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3 years ago

Determine the solution x^2 - 3x -28 ≥ 0

Mathematics

2 answers:

KengaRu [80]3 years ago

8 0

Answer:

(Infinity)],-4 or 7, ( Infinity)

IgorC [24]3 years ago

4 0

x^2 - 3x -28 ≥ 0

factor

(x-7) (x+4) ≥ 0

x=7 x=-4

we have three regions where the answers can lie

x<-4 between -4 and 7 and x>7

pick a point and see if it works

x=-10

(-10-7) (-10+4) ≥ 0

negative * negative is greater than 0 so this is a solution x< -4

x=0

(0-7) (+4) ≥ 0

negative * positive is less than 0 so this is not a solution

x=10

(10-7) (10+4) ≥ 0

positive * positive is greater than 0 so this is a solution x>7

We have two regions that work

x<-4 and x>7

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Give an example of a set of five ordered pairs. Determine the domain and range of the set.

sergey [27]

DOMAIN:

Set of allowable inputs (set of 1st elements of the ordered pairs in relation)
X point
AKA the input

DEFINE INPUT:

A number that'll be used in a function rule to determine the value of the output

RANGE:

Set of possible outputs (set of 2nd elements of the ordered pairs in relation)
Y point
AKA the output

DEFINE OUTPUT:

Result of applying a function rule to the value of ani input

ORDERED PAIRS:

(-1, 7): -1 is the domain & 7 is the range

(3, 5): 3 is the domain & 5 is the range

(4, 9): 4 is the domain & 9 is the range

(-6, -10): -6 is the domain & -10 is the range

(-2, 8): -2 is the domain & 8 is the range

Hope this helps you!!! :)

8 0

3 years ago

12 one half times one nine

pshichka [43]

Answer:

$1\frac{7}{18}$ or one seven eighteenths

Step-by-step explanation:

we know that

12 one half times one nine is the same that multiply 12 one half by one nine

$(12\frac{1}{2})(\frac{1}{9})$

Convert mixed number to an improper fraction

$12\frac{1}{2}=12+\frac{1}{2}=\frac{12*2+1}{2}=\frac{25}{2}$

substitute

$(\frac{25}{2})(\frac{1}{9})=\frac{25}{18}$

Convert to mixed number

$\frac{25}{18}=\frac{18}{18}+\frac{7}{18}=1\frac{7}{18}$

therefore

12 one half times one nine is one seven eighteenths

6 0

3 years ago

Inverse of f(x) = 3 + 2 In x?

JulsSmile [24]

This is the inverse of f(x)= -3 - (-2)

4 0

3 years ago

A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20

irakobra [83]

Answer:

a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that $p = 0.7$

20 adult women

This means that $n = 20$

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

$P(X < 14) = 1 - P(X \geq 14)$

In which

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916$

$P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789$

$P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304$

$P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716$

$P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278$

$P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068$

$P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008$

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079$

$P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921$

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

From the values found in item (a).

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107$

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

6 0

2 years ago

Solve |2x - 2| < 8

Maurinko [17]

C
You split inequality into
2x-2>8 and 2x—2<-8
X>5. X<-3

5 0

3 years ago