Which statement is true of normally distributed data?

Mathematics

2 answers:

mylen [45]2 years ago

8 0

have a nice day!!!!!!!!!!!!!!!!!!!!!!!

gogolik [260]2 years ago

5 0

C

Approximately 95% of data falls within 2 standard deviations (±2) of the mean.

Explanation

According to the empirical rule of normal distribution:

Approximately 68% of the data falls within ±1 standard deviation of the mean

2. Approximately 95% of the data falls within ±2 standard deviations of the mean

3. Approximately 99.7% of the data falls within ±3 standard deviations of the mean.

Therefore, among the given options, only option C adheres to the empirical rule of the normal distribution. Therefore, the option C is correct

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3. f(x) = 2x + 5 Find f(0)

Phantasy [73]

Answer:

Step-by-step explanation:

$f(x) = 2x + 5 \\ pluggng \: x = 0 \\ f(0) = 2 \times 0 + 5 = 0 + 5 = 5 \\ f(0) =5 \\$

8 0

3 years ago

Find the x- and y- intercepts of parabola y=5x^2-16x+10

____ [38]

Y-INTERCEPT

$y = 5x^2 - 16x + 10$

The y-intercept is where the equation/curve/parabola cosses the y-axis.

The y-axis is where x = 0. (The x-axis is where y = 0)

To find the y-intercept:

$\text{y-axis} \rightarrow \text{x = 0} \rightarrow y = 5(0)^2 -16(0) + 10 = 10$

The y-intercept must be at (0, 10)

X-INTERCEPT (ROOTS/SOLUTIONS)

$y = 5x^2 - 16x + 10\\\text{make it equal 0}\\y = 0\\\therefore 5x^2 - 16x + 10 = 0$

We need to use the quadratic formula

The quadratic formula helps us find what values of $x$ make the equation = 0

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-16) + \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\\x = \frac{16 + \sqrt{256-200}}{10}\\x = \frac{16 + \sqrt{56}}{10}\\x = \frac{16 + 2\sqrt{14}}{10}\\x = \frac{8 + \sqrt{14}}{5}\\\\\\x=\frac{-(-16) - \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\text{doing the same thing...}\\\text{end up with...}\\x = \frac{8 - \sqrt{14}}{5}\\$