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2 years ago

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Test scores for a Statistics class have a mean of 78 with a standard deviation of 6. Suppose a student gets an 81 on that test.

What is the z-score for that grade?

1 answer:

Ostrovityanka [42]2 years ago

5 0

Answer:

0.3209

in case you dont know A z-score tells you if the distribution it comes from is normal.

Step-by-step explanation:

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What is the measure of an angle if the measures of its two adjacent supplementary angles add up to 100°?

Sindrei [870]

<h3>Answer: 130</h3>

Explanation:

Let x be the unknown angle we want to find.

Let y be adjacent and supplementary to x. This means x+y = 180

Let z also be adjacent and supplementary to x. So x+z = 180 also

Subtracting the two equations leads to y-z = 0 and y = z. So effectively we've proven the vertical angle theorem.

Since the supplementary angles to x add to 100, we know that y+z = 100. Plug in y = z and solve for z

y+z = 100

z+z = 100

2z = 100

z = 100/2

z = 50

Therefore,

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3 years ago

What’s 437727x9018 ?

makvit [3.9K]

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Find (12x+12)−(−5x+8). step by step answers please.

Bas_tet [7]

Answer:

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2 years ago

Can someone please explain this to me? Thanks!

Makovka662 [10]

Answer: Choice D

$\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\$

==========================================================

Explanation:

Let g(t) be the antiderivative of $g'(t) = \sqrt{1+t^3}$ . We don't need to find out what g(t) is exactly.

Recall by the fundamental theorem of calculus, we can say the following:

$\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)$

This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).

So,

$\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\$

From here, we apply the derivative with respect to x to both sides. Note that the $g(\pi)$ portion is a constant, so $g'(\pi) = 0$

$\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\$

$\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\$

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