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3 years ago

24 written as a percentage

1 answer:

5 0

Any number is equal to that number over 1

So x is equal to x/1

Same thing with 24. 24 = 24/1

To get it into percent we multiply by 100 so:

$\sf{\frac{24}{1}=\frac{24\times100}{1\times100}}$

And simplifying that we get

$\sf{\frac{2400}{100}}$

And x/100 = x%

so 2400/100 = 2400%

So your answer is <em><u>2400%</u></em> :)

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74% of freshmen entering public high schools in 2006 graduated with their class in 2010. A random sample of 81 freshmen is selec

Assoli18 [71]

Answer: Probability that the proportion of students who graduated is greater than 0.743 is P = 0.4755

Step-by-step explanation:

Given that,

Probability of freshmen entering public high schools in 2006 graduated with their class in 2010, p = 0.74

Random sample of freshman, n = 81

Utilizing central limit theorem,

$P(\hat{p}$

So,

$(P(\hat{p}>0.743) = P(Z>0.743 - \frac{0.74}{\sqrt{\frac{0.74(1-0.74)}{81} } } )$

= P( Z > 0.0616)

= 0.4755 ⇒ probability that the proportion of students who graduated is greater than 0.743.

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3 years ago

Please take your time but I really need anyone's help!!!!

stiv31 [10]

Answer:

C) 5 3√2

Step-by-step explanation:

5 0

2 years ago

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Two minor league baseball players got a total of 210 hits. Washington had 12 more hits than Sanchez. find the number of hits fro

mote1985 [20]

Washington: 117 Hits
Sanchez: 93 Hits

Hope this helps, have a nice day!

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2 years ago

Need help on this question

bazaltina [42]

Hello,
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3 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 $g'(\pi) = 0$

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

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

Answer is choice D

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2 years ago

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