of exponentiation:

$0.52^q=4\\\\\log (0.52^q)=\log 4\\\\q\log 0.52=\log 4\\\\\\q=\dfrac{\log 4}{\log 0.52}\\\\\\q\approx-2.12$

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Answer:

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3 years ago

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balu736 [363]

Answer:

(a) The sample size required is 43.

(b) The sample size required is 62.

Step-by-step explanation:

The (1 - α) % confidence interval for population mean is:

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The margin of error for this interval is:

$MOE=z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}$

(a)

The information provided is:

<em>σ</em> = 4 minutes

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Confidence level = 95%

α = 5%

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$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

Compute the sample size required as follows:

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$n=[\frac{z_{\alpha/2}\times \sigma}{MOE} ]^{2}$

$=[\frac{1.96\times 4}{1.2}]^{2}\\\\=42.684\\\\\approx 43$

Thus, the sample size required is 43.

(b)

The information provided is:

<em>σ</em> = 4 minutes

MOE = 1 minute

Confidence level = 95%

α = 5%

Compute the critical value of z for α = 5% as follows:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

Compute the sample size required as follows:

$MOE=z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times \sigma}{MOE} ]^{2}$

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3 years ago

(Essay Question 7 Points)

Irina18 [472]

Well I don't know.
Let's think about it:

-- There are 6 possibilities for each role.
So 36 possibilities for 2 rolls.
Doesn't take us anywhere.

New direction:
-- If the first roll is odd, then you need another odd on the second one.
-- If the first roll is even, then you need another even on the second one.
This may be the key, right here !

-- The die has 3 odds and 3 evens.

-- Probability of an odd followed by another odd = (1/2) x (1/2) = 1/4
-- Probability of an even followed by another even = (1/2) x (1/2) = 1/4

I'm sure this is it. I'm a little shaky on how to combine those 2 probs.

Ah hah !
Try this:

Probability of either 1 sequence or the other one is (1/4) + (1/4) = 1/2 .

That means ... Regardless of what the first roll is, the probability of
the second roll matching it in oddness or evenness is 1/2 .

So the probability of 2 rolls that sum to an even number is 1/2 = 50% .

Is this reasonable, or sleazy ?

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3 years ago