I think k equals negative 2, or -2.<span>
</span>
<span>a regular hexagon inscribed in a circle </span>
Answer:
The area between z = 1.74 and z = 1.25 is of 0.065.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
The area between two values of Z is given by the subtraction of the pvalue of the larger value by the smaller.
The area between z = 1.74 and z = 1.25.
This is the pvalue of z = 1.74 subtracted by the pvalue of z = 1.25.
z = 1.74 has a pvalue of 0.959
z = 1.25 has a pvalue of 0.894
0.959 - 0.894 = 0.065
The area between z = 1.74 and z = 1.25 is of 0.065.
Since each vector is a member of

, the vectors will span

if they form a basis for

, which requires that they be linearly independent of one another.
To show this, you have to establish that the only linear combination of the three vectors

that gives the zero vector

occurs for scalars

.

Solving this, you'll find that

, so the vectors are indeed linearly independent, thus forming a basis for

and therefore they must span

.
Answer:
It is number Five. Explanation: I know there congruent because if any two sides of triangle are the same then both triangles are congruent
Step-by-step explanation: