How to solve 6^x-2=1/4

1 answer:

4 0

You can solve it by take log( ) for both sides
6^x-2= 1/4
(x-2) log(6) = log(1/4)
(x-2) = log(1/4) / (log(6))
x-2= -0.774
x= -0.774 +2
x= 1.226

You might be interested in

ABCD is a rectangle. Its diagonals are AC and BD. If AC = 12cm, then BD = _______cm

defon

Answer:

$BD = 12\ cm$

Step-by-step explanation:

Given

Shape: Rectangle

Diagonals: AC and BD

$AC = 12\ cm$

Required

Determine BD

From the question, we understand that the shape is a rectangle and its diagonals are AC and BD

Every rectangle has 2 diagonals, both of which are equal.

This implies that:

$BD = AC$

Recall that:

$AC = 12\ cm$

Hence,

$BD = 12\ cm$

8 0

3 years ago

A certain change in a process for manufacture of component parts is being considered. Samples are taken using both the existing

Katyanochek1 [597]

Answer:

a) The confidence interval is $0.00\leq\pi_1-\pi_2\leq0.02$ .

Step-by-step explanation:

We have to calculate a confidence interval (CI) of a difference of proportions.

For the existing procedure, the proportion is:

$p_1=75/1500=0.05$

For the new procedure, the proportion is:

$p_2=80/2000=0.04$

To calculate the CI, we need to estimate the standard deviation

$s=\sqrt{\frac{p_1(1-p_1)}{n_1} +\frac{p_2(1-p_2)}{n_2} }\\\\s=\sqrt{\frac{0.05(1-0.05)}{1500} +\frac{0.04(1-0.04)}{2000} }=\sqrt{ 0.000032 + 0.000019 }=\sqrt{ 0.000051 }\\\\s=0.007$

For a 90% CI, the z-value is 1.64.

Then, the CI is:

$(p_1-p_2)-z*\sigma \leq\pi_1-\pi_2\leq(p_1-p_2)+z*\sigma \\\\(0.05-0.04)-1.64*0.007\leq\pi_1-\pi_2\leq(0.05-0.04)+1.64*0.007\\\\0.01-0.01\leq\pi_1-\pi_2\leq0.01+0.01\\\\0.00\leq\pi_1-\pi_2\leq0.02$