All categories

2 years ago

Find the type and number of solution for g(x) = x2 -14x = -50

Mathematics

1 answer:

inysia [295]2 years ago

3 0

Answer:

Step-by-step explanation:

x² -14x+50= 0

this equation has 2 solutions because is a quadratic

the solurions are imaginary roots because the discriminant is less then 0

b²-4ac = (-14)²-4*1*50 = 194-200= -6

to find the actual roots use the quadratic formula

You might be interested in

How many tens are there in 10 ones?

AlexFokin [52]

Answer:

1 ten.

Step-by-step explanation:

1 ten = 10 ones/units

7 0

2 years ago

Read 2 more answers

Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they

Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

$n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276$

Number of ways of choosing 3 students From all males

$n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220$

The probability that all are male of choosing '3' students

$P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }$

$P(E) = \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}$

P(E) = 0.067 = 6.71%

Final answer:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

3 0

3 years ago

adults say that they have cheated on a test or exam before. You randomly select six adults. Find the probability that the number

VLD [36.1K]

Answer: a. 1/6. b. 2/3. c. 5/6

Step-by-step explanation:

Number of adults randomly selected= 6

a. Finding the probability of adults that have cheated on an examination before is exactly 4.

The numbers are 1, 2, 3, 4, 5 and 6.

Probability of chosen exactly 4 will be 1/6 since 4 can only occur once.

b. More than 2means 3,4, 5 and 6.

Probability that the number of adults that have cheated on a test or exam is more than 2 will be:

4/6 = 2/3

c. At most 5 means 1,2,3,4 or 5

This will give 5/6

5 0

2 years ago

4.25 rounded to the nearest tenths place

musickatia [10]

4.3

5 or more raise the score

4 or less let it rest

4 0

2 years ago

Read 2 more answers

Verify that the following function is a probability mass function, and determine the requested probabilities. f left-parenthesis

Licemer1 [7]

Answer:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

1) $f(x_i) \geq 0, \forall x_i$

2) $sum_{i=1}^n P(X_i) =1$

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

Step-by-step explanation:

For this case we have the following density function:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

In order to satisfty that this function is a probability mass function we need to check two conditions:

1) $f(x_i) \geq 0, \forall x_i$

2) $sum_{i=1}^n P(X_i) =1$

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

And if we want to find the following probabilities:

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

7 0

2 years ago