Find the Equivalent Ratio 3:6 x:3

Given f(x) = 3x- 1 and 9(x) = 2x-3, for which value of

Montano1993 [528]

Answer:

x = 4

Step-by-step explanation:

$f(x)=3x-1\\g(x)=2x-3$

substitute x = 2 into f(x):

$f(2)=(3 \times2)-1=5$

equate g(x) with found value for f(2) and solve for x:

$g(x)=f(2)\\2x-3=5\\2x=8\\x=4$

5 0

2 years ago

What’s number 2 ??? Please help

labwork [276]

Answer:

c

Step-by-step explanation:

i think

8 0

3 years ago

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Assume that college students have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15. Fin

jolli1 [7]

Answer:

0.018 is the probability that a randomly selected college student has an IQ greater than 131.5

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 15

We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.

Formula:

a) P(IQ greater than 131.5)

P(x > 131.5)

Calculation the value from standard normal z table, we have,

0.018 is the probability that a randomly selected adult has an IQ greater than 131.5

3 0

3 years ago

Suppose cot(theta) = 5/12, where pi < theta < 3pi/2. What is sin(theta)?

trasher [3.6K]

Answer:

- 12 /13

Step-by-step explanation:

Given that:

Cot(θ) = 5 /12 ;

Recall ; cot θ = 1 / tan θ ; tan θ = sin θ / cos θ

Hence ;

tan θ = 1 / cot θ

tan θ = 1 ÷ (5 / 12)

tan θ = 1 * 12 /5 = 12/5

Tan θ = opposite / Adjacent = 12 /5

Sin θ = opposite / hypotenus

Hypotenus = √(opposite² + adjacent²)

Hypotenus = √12² + 5²

Hypotenus = √(144 + 25)

Hypotenus = √169

Hypotenus = 13

Hence,

Sin θ = opposite / hypotenus = 12 / 13

pi < theta < 3pi/2 lies in the 3rd quadrant ; SinΘ will be negative ;

Sin θ = - 12 /13

6 0

3 years ago

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|x-3|<2 |4x+1|>0 |x-1|<5 Ayudaaaaaaa pleaseeeee

Dafna11 [192]

Recuerda que

• |x| = x si x ≥ 0

• |x| = -x si x < 0

Necesitas considerar dos casos:

• si x - 3 ≥ 0,

|x - 3| < 1 ⇒ x - 3 < 1 ⇒ x < 4

• si x - 3 < 0,

|x - 3| < 1 ⇒ -(x - 3) = 3 - x < 1 ⇒ -x < -2 ⇒ x > 2

Entonces la solución consta de todos los números reales x tales que x > 2 y x < 4, o simplemente 2 < x < 4.

El método para resolver las otras desigualdades es el mismo.

|4x + 1| > 0 ⇒ 4x + 1 > 0 o -(4x + 1) > 0

… ⇒ 4x + 1 > 0 o -4x - 1 > 0

… ⇒ 4x > -1 o -4x > 1

… ⇒ x > -1/4 o x < -1/4

⇒ x ≠ -1/4

|x - 1| < 5 ⇒ x - 1 < 5 o -(x - 1) < 5

… ⇒ x - 1 < 5 o -x + 1 < 5

… ⇒ x < 6 o -x < 4

… ⇒ x < 6 o x > -4

⇒ -4 < x < 6

7 0

3 years ago