PLEASE HELP! What is the value of the expression, written in standard form?

A vehicle arriving at an intersection can turn right, turn left, or continue straight ahead. The experiment consists of observin

Temka [501]

Answer:

a) Sample space (S) = {TR, TL, Cs}

b) Pr(Vehicle Turns) = 2/3.

Step-by-step explanation:

a) By sample space (S) we mean the list all possible outcome of an event. From the illustration in the question, only three (3) outcome is possible:

i) Vehicle Turn Right (TR)

ii) Vehicle Turn Left (TL)

iii) Continue Straight ahead (Cs)

And since the experiment consists of observing the movement of a single vehicle through the intersection, then, the sample space (S) is:

==> {TR, TL, Cs}

b) Since we are assuming that all sample points are equally likely, it imply that they have equal probability of occurring. And by probability:

Pr(TR) = 1/3

Pr(TL) = 1/3

Pr(Cs) = 1/3

Meanwhile, the question wants us to find the probability that the vehicle turns. This means, the vehicle turning either right or left. Thus;

Pr(vehicle turns) = Pr(TR) or Pr(TL) = Pr(TR) + Pr(TL)

Pr(vehicle turns) = (1/3) + (1/3) = 2/3

4 0

3 years ago

How do you do this fraction

JulsSmile [24]

The answer to the question is 11/14

7 0

2 years ago

Read 2 more answers

PLEASE HELP!!

sp2606 [1]

Standard form is just putting (in this case) variables in alphabetical order. First we simplify- 2x+4=4y would become 1/2x+1=y. This is already in standard form, as numbers w/o variables come at the end. simplifying the next one is more tricky. first you get a variable/number alone-
9-2x-2y=4x+3
9-2y=6x+3
6-2y=6x
1-1/3y=x
1=x+1/3y
3=x+y
x+y=3
sorry if I was wrong and not of any help, But I do believe this is correct.

5 0

3 years ago

Read 2 more answers

NEED MAJOR HELP WITH HW!!! ps: will mark brainliest :)<img src="https://tex.z-dn.net/?f=%20" id="TexFormula1" title=" " alt=" "

TiliK225 [7]

1.5
2.5
45/66
35/90
70/120
81/90
12/45
35/70
rahul got 4
15minutes
4/15
43
I did this in class today i have the worksheet with answers Your welcome no work needed put that on the paper

6 0

3 years ago

Int(1 \(1 + {e}^{x} )

Andreyy89

Answer:

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx}= x - \ln(1 + e^{x}) + C\end{aligned}$ .

Step-by-step explanation:

The first derivative of the denominator $1 + e^{x}$ is $e^{x}$ . Rewrite the fraction to obtain that expression on the numerator.

$\begin{aligned}\frac{1}{1 + e^{x}} &= \frac{1 + e^{x}}{1 + e^{x}} - \frac{e^{x}}{1+e^{x}}\\&=1-\frac{e^{x}}{1+e^{x}}\end{aligned}$ .

In other words,

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\end{aligned}$ .

Apply $u$ -substitution on the integral $\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx}$ :

Let $u = 1 + e^{x}$ . $u > 1$ .

$du = e^{x}\cdot dx$ .

$\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx} = \int{\frac{du}{u}} = \ln{|u|} = \ln{u} +C = \ln{(1+e^{x})}+C$ .

Therefore

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\\ & = x - \ln{(1 + e^{x})}+C\end{aligned}$ .

7 0

3 years ago