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3 years ago

8

Solve the quadratic equation by completing the square x^2+5x-3=0

1 answer:

PIT_PIT [208]3 years ago

8 0

${\tt{x = \frac{ -b± \sqrt{ {b}^{2} - 4ac } }{2a} }}$

${\tt{ x = \frac{ - 5±\sqrt{{5}^{2} - 4 \times 1 \times ( - 3)} }{2 \times 1} }}$

${\tt{x = \frac{ - 5±\sqrt{25 + 12} }{2} }}$

${\tt{x = \frac{ - 5 ±\sqrt{37} }{2} }}$

$\: \:$

${\tt{x = \frac{ - 5 + \sqrt{37} }{2} }}$

${\tt{x = \frac{ - 5 - \sqrt{37} }{2} }}$

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Geometry question?

IRINA_888 [86]

J is your andswer. Area is side x side so multiply

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4 years ago

A single die is rolled twice. The set of 36 equally likely outcomes is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),

lana [24]

Answer:

(9) $\frac{1}{12}$ (10) $\frac{1}{12}$ (11) $\frac{5}{12}$ (12) $\frac{1}{4}$ (13) $\frac{1}{6}$ 14) $\frac{5}{36}$ (15) $\frac{1}{12}$ (16)0

Step-by-step explanation:

The sample Space of the single die rolled twice is presented below:

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6),}.

n(S)=36

(9)Probability of getting two numbers whose sum is 9.

The possible outcomes are: (3, 6), (4, 5), (5, 4)

$P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}$

10) Probability of getting two numbers whose sum is 4.

The possible outcomes are: (1, 3),(2, 2),(3, 1),

$P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}$

11.)Find the probability of getting two numbers whose sum is less than 7.

The possible outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)

$P(\text{two numbers whose sum is less than 7})=\frac{15}{36}=\frac{5}{12}$

12.Probability of getting two numbers whose sum is greater than 8

The possible outcomes are:(4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)

$P(\text{two numbers whose sum is greater than 8})=\frac{9}{36}=\frac{1}{4}$

(13)Probability of getting two numbers that are the same (doubles).

The possible outcomes are:(1, 1)(2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

$P(\text{two numbers that are the same})=\frac{6}{36}=\frac{1}{6}$

14.Probability of getting a sum of 7 given that one of the numbers is odd.

The possible outcomes are: (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

$P(\text{getting a sum of 7 given that one of the numbers is odd.})=\frac{5}{36}$

(15)Probability of getting a sum of eight given that both numbers are even numbers.

The possible outcomes are: (2, 6), (4, 4), (6, 2)

$P(\text{getting a sum of eight given that both numbers are even numbers.})=\frac{3}{36}\\=\frac{1}{12}$

16.Probability of getting two numbers with a sum of 14.

$P(\text{getting two numbers with a sum of 14.})=\frac{0}{36}=0$

4 0

4 years ago

A friend of mine is giving a dinner party. His current wine supply includes 9 bottles of zinfandel, 10 of merlot, and 12 of cabe

e-lub [12.9K]

Answer:

Step-by-step explanation:

From the given information:

The total number of wine = 9 + 10 + 12 = 31

(1)

The number of distinct sequences used for serving any five wines can be estimated by using the permutation of the number of total wines with the number of wines served.

i.e

= $^{31}P_5$

$=\dfrac{31!}{(31-5)!}$

$=\dfrac{31!}{(26)!}$

$=\dfrac{31\times 30\times 29\times 28\times 27\times 26!}{(26)!}$

= 20389320

(2)

If the first two wines served = zinfandel and the last three is either merlot or cabernet;

Then, the no of ways we can achieve this is:

= $^9P_2\times ^{22}P_3$

$= \dfrac{9!}{(9-2)!}\times \dfrac{22!}{(22-3)!}$

$= \dfrac{9!}{(7)!}\times \dfrac{22!}{(19)!}$

$= \dfrac{9*8*7!}{(7)!}\times \dfrac{22*21*20*19!}{(19)!}$

= 665280

(3)

The probability that no zinfandel is served is computed as follows:

Total wines (with zinfandel exclusion) = 31 - 9 = 22

Now;

the required probability is:

$= \dfrac{^{22}P_5 }{^{31}P_5}$

$= \dfrac{\dfrac{22!}{(22-5)!} } {\dfrac{31!}{(31-5)!} }$

$= \dfrac{\dfrac{22!}{17!} } {\dfrac{31!}{(26)! }}$

$= \dfrac{(\dfrac{22*21*20*19*18*17!}{17!})} {(\dfrac{31*30*29*28*27*26!}{(26)! })}$

= 0.1549

≅ 0.155

4 0

3 years ago

A 5-lb bag of apples costs 4.50 and an 8-lb bag of the same type of apple costs 7.52. Greg found the unit price which is the con

tester [92]

Answer:

The unit price of 5-lb bag is less than the unit price of 8lb bag.

Step-by-step explanation:

Consider the provided information.

We need to calculate the unite price,

It is given that A 5-lb bag of apples costs 4.50

Calculate the unit price as shown:

$\frac{Cost}{weight} =\frac{4.50}{5}$

We need the cost of 1 l b .

$\frac{Cost}{weight} =\frac{4.50\div5 }{5\div 5}$

$\frac{Cost}{weight} =\frac{0.90}{1}$

So, the cost of 1lb of apple is $0.90.

8-lb bag of the same type of apple costs 7.52.

Calculate the unit price as shown:

$\frac{Cost}{weight} =\frac{7.52}{8}$

We need the cost of 1 l b .

$\frac{Cost}{weight} =\frac{7.52\div8 }{8\div 8}$

$\frac{Cost}{weight} =\frac{0.94}{1}$

So, the cost of 1lb of apple is $0.94.

Therefore, the unit price of 5-lb bag is less than the unit price of 8lb bag.

8 0

3 years ago

Do-nothing #1 can count 5 more ceiling dots

klasskru [66]

Answer:

2

Step-by-step explanation:

5 0

2 years ago