True or false, areas under the standard normal curve cannot be negative, whereas z-scores can be positive or negative.

Mathematics

1 answer:

Veseljchak [2.6K]3 years ago

5 0

Answer:

True.

Step-by-step explanation:

We have been given a statement. We are asked to determine the given statement is true or false.

Statement: Areas under the standard normal curve cannot be negative, whereas z-scores can be positive or negative.

We know that area under the standard normal curve is equal to 1, so area under normal curve cannot be negative.

We know that a z-score represents that a data point is how many standard deviation above or below mean.

When a data point is less than mean, then z-score will be negative and when a data point is greater than mean, the z-score will be positive.

Therefore, our given statement is true.

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a point on a perpendicular bisector is 7cm from each endpoint of the bisected segment and 5cm from the point of intersection. Wh

love history [14]

9514 1404 393

Answer:

about 9.80 cm

Step-by-step explanation:

The length of half the segment (h) can be found from the Pythagorean theorem:

h² +5² = 7²

h² = 7² -5² = 49 -25 = 24

h = √24 = 2√6

This is half the segment length, so the whole segment length is ...

L = 2h = 2(2√6)

L = 4√6 ≈ 9.7980

The length of the segment is 4√6 ≈ 9.80 cm.

8 0

3 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1