<h2>Answer :-</h2>
As we know that,
Pythagoras triplet
1) a² + b² = c²
Let
<h3>Hence, A can't be Pythagoras triplet</h3>
2) a² + b² = c²
<h3>Therefore, B can be Pythagoras triplet</h3>
3)a² + b² = c²
<h3>Hence, C can't be Pythagoras triplet</h3>
4) a² + b² = c²
<h3>Hence, D can't be Pythagoras triplet</h3>
<h2 /><h2>Therefore :-</h2>
Only B can be Pythagoras triplet.
It is .00019
hope it helps!
Option D: are the coordinates of the point P
Explanation:
It is given that "A coordinate grid shown from negative 2 to positive 2 on x-axis and negative 2 to positive 2 on y-axis. There are 4 grid lines between a whole unit of the grid".
Thus, the value of each grid is
It is also given that, "A point labeled P is shown at the intersection of 8 grid lines to the left of the y-axis and 4 grid lines above the x-axis".
Since the point P lies on the left of the y-axis and above the x-axis, the x-value will be negative and y-value will be positive.
Thus, the value of x-axis is given by
And the value of y-axis is given by
Thus, the coordinates of the point P is
Hence, Option D is the correct answer.
First, make sure that all the variables are in one side. The first equation is gonna be x-y=3
Second, we are going to eliminate the y because it is much easier, but you can also eliminate the x
We have to add both equation because the y in the first equation is negative and in the second equation it’s positive.
We are left with 3x=9
Therefore, x=3
Last, substitute x= 3 in one of the equations
Y=6
P.O.I is (3,6)
Answer:
0.16
Step-by-step explanation:
This probability is the area under the standard normal curve to the right of 82. Note that 82 is ONE standard deviation above the mean: 75 + 7 = 82.
According to the Empirical Rule, 68% of data lies within one standard deviation of the mean. In other words, 34% lies between the mean and one standard deviation above the mean. The remaining area under the standard normal curve is thus (100% - 68%), or 32%. Half of that is 16%.
The probability that the temperature is above 82, or above 1 standard deviation above the mean, is 16%, which we can also express as 0.16.
Alternatively, use a calculator with built-in statistical functions. In this case the desired result is found by executing the following command:
normcdf(82, 10000, 75, 7). The desired result is 0.1587, or approximately 0.16.