Mathematics, 16.09.2019 14:30, andrewjsoto

John is playing a game of darts. the probability that he throws a dart into the center of the dart board (the bull’s eye) is 1/10. the probability that he throws the dart into the 10-point ring is 3/10.
what is the probability that he either hits a bull’s eye or scores 10 points?
a. 1/3
b. 2/3
c. 3/5
d. 2/5
e. 1/4

Answers: 2

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answered: gmac24

The correct answer for the question that is being presented above is this one: "d. 2/5." John is playing a game of darts. The probability that he throws a dart into the center of the dart board (the Bull’s eye) is 1/10. The probability that he throws the dart into the 10-point ring is 3/10. The probability that he either hits a Bull's eye or scores 10 points is 2/5

answered: toxsicity

The correct option is (d). $\dfrac{2}{5}.$

Step-by-step explanation: Given that John is playing a game of darts. The probability that he throws a dart into the centre of the dart board (the Bull’s eye) is $\dfrac{1}{10}$ and the probability that he throws the dart into the 10-point ring is $\dfrac{3}{10}.$

We are to find the probability that he either hits a Bull’s eye or scores 10 points.

Let, 'A' and 'B' represents the events that John throws the dart into a Bull's eye and 10-point ring respectively.

Then, according to the given information, we have

$P(A)=\dfrac{1}{10},~~P(B)=\dfrac{3}{10},~~~P(A\cup B)=?$

Since John cannot throw the dart into the Bull's eye and 10 point ring together, both the events are independent of each other.

Therefore,

$P(A\cap B)=0$

From the theorems of probability, we have

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=\dfrac{1}{10}+\dfrac{3}{10}-0=\dfrac{4}{10}=\dfrac{2}{5}.$

Therefore, the probability that John either hits a Bull’s eye or scores 10 points is $\dfrac{2}{5}.$

Thus, (d) is the correct option.

answered: Guest

$y> \frac{1}{3}x$ and y≤ -x +4 .

step-by-step explanation:

we are given a graph for system of inequalities.

in order to write the system of inequalities for the given graph, we need to find the inequality for each of the line separately.

let us find the slope of dotted line first.

taking two coordinates on the graph.

let us say (0,0) and (3,1).

let us find the slope of the line now.

slope of the dotted line would be = $\frac{1-0}{3-0} = \frac{1}{3}$ .

y-intercept of dotted line is y=0 only.

plugging slope m= $\frac{1}{3}$ and b=0 in slope-intercept form y=mx+b, we get

$y=\frac{1}{3}x$ .

the line is shaded up of that dotted line.

therefore, for shading up we would take greater than inequality symbol > .

and because it's a dotted line we would take only > symbol in place of = sign in above equation of line.

therefore, inequality for dotted line would be

$y> \frac{1}{3}x$ .

let us take two points (3,1) and (0,4) on the solid line.

slope of the solid line would be = $\frac{4-1}{0-3} = \frac{3}{-3}=-1$ .

for the solid line we have y-intercept b= 4.

plugging slope m= -1 and b=4 in slope-intercept form y=mx+b, we get

y = -1x + 4.

the line is shaded down of that solid line.

therefore, for shading down we would take less than inequality symbol < .

and because it's a solid line we would take ≤ symbol in place of = sign in above equation of line.

therefore, inequality for solid line would be

y≤ -x +4 .
Write a system of inequalities to represent the shaded portion of the graph.

answered: Guest

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in $\delta bea$ and $\delta dea$ ;

$ba=da$ , as these are radii of our given circle.

$ea=ea$

therefore, $\delta bea\cong \delta dea$ by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

$3x-24=x+10$