I need help with 1-8.

Mathew rolls a number cube labeled with the numbers 1−6. He then flips a fair coin. What is the probability that he rolls a 4 an

KATRIN_1 [288]

Answer:

0.083

Step-by-step explanation:

Numbers of Cube = 6

Faces on coins = 2

Therefore, The total outcomes = $6 \times 2 = 12$

Now the favourable outcome that he rolls a 4 & flips a head = 1

The probability that he rolls a 4 & flips a head = 1

$\frac{Favourable \: Outcome}{Total \: Outcome}$

=> The probability that he rolls a 4 & flips a head = $\frac{1}{2} = 0.83333/ approx. 0.83$

Therefore, The probability that he rolls a 4 & flips a head = 0.083

7 0

3 years ago

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4 = 3x -x-4<br> I need steps to how to do this

vesna_86 [32]

Step-by-step explanation:

so first you combine like terms , 3x-x-4=4 turns into 4x-4=4 then u add 4 to both sides [inverse operations]

4x-4=4

+4 +4

--------------

4x= 8

----------

4. 4

×=2

then u use inverse operations. agian and divide 4 into 8 and you get 2 !! x = 2

4 0

3 years ago

Read 2 more answers

Select all of the equations that are not functions.

Kryger [21]

Answer:

B

Step-by-step explanation:

x² + y² = 3 creates a circle and circles are not functions because one x-value relates to two y-values instead of one

7 0

3 years ago

Which graph below has a “period” of 4?

Inessa [10]

Answer:

c

Step-by-step explanation:

6 0

3 years ago

16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.

Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$

3 0

3 years ago