If two dice are rolled, what is the probability that their sum will be 10, given that both dice are odd?

Mathematics

1 answer:

8090 [49]2 years ago

5 0

Answer:

I think it is either 1/6 or 1/8

Hope it helps :)

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2295 divided by blank equals 51

tekilochka [14]

It would be 117,045. 117,045/ 2295= 51

8 0

3 years ago

Bengus the expert blacksmith forged 32 swords from copper, twice as many swords from iron, and 19 swords from mythril. Estimate

Anna35 [415]

Answer: 115 swords

Step-by-step explanation:

He forged 32 swords from copper, 2*32 from iron and 19 from mythril. Thus, he forged 32+(32*2)+19, or 32+64+19, or 115 swords.

<em>Hope it helps <3</em>

8 0

3 years ago

A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) A

julsineya [31]

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

$D^{2} = (90 - x)^{2} + 90^{2}$ (1)

$\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}$

$2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}$

$D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}$ (2)

Since:

$D = \sqrt{(90 -x)^{2} + 90^{2}}$

When x = 45 (the batter is halfway to first base), D is:

$D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62$

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

$100.62 \frac{d(D)}{dt} = -(90 - 45)*31$

$\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s$

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

$D^{2} = 90^{2} + x^{2}$