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3 years ago

13

I really need help with this statistical/variable question, i am really struggling and would mean alot if someone could answerr

1 answer:

inna [77]3 years ago

6 0

Answer:

Step-by-step explanation:

1) quantitative

2)categoreal

3)neither

4) quantitative

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Write the word name for each decimals of 2,010.8

Goshia [24]

two thousand ten and eight tenths

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4 years ago

WILL MARK BRAINLIEST what is the standard deviation for the normal distribution shown at the right?

MrRissso [65]

Answer: 916. Because if you do 856+60=916.

Step-by-step explanation: The standard normal distribution is a normal distribution with a mean of zero and standard deviation of 1. Formula: and natural statistics x and x2. The dual expectation parameters for normal distribution are η1 = μ and η2 = μ2 + σ2.

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3 years ago

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

$6x^{2}+6x-336=0$

$x^{2}+x-56=0$

$x^{2}+8x-7x-56=0$

$(x+8)-7(x+8)=0$

$(x+8)(x-7)=0$

$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

(b) Since $x=-8,7$ are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

$f(-8)=1856$

$f(7)=2(7)^3+3(7)^2-336(7)$

$f(7)=-1519$

Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

3 0

3 years ago

A carpenter designs an A-frame chalkboard. Its support bars form the legs of an isosceles triangle.

ahrayia [7]

angle for triangle is 180°

angle ACB is 75° because it has mention that is an isosceles triangle

So that angle of BAC is 180°-75°-75°=30°

6 0

4 years ago

Given g(x) = -x + 4, find g(5).

lilavasa [31]

Answer:

g(5) = -1

Step-by-step explanation:

Let x = 5.

$g(5)=-(5)+4\\g(5)=-5+4\\g(5)=-1$

6 0

3 years ago

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