All categories

3 years ago

You roll a number cube twice. Find the probability of the events. Rolling an odd number 2 or a 4.

Mathematics

1 answer:

Digiron [165]3 years ago

5 0

Answer:

5/12

Step-by-step explanation:

If you roll it twice there are 12 options. The odd numbers are 1, 3, and 5. Add 2 and 4, and that's a 5/12 chance.

You might be interested in

Use the interactive number line to find each sum to complete the table

Likurg_2 [28]

-1 - 2 = -3

R = -3

-4 + 1 = -3

S = -3

-6 - 3 = -9

T = -9

6 0

3 years ago

Read 2 more answers

3 connecting lines are shown. Line D F is horizontal. Line D E is about half the length of line D F. Line F E is about one-third

mars1129 [50]

The inequality that explains why the three segments cannot be used to construct a triangle is ED + EF < DF

<h3>Inequalities </h3>

From the question, we are to determine which of the given inequalities explains why the three segments cannot be used to construct a triangle

From the given information,

Line DE is about half the length of line DF

That is,

ED = 1/2 DF

Also,

Line FE is about one-third of the length of line DF

That is,

EF = 1/3 DF

Then, we can write that

ED + EF = 1/2DF + 1/3DF

ED + EF = 5/6 DF

Since,

5/6 DF < DF

Then,

ED + EF < DF

Hence, the inequality that explains why the three segments cannot be used to construct a triangle is ED + EF < DF

Learn more on Inequalities here: brainly.com/question/1447311

#SPJ1

5 0

2 years ago

Does 0.4 equal 0.4%? Explain.

insens350 [35]

Answer:

its 40%

Step-by-step explanation:

ok so its 0.4 and there's a 0 in it so you swop the number and then you get it and. I hope you enjoy.

8 0

3 years ago

Please open the image and help me

Kay [80]

Looks like

$y=2\sqrt[3]{x}+\dfrac1{x^2}+\pi$

Write $\sqrt[3]{x}$ as a fractional power, $x^{1/3}$ . This makes it more obvious that the power rule should be used here.

$y'=(2x^{1/3})'+(x^{-2})'+\pi'$

$y'=\dfrac23x^{-2/3}-2x^{-3}$

$y'=\dfrac2{3\sqrt[3]{x^2}}-\dfrac2{x^3}$

$y=(\sin(2x)+\tan(3x))^e$

Power and chain rule:

$y'=\left((\sin(2x)+\tan(3x))^e\right)'$

$y'=e(\sin(2x)+\tan(3x))^{e-1}(\sin(2x)+\tan(3x))'$

$y'=e(\sin(2x)+\tan(3x))^{e-1}(\cos(2x)(2x)'+\sec^2(3x)(3x)')$

$y'=e(\sin(2x)+\tan(3x))^{e-1}(2\cos(2x)+3\sec^2(3x))$

8 0

3 years ago

What is it need answer quick!!

Pavlova-9 [17]

Answer:

your answer will be option D)224

Step-by-step explanation:

have a nice day/night!!

7 0

3 years ago

Read 2 more answers