3 years ago

What is the probability of getting a composite number on the first roll of a die and getting a prime number on the second roll?

1 answer:

7 0

for a composite number its 1/3 (2/6)

for a prime number its 1/2

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Alex73 [517]

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3 years ago

Alekssandra [29.7K]

It should be in the third quadrant

5 0

3 years ago

Levart [38]

we have been asked to find the sum of the series

$\sum _{n=1}^5\left(\frac{1}{3}\right)^{n-1}$

As we know that a geometric series has a constant ratio "r" and it is defined as

$r=\frac{a_{n+1}}{a_n}=\frac{\left(\frac{1}{3}\right)^{\left(n+1\right)-1}}{\left(\frac{1}{3}\right)^{n-1}}=\frac{1}{3}$

The first term of the series is $a_1=\left(\frac{1}{3}\right)^{1-1}=1$

Geometric series sum formula is

$S_n=a_1\frac{1-r^n}{1-r}$

Plugin the values we get

$S_5=1\cdot \frac{1-\left(\frac{1}{3}\right)^5}{1-\frac{1}{3}}$

On simplification we get

$S_5=\frac{121}{81}$

Hence the sum of the given series is $\frac{121}{81}$

5 0

3 years ago

seraphim [82]

2^5/2^2 = 2^(5-2)
= 2^3
= 8

8 0

3 years ago

Andre45 [30]

Answer:

C. $4\sqrt{7} -4x\sqrt{7}$ is correct

Steps:

$4\sqrt{7} -3\sqrt{7}x-x\sqrt{7} \\\\4\sqrt{7} -3\sqrt{7} x-\sqrt{7} x\\\\4\sqrt{7} +(-3\sqrt{7} x-\sqrt{7} x)\\\\4\sqrt{7} -4\sqrt{7} x$

7 0

3 years ago