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3 years ago

Two dice are tossed. The probability that the total score is a prime number is:

Mathematics

2 answers:

andrew11 [14]3 years ago

8 0

Step-by-step explanation:

50/50 it's random I think but I learned that in science

BigorU [14]3 years ago

5 0

50/50 because there are 3 prime numbers and 3 composite numbers on a dice

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Identify the correct solution for 3/6 x 5.

Setler79 [48]

Answer:

suggestions use algebra calculator or mathaway

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3 years ago

Add the equations. 2x - 3y = -1 + 3x+3y= 26

Basile [38]

5x=25

Because negative 3y and positive 3y got cancelled

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3 years ago

Read 2 more answers

One quart of water-resistant paint covers 75 square feet. A swimming pool is in the shape of a rectangular prism with a length o

docker41 [41]

18.66 quarts of paint is needed to paint the swimming pool with two coats of paint

Solution:

Given that,

A swimming pool is in the shape of a rectangular prism

Length = 20 feet

Width = 10 feet

Height = 5 feet

Find the surface area of prism

$A=2(wl+hl+hw)$

Where, "l" is the length and "w" is the width and "h" is the height

$A = 2(10 \times 20 + 5 \times 20 + 5 \times 10)\\\\A = 2(200+100+50)\\\\A = 2 \times 350\\\\A = 700$

Thus surface area area of prism is 700 square feet

Given that,

One quart of water-resistant paint covers 75 square feet

Therefore,

1 quart paint = 75 square feet

x quart = 700 square feet

This forms a proportion

$1 \times 700 = 75 \times x\\\\x = \frac{700}{75}\\\\x = 9.33$

Thus 9.33 quarts of paint is needed to cover 700 square feet

How many quarts should you buy to paint the swimming pool with two coats of paint

For two coats of paint we need = 9.33 + 9.33 = 18.66 quarts of paint

4 0

2 years ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

3 years ago

F(n)=x2+1 g(n)=2x-5 Find f(n)-g(n)

Artist 52 [7]

Answer:

x^2 - 2x + 6

Explanation:

(x^2 + 1) - (2x - 5)
x^2 + 1 - 2x + 5
x^2 - 2x + 6

8 0

3 years ago

Two dice are tossed. The probability that the total score is a prime number is:​

Two dice are tossed. The probability that the total score is a prime number is: