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3 years ago

The probability that the ticket drawn has a number which is a multiple of 3 or 5?

Mathematics

1 answer:

Dvinal [7]3 years ago

4 0

Answer:

Unfortintaly, more information is needed to get a proper/correct answer to your question. Either attach a screenshot of the question or re-word your typing so it sounds clearer and easier to understand. You might have to do this in another question rather than editing this one. If that is the case, be sure to tag me so I can help. Sorry I oculd not help you. (Yet.)

_Sophia :)

You might be interested in

5. Find the value(s) of x so that the line containing the points (2x + 3, x + 2) and (0, 2) is

Dvinal [7]

Answer:

x = -2 or -9

Step-by-step explanation:

You want the values of x such that the line defined by the two points (2x+3, x+2) and (0, 2) is perpendicular to the line defined by the two points (x+2, -3-3x) and (8, -1).

<h3>Slope</h3>

The slope of a line is given by the slope formula:

m = (y2 -y1)/(x2 -x1)

Using the formula, the slopes of the two lines are ...

m1 = (2 -(x+2))/(0 -(2x+3)) = (-x)/(-2x-3) = x/(2x +3)

and

m2 = (-1 -(-3-3x))/(8 -(x+2)) = (2+3x)/(6 -x)

<h3>Perpendicular lines</h3>

The slopes of perpendicular lines have product of -1:

$\dfrac{x}{2x+3}\cdot\dfrac{2+3x}{6-x}=-1\\\\x(3x+2)=(2x+3)(x-6)\qquad\text{multiply by $(2x+3)(6-x)$}\\\\3x^2+2x=2x^2-9x-18\qquad\text{eliminate parentheses}\\\\x^2+11x+18=0\qquad\text{put in standard form}\\\\(x+2)(x+9)=0\qquad\text{factor}$

<h3>Solutions</h3>

The values of x that satisfy this equation are x = -2 and x = -9. The attached graphs show the lines for each of these cases.

4 0

1 year ago

The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas

saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by $s(t) = -t^3+3t^2+7t +4$ , where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

$s(t) = -t^3+3t^2+7t +4$

Derivate w.r.t. time,

$v(t)=s'(t) = -3t^2+6t+7$

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

$x=-\frac{b}{2a}$

Where, a=-3, b=6 and c=7

$t=-\frac{6}{2(-3)}$

$t=-\frac{6}{-6}$

$t=1$

As 1 lie between interval [0,4]

Substitute t=1 in the function,

$v(t)= -3(1)^2+6(1)+7$

$v(t)= -3+6+7$

$v(1)=10$

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

8 0

3 years ago

Read 2 more answers

What is the y- intercept of: y=-2x

shusha [124]

Answer:

it is Zero 0

Step-by-step explanation:

7 0

3 years ago

10 POINTS!!! FULL ANSWER IN STEP BY STEP FORMAT!!

Lilit [14]

After a little manipulation, the given diff'l equation will look like this:

e^y * dy = (2x + 1) * dx.
x^2
Integrating both sides, we get e^y = 2------- + x + c, or e^y = x^2 + x + c
2

Now let x=0 and y = 1, o find c:

e^1 = 0^2 + 0 + c. So, c = e, and the solution is e^y = x^2 + x + e.

4 0

3 years ago

Can you help me please with this question

GREYUIT [131]

Answer:

132

Step-by-step explanation:

You can tell it is 132 from <ABC because the whole angle is a 180 degree angle now subtract 180-48 which is 132. please give me brainiest

6 0

3 years ago