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3 years ago

Please help i have no clue how to do this please solve and explain

Mathematics

1 answer:

Mariulka [41]3 years ago

6 0

The answer is c. 9z^7/y^6

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The weekly earnings of students in one age group are normally distributed with a standard deviation of 47 dollars. A researcher

ELEN [110]

Answer:

Option D) 340

Step-by-step explanation:

We are given the following in the question:

Alpha, α = 0.05

Population standard deviation, σ = $47

Margin of error = 5

95% Confidence Interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

$\text{Margin of error} = z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$5 = 1.96(\dfrac{47}{\sqrt{n}} )\\\\\sqrt{n} = \dfrac{47\times 1.96}{5}\\\\n = 339.443776\\\Rightarrow n \approx 340$

Thus, the correct answer is

Option D) 340

7 0

3 years ago

A trapezoid has an area of 266.4 square feet. One base is 8.5 feet long. The height measures 24 feet. What is the length of the

exis [7]

266.4 - 8.5 - 24 will give you the answer of 233.9

6 0

4 years ago

Which of these scales are equivalent to 3 cm to 4 km?

satela [25.4K]

Answer: The answer is A and D

3 0

3 years ago

f(x)=3x^(4)-5x^(3)-5x^(2)+5x+2 The number 3 is an upper bound for the set of roots of this polynomial function True or False

Leona [35]

Answer: TRUE

Step-by-step explanation: The number 3 is an upper bound because use 3 as a multiplyer of each bcoefficient of the polynomial and at each multiple,subtract the next coefficient from it. if all result after subtraction show positive, then the number used is an upper bound. This method is known as the synthetic division.

3 3 -5 -5 5 2

0 9 12 21 78

3 4 7 26 80

7 0

3 years ago

How to solve this?<br>\int \frac { 4 - 3 x ^ { 2 } } { ( 3 x ^ { 2 } + 4 ) ^ { 2 } } d x

ivanzaharov [21]

$\Large \mathbb{SOLUTION:}$

$\begin{array}{l} \displaystyle \int \dfrac{4 - 3x^2}{(3x^2 + 4)^2} dx \\ \\ = \displaystyle \int \dfrac{4 - 3x^2}{x^2\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ = \displaystyle \int \dfrac{\dfrac{4}{x^2} - 3}{\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ \text{Let }u = 3x + \dfrac{4}{x} \implies du = \left(3 - \dfrac{4}{x^2}\right)\ dx \\ \\ \text{So the integral becomes} \\ \\ = \displaystyle -\int \dfrac{du}{u^2} \\ \\ = -\dfrac{u^{-2 + 1}}{-2 + 1} + C \\ \\ = \dfrac{1}{u} + C \\ \\ = \dfrac{1}{3x + \dfrac{4}{x}} + C \\ \\ = \boxed{\dfrac{x}{3x^2 + 4} + C}\end{array}$

5 0

3 years ago