All categories

3 years ago

a fair (6 sided) die is rolled TWO times. What is the probability that the first die is a 2 or the second die is a prime

Mathematics

1 answer:

alexdok [17]3 years ago

4 0

Answer:

The probability that the first die is a 2 or the second die is a prime = 1/12

Step-by-step explanation:

It is given that,a fair (6 sided) die is rolled TWO times.

The sample space for rolling 2 dies

(1, 1), (1, 2) .......... (1,6)

(2,1) ..........................

(6, 1) ..........................(6, 6)

There are total of 6^2 = 36 sample space

To find the probability

The prime numbers from 1 to 6 are 2, 3 and 5

the possible outcomes are(2,1), (2, 3) and (2, 5)

The probability that the first die is a 2 or the second die is a prime = 3/36 = 1/12

You might be interested in

What is the value of X? Remember to show calculations and formulas used for credit.

finlep [7]

(n-2) × 180° = 4 x 180 = 720
so sum of interior angles = 720
x = 720 - (112+133+128+ 100+120)
x = 720 - 593
x = 127

answer
x = 127

8 0

3 years ago

Read 2 more answers

Solve V- E = 2 for E; E = A.) V - 2 B.) V + 2 C.) - V - 2 D.) - V + 2

zhuklara [117]

Add v to both sides: -E=2+V
Divide both sides by -1: E= -2-V

Answer: -V-2

3 0

4 years ago

P^3×q^2(p^4×q^n×r^3/r^-4)=p^7q^5r^7 What does n=

Juli2301 [7.4K]

Answer:

$n = 3$

Step-by-step explanation:

Given:

The equation given is:

$p^3q^2(p^4q^n\times \frac{r^3}{r^{-4}})=p^7q^5r^7$

We simplify the given equation using the following law of indices:

$a^m\times a^n=a^{m+n}\\\frac{a^m}{a^n}=a^{m-n}$

Therefore,

$p^{3+4}q^{n+2}\times r^{3-(-4)})=p^7q^5r^7\\p^7q^{n+2}r^7=p^7q^5r^7\\q^{n+2}=q^5$

Since the base on both the sides is same, therefore, their exponents must also be same. This gives,

$n+2=5\\n=5-2\\n=3$

Therefore, the value of 'n' is 3.

5 0

3 years ago

Your third-grade classroom has 32 boys and girls. Two-thirds of the boys and three-fourths of the girls are going on a field tri

adelina 88 [10]

Answer: The ratio of boys to girls would be 3 : 5 .

Step-by-step explanation:

Since we have given that

Number of boys and girls = 32

Fraction of boys are going on a field trip = $\dfrac{2}{3}$

Fraction of girls are going on a field trip = $\dfrac{3}{4}$

Number of children left = 9

Let the number of boys be 'b'.

Let the number of girls be 'g'.

According to question, it becomes ,

$b+g=32------------(1)\\\\32-9=\dfrac{2}{3}b+\dfrac{3}{4}g\\\\23=\dfrac{2b}{3}+\dfrac{3g}{4}-------------(2)$

From eq(1), we get that g = 32-b

So, it becomes,

$\dfrac{2}{3}b+\dfrac{3}{4}(32-b)=23\\\\\dfrac{2}{3}b+24-\dfrac{3}{4}b=23\\\\\dfrac{2}{3}b-\dfrac{3}{4}b=23-24=-1\\\\\dfrac{8b-9b}{12}=-1\\\\\dfrac{-b}{12}=-1\\\\b=-1\times -12\\\\b=12$

so, number of girls would be 32 - b = 32 - 12 = 20

So, Ratio of boys to girls in class would be 12 : 20 = 3 : 5.

Therefore, the ratio of boys to girls would be 3 : 5 .

7 0

4 years ago

Evaluate the function g(t)=2/3t when t = 30

Irina-Kira [14]

Plug in the value of t in the function
g(30)=2/3*(30)
=2*10
=20 (answer)

5 0

3 years ago

Read 2 more answers