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2 years ago

The 7 a.m. flight from Dallas to Chicago is on time 75% of the time. Fran created this spinner to simulate the scenario.

Mathematics

1 answer:

Elena-2011 [213]2 years ago

4 0

Answer:

CORRECT OPTION is (C) The flight will be late on one of the three days.

Step-by-step explanation:

The 7am flight from Dallas to Chicago is on time 75% of the time.

A spinner with four sections was created by Fran to simulate this scenario.

Since the 4 sections were equal, it means each section was 25% of the spinner.

3 sections were shaded. That implies 25 × 3 = 75% of the spinner was shaded area.

This coincides with the 75% probability that the flight is on time on a given day.

In other words, if the spinner's pointer lands on shaded area, the flight will be on time. If the pointer lands on an unshaded portion, the flight will be late.

Now Fran spun the spinner 3 times so she has 3 outcomes. The outcomes are:

- Shaded - Shaded - Unshaded

CORRECT OPTION is (C) The flight will be late on one of the three days.

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The graphs of f(x) and g(x) are shown below:

rosijanka [135]

<h2>Answer:</h2>

Option: C is the correct answer.

C.) x = −3.8, 3

<h2>Step-by-step explanation:</h2>

The function f(x) is given by:

$f(x)=x^2-x-12$

and the function g(x) is given by:

$g(x)=-1.8x-0.6$

Now, we are asked to find the solution of the equation:

$f(x)=g(x)$

i.e. we have to find the value of x such that both the functions are equal i.e.

$x^2-x-12=-1.8x-0.6\\\\i.e.\\\\x^2-x+1.8x-12+0.6=0\\\\i.e.\\\\x^2+0.8x-11.4=0\\\\i.e.\\\\10x^2+8x-114=0\\\\i.e.\\\\5x^2+4x-57=0$

Now, on solving the equation using the quadratic formula i.e. the solution of the equation:

$ax^2+bx+c=0$

is given by:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here we have:

$a=5,\ b=4\ and\ c=-57$

Hence, the solution is given by:

$x=\dfrac{-4\pm \sqrt{4^2-4\times 5\times (-57)}}{2\times 5}\\\\i.e.\\\\x=\dfrac{-4\pm \sqrt{16+1140}}{10}\\\\i.e.\\\\x=-3.8\ and\ x=3$

6 0

3 years ago

which of the functions have a range of real numbers greater than or equal to 1 or less than or equal to-1

lilavasa [31]

The question is incomplete. Here is the complete question:

Which of the functions have a range of all real numbers greater than or equal to 1 or less than or equal to -1? check all that apply.

A. $y=\sec x$

B. $y= \tan x$

C. $y= \cot x$

D. $y= \csc x$

Answer:

A. $y=\sec x$

D. $y=\csc x$

Step-by-step explanation:

Given:

The range is greater than or equal to 1 or less than or equal to -1.

The given choices are:

Choice A: $y=\sec x$

We know that, the $\sec x=\frac{1}{\cos x}$

The range of $\cos x$ is from -1 to 1 given as [-1, 1]. So,

$|\cos x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\cos x|}\geq 1\\|\sec x|\geq 1$

Therefore, on removing the absolute sign, we rewrite the secant function as:

$\sec x\leq -1\ or\ \sec x\geq 1\\$

Therefore, the range of $y=\sec x$ is all real numbers greater than or equal to 1 or less than or equal to-1.

Choice B: $y= \tan x$

We know that, the range of tangent function is all real numbers. So, choice B is incorrect.

Choice C: $y= \cot x$

We know that, the range of cotangent function is all real numbers. So, choice C is incorrect.

Choice D: $y=\csc x$

We know that, the $\csc x=\frac{1}{\sin x}$

The range of $\sin x$ is from -1 to 1 given as [-1, 1]. So,

$|\sin x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\sin x|}\geq 1\\|\csc x|\geq 1$

Therefore, on removing the absolute sign, we rewrite the cosecant function as:

$\csc x\leq -1\ or\ \csc x\geq 1\\$

Therefore, the range of $y=\csc x$ is all real numbers greater than or equal to 1 or less than or equal to-1.

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