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3 years ago

Which of the following gives all of the sets that contain √9?

1 answer:

7 0

Soldiers for the nation of Gilead. “Blessed Be the Fruit”: The standard greeting amongst Gilead residents. The traditional reply is “May the Lord open.” “Under His Eye”: The Gileadean equivalent of “Aloha” — it works as both a hello and a goodbye.

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Christy draws 3 cards from the deck one at a time without replacement. What is the probability of selecting the cards in the ord

Inga [223]

Start with a deck of 52 cards. Draw 1 card. Probability of drawing a diamond first:

$\dfrac{\binom{13}1}{\binom{52}1}$

Now draw another single card from the deck, which now contains 51 cards. Probability of drawing a spade:

$\dfrac{\binom{13}1}{\binom{51}1}$

Draw 1 last card. The deck has 50 cards left, and 12 of these are diamonds. Probability of drawing a diamond:

$\dfrac{\binom{12}1}{\binom{50}1}$

So the probability of drawing a diamond, then a spade, then another diamond is

$\dfrac{\binom{13}1}{\binom{52}1}\cdot\dfrac{\binom{13}1}{\binom{51}1}\cdot\dfrac{\binom{12}1}{\binom{50}1}=\dfrac{13}{850}$

5 0

3 years ago

1. Find the sum of the series (IMAGE) . Show your work.

pav-90 [236]

QUESTION 1 ANSWER

The series is

$\sum_{n=1}^{15}{ (2n-1)}$

So the nth term of this sequence is

$U_n=2n-1$

We generate some few terms of the sequence as follows:

When $n=1$ ,

The first term of the sequence becomes,

$U_1=2(1)-1$

$U_1=1$

When $n=2$ ,

$U_2=2(2)-1$

$U_2=3$

When $n=3$ ,

$U_2=2(3)-1$

$U_2=5$

The constant difference is $d=2$ .

The sum of the series is given by,

$S_n=\frac{n}{2}(2a+(n-1)d)$

There are 15 terms in the series, therefore $n=15$

$S_{15}=\frac{15}{2}(2(1)+(15-1)2)$

$S_{15}=\frac{15}{2}(2+14(2))$

$S_{15}=\frac{15}{2}(30)$

$S_{15}=15\times 15$

$S_{15}=225$

ANSWER TO QUESTION 2

The bottom row has $23$ boxes. This means $a=U_1=23$

a) Since each row has 3 fewer boxes $d=-3$ .

The nth row is given by,

$U_n=23+(n-1)\times(-3)$

$U_n=23+-3n+3$

$U_n=23-3n$

The top row is the 6th row, meaning $n=6$ .

$U_6=23-3(6)$

$U_6=23-18$

$U_6=5$

b) The number of boxes in the entire display is given by

$S_n=\frac{n}{2} (a+l)$

Where $a=23$ , the first term(row) and $l=5$ , the last term(row).

Since there are 6 rows,

$S_6=\frac{6}{2} (23+5)$

$S_6=3 (28)$

$S_6=84$

ANSWER TO QUESTION 3

The number of visitors in the first week is $a=5$ .

Since the number of visitors each week is doubled the number of visitors in the subsequent weeks, the sequence is a geometric progression with a common ratio of $r=2$ .