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2 years ago

13

A pair of dice is rolled. What is the probability that the sum of the two dice will less than 4 given that the first die rolled

is a 2?

1 answer:

leonid [27]2 years ago

5 0

1/6 because the only way the sum can be less than 4 if the first die is a 2 is if the second die is a 1.

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19. You roll two standard number cubes. What is the probability that the sum is odd, given than one of the number cubes shows a

In-s [12.5K]

6 numbers per cube

if one is a 1, then the other one has to be a 2, 4 or 6 to be an odd sum

1+2=3, 1+4=5,1+6=7

since there are 3 numbers you can get that would be a 3/6 ( reduce to 1/2) probability

so 1/2 probability

5 0

3 years ago

Read 2 more answers

Find x <br> Help me plz!!!

mariarad [96]

9514 1404 393

Answer:

x = 8

Step-by-step explanation:

Assuming the figure is a parallelogram, the diagonals bisect each other. That means ...

3x -11 = x +5

2x = 16 . . . . . . . add 11-x to both sides

x = 8 . . . . . . divide by 2

3 0

3 years ago

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Can someone help me with both of these? You can answer only one, however, If you answer both that is who gets brainliest. don't

damaskus [11]

Answer:

20x+80

10/21

Step-by-step explanation:

3 0

2 years ago

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9. Alicia Martin's savings account has a principle of $1,200. It earns 6% interest compounded quartly

zavuch27 [327]

Answer:

$9) \$1236.27\,10)\,\$6451.07\, 11)\,\$10,152.87 \,12)\,\$907.95 \,13)\,\$4957.69$

Step-by-step explanation:

9) Since Alicia Martin's savings earns 6% quarterly for two quarters then:

$A=P(1+\frac{r}{n})^{nt}$ ⇒ Amount (A), Principle (P), rate (r) in decimal form, number of compoundings (n) a year and t, in year or its fractions.

$A=P(1+\frac{r}{n})^{nt}\Rightarrow A=1200(1+\frac{0.06}{4})^{4*\frac{1}{2}}\Rightarrow A=\$1236.27$

10) Aubrey Daniel's case:

$A=P(1+\frac{r}{n})^{nt}\Rightarrow A=5725(1+\frac{0.04}{4})^{4*3}\Rightarrow A\approx \$6451.07$

11) As for Angelo, similarly to Alicia.

$A=P(1+\frac{r}{n})^{nt}\Rightarrow A=9855(1+\frac{0.06}{4})^{4*\frac{1}{2}}\Rightarrow A\approx \$10,152.87$

12) Simpson's. For semiannual n=2

$A=P(1+\frac{r}{n})^{nt}\Rightarrow A=860(1+\frac{0.055}{2})^{2*1}\Rightarrow A\approx \$907.95$

13) Jana Lacey amount:

$A=P(1+\frac{r}{n})^{nt}\Rightarrow A=4860(1+\frac{0.04}{4})^{4*\frac{1}{2}}\Rightarrow A\approx \$4957.69$

6 0

2 years ago

Compute <br><br> I need help

nataly862011 [7]

$(\sqrt{3}-\sqrt{6}+\sqrt{12}-\sqrt{24})\times \frac{\sqrt{6}}{2} \\ (\sqrt{3}-\sqrt{6}+2\sqrt{3}-\sqrt{24})\times \frac{\sqrt{6}}{2} \\ (\sqrt{3}-\sqrt{6}+2\sqrt{3}-2\sqrt{6})\times \frac{\sqrt{6}}{2} \\ ((\sqrt{3}+2\sqrt{3})+(-\sqrt{6}-2\sqrt{6}))\times \frac{\sqrt{6}}{2} \\ (3\sqrt{3}-3\sqrt{6})\times \frac{\sqrt{6}}{2} \\ \frac{(3\sqrt{3}-3\sqrt{6})\sqrt{6}}{2} \\ \frac{3(\sqrt{3}-\sqrt{6})\sqrt{6}}{2}$

7 0

2 years ago