Asix-sided number cube is tossed and a coin is flipped. the sample space is {1h, 2h, 3h, 4h, 5h, 6h, 1t, 2t, 3t, 4t, 5t, 6t}.what is the probability of rolling a number greater than 2 and flipping heads? enter your answer, as a fraction in simplest form, in the box.

Since you're already given the sample space, we can simply count how many outcomes satisfy the request, and divide that number by the cardinality of the sample space.

In other words, we're using the formula

So, the outcomes with a number higher than two are

3H, 4H, 5H, 6H, 3T, 4T, 5T, 6T

out of these outcomes, we can filter those with heads flipped:

3H, 4H, 5H, 6H

So, there are 4 cases out of 12, and the probability is thus

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

me! what did you put for this question?

step-by-step explanation: