Is (-1,2) a solution of the system y>x and x+y=>

1 answer:

8 0

(-1,2) is a solution

y>x

2>-1 is correct

in this case yes

x+y≥x

-1+2≥-1

1≥-1 is correct: Yes

in this case no

x+y≥y

1≥2

is not correct. No

You might be interested in

1) Suppose the heights of 18-year-old men are approximately normally distributed, with mean 71 inches and standard deviation 5 i

Luba_88 [7]

Answer:

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 71 inches and standard deviation 4 inches.

(a) What is the probability that an 18-year-old man selected at random is between 70 and 72 inches tall? (Round your answer to four decimal places.)

z1 = (70-71)/4 = -0.25

z2 = (72-71/4 = 0.25

P(70<X<72) = p(-0.25<z<0.25) = 0.1974

Answer: 0.1974

(b) If a random sample of thirteen 18-year-old men is selected, what is the probability that the mean height x is between 70 and 72 inches? (Round your answer to four decimal places.)

z1 = (70-71)/(4/sqrt(13)) = -0.9014

z2 = (72-71/(4/sqrt(13)) = 0.9014

P(70<X<72) = p(-0.9014<z<0.9014) = 0.6326

Answer: 0.6326

please mark me the brainiest

3 0

2 years ago

How do you convert 0.6 into a fraction.

miss Akunina [59]

Answer:

since it is in the tenths place, it'd be 6/10, or 6 tenths.

7 0

3 years ago

Read 2 more answers

How to factor the gcf out of a polynomial

Ann [662]

Step-by-step explanation:

$15x^6+6x^4+9x^3\\\\\text{Find GCF of 15, 6 and 9:}\\\\15=\boxed3\cdot5\\6=\boxed3\cdot2\\9=\boxed3\cdot3\\\\GCF(15,\ 6,\ 9)=\boxed3\\\\\text{Find GCF of}\ x^6,\ x^4\ \text{and}\ x^3:\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\x^6=x^{3+3}=\boxed{x^3}\cdot x^3\\\\x^4=x^{3+1}=x^3\cdot x^1=\boxed{x^3}\cdot x\\\\x^3=\boxed{x^3}\cdot1\\\\GCF(x^6,\ x^4,\ x^3)=\boxed{x^3}\\\\\text{Therefore}\\\\GCF(15x^6+6x^4+9x^3)=\boxed{3}\cdot\boxed{x^3}=3x^3\\\\15x^6+6x^4+9x^3=3x^3(5x^3+2x+3)$