Cardinalidad de: U = { a, b, c, d, e, f, g, h, i, j, k }

Find two consecutive integers such that the sum of their squares added to the lesser number equals 21

Sati [7]

Answer:

-4 and -3.

Step-by-step explanation:

Let the numbers be x and x+1.

From the given information:

x^2 + (x + 1)^2 + x = 21

x^2 + x^2 + 2x + 1 + x - 21 = 0

2x^2 + 3x - 20 = 0

(2x - 5 )(x + 4 ) = 0

x = 5/2 or -4 We ignore 5/2 as its not an interger

So the required integers are -4 and -3.

7 0

4 years ago

Read 2 more answers

HELP ASAP NOW!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Sedbober [7]

Answer:

x +2y = 36

2x+3y =64

Step-by-step explanation:

x = price of a T-shirt

y = price of a puck

Adam buys a T-shirt and two souvenir pucks for $36.00

x +2y = 36

Luke spends $64.00 on two T-shirts and three pucks

64 = 2x+3y

Our system of equations is

x +2y = 36

2x+3y =64

6 0

4 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

4 years ago

Make a conjecture about the next item in the sequence.<br> 6, 9, 7, 10, 8

d1i1m1o1n [39]

Answer:

11

Step-by-step explanation:

6 to 9 = +3,9 to 7= -2,7 to 10= +3,10 to 8 = -2 and the answer have to + 3 with 8

5 0

4 years ago

Read 2 more answers

Tim bought 3 hat for 120/-. how much can he buy from 180/-<br><br> ASAP plz

Leni [432]

Answer: 16

Step-by-step explanation:

5 0

3 years ago