All categories

3 years ago

Rounded to the thousands place :) thank you!! 5e^3x – 4 = 31

Mathematics

1 answer:

Lerok [7]3 years ago

7 0

Step-by-step explanation:

Add 4 to both sides: 5e^3x = 35, or

You might be interested in

The bricklayer bought 7500 bricks to brick a wall. He calculated that every horizontal layer of bricks will contain 321 bricks.

statuscvo [17]

There will be 23 layers of brick, because 7500/321 = 23 with 117 leftover.

6 0

3 years ago

A 90% confidence interval for the mean height of students

Westkost [7]

Answer:

$ME= \frac{69.397-60.128}{2}= 4.6345 \approx 4.635$

And the best answer on this case would be:

b) m = 4.635

Step-by-step explanation:

Let X the random variable of interest and we know that the confidence interval for the population mean $\mu$ is given by this formula:

$\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

The confidence level on this case is 0.9 and the significance $\alpha=1-0.9=0.1$

The confidence interval calculated on this case is $60.128 \leq \mu \leq 69.397$

The margin of error for this confidence interval is given by:

$ME =t_{\alpha/2} \frac{s}{\sqrt{n}}$

Since the confidence interval is symmetrical we can estimate the margin of error with the following formula:

$ME = \frac{Upper -Lower}{2}$

Where Upper and Lower represent the bounds for the confidence interval calculated and replacing we got:

$ME= \frac{69.397-60.128}{2}= 4.6345 \approx 4.635$

And the best answer on this case would be:

b) m = 4.635

4 0

3 years ago

Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A<br>What is the value of the A^2?<br><br>

Alla [95]

$\large \mathbb{PROBLEM:}$

$\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}$

$\large \mathbb{SOLUTION:}$

$\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}$

$\boxed{ \tt \red{C}arry \: \red{ O}n \: \red{L}earning} \: \underline{\tt{5/13/22}}$

4 0

2 years ago

Please help

Aloiza [94]

A=10x(15x)-p(4x)^2

A=150x^2-16px^2

A=(150-16p)x^2

6 0

3 years ago

Iliana cubed the number of flowering plants in her garden, then added 2 vegetable plants. Let p represent the original number of

adelina 88 [10]

When she cubed the number of flowering plants that means p^3 so the answer is not A or B
now take a look at C or D
both are p^3 + 2 which is right now lets solve it:
3^3 + 2 = 27 +2 = 29
so the answer is D :)))
i hope this is helpful
have a nice day

3 0

4 years ago

Read 2 more answers