<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u> </u>√15 units
Step-by-step explanation:
Let (6,1) be (x^1,y^1) and (1,-9) be (x^2,y^2) .
As we know ,
Distance(D) = √(x^1-x^2) +(y^1-y^2)
Now,
D= √(x^1-x^2) +(y^1-y^2)
= √(6-1) +(1+9)
= √5+10
= √15 units
: Therefore the distance between (6,1) and (1,-9) is √15 units.
The original number to the nearest tenth is 23.8
Answer:
x=4
Step-by-step explanation:
(whole secant) x (external part) = (tangent)^2
(x+5) *x = 6^2
x^2 +5x = 36
Subtract 36 from each side
x^2 +5x - 36 = 0
Factor
( x-4) (x+9) = 0
Using the zero product property
x-4 = 0 x+9 =0
x = 4 x=-9
Cannot be negative since that is negative length
x=4
Answer:
Step-by-step explanation:
y= 3 (-4)+5
y= - 12+5
y= -7
Correct Ans:Option A. 0.0100
Solution:We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.
First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.
Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.