The vertices of AMNO are M (1,3), N (4,9), and O (7,3). The vertices of APQR are P (3,0), Q (4,2), and R (5,0) Which conclusion
Pavel [41]
Answer:
the correct answer should be C
Step-by-step explanation:
hope this helps you
Answer:9x-9=7 or 9x-9=-7
Step-by-step explanation:
An absolute value is the distance from zero, therefore the answer can not be negative. In this case, 9x-9=7 and 9x-9=-7 both work, because 7 and -7 are the same distance from zero.
Answer: Hello there!
this type of equations in one dimension (when all the factors are constants) are written as:
h = initial position + initial velocity*t + (acceleration/2)*t^2
First, let's describe the hunter's equation:
We know that Graham moves with a velocity of 1.5 ft/s, and when he is 18 ft above the ground, Hunter throws the ball, and because Graham is pulled with a cable, he is not affected by gravity.
If we define t= 0 when Graham is 18 ft above the ground, the equation for Graham height (in feet) is:
h = 18 + 1.5t
where t in seconds.
Now, the equation for the ball:
We know that at t= 0, the ball is thrown from an initial distance of 5ft, with an initial velocity of 24ft/s and is affected by gravity acceleration g, where g is equal to: 32.2 ft/s (notice that the gravity pulls the ball downwards, so it will have a negative sign)
the equation for the ball is:
h = 5 + 24t - (32.2/2)t^2 = 5 + 24t - 16.1t^2
So the system is:
h = 18 + 1.5t
h = 5 +24t - 16.1t^2
so the right answer is A
Minus 6 both sides
4>x/3
times3 both sides
12>x
Answer:
z= 0.278
Step-by-step explanation:
Given data
n1= 60 ; n2 = 100
mean 1= x1`= 10.4; mean 2= x2`= 9.7
standard deviation 1= s1= 2.7 pounds ; standard deviation 2= s2 = 1.9 lb
We formulate our null and alternate hypothesis as
H0 = x`1- x`2 = 0 and H1 = x`1- x`2 ≠ 0 ( two sided)
We set level of significance α= 0.05
the test statistic to be used under H0 is
z = x1`- x2`/ √ s₁²/n₁ + s₂²/n₂
the critical region is z > ± 1.96
Computations
z= 10.4- 9.7/ √(2.7)²/60+( 1.9)²/ 100
z= 10.4- 9.7/ √ 7.29/60 + 3.61/100
z= 0.7/√ 0.1215+ 0.0361
z=0.7 /√0.1576
z= 0.7 (0.396988)
z= 0.2778= 0.278
Since the calculated value of z does not fall in the critical region so we accept the null hypothesis H0 = x`1- x`2 = 0 at 5 % significance level. In other words we conclude that the difference between mean scores is insignificant or merely due to chance.
