<img src="https://tex.z-dn.net/?f=2x%20%7B%20%7D%5E%7B2%7D%20%20%2B%205x%20-%203" id="TexFormula1" title="2x { }^{2} + 5x

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NeX [460]

3 years ago

alt="2x { }^{2} + 5x - 3" align="absmiddle" class="latex-formula">
how do I solve this

Mathematics

1 answer:

Kaylis [27]3 years ago

5 0

If you’re factoring it, it’s

(2x - 1) (x+3)

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In 2009, the world's largest pumpkin weighed 784 kilograms. An average-sized pumpkin weighs 5,000 grams. The 2009 world-record p

denpristay [2]

First you need to convert grams into kilograms. You know that 1kg=1000g
if 1kg=................. 1000g
x kg=.................5000g

x=5000/1000 = 5kg
Now you need to find the difference between an average-sized pumpkin and the <span>world's largest pumpkin which is:
784-5=779

So the right answer is 779.</span>

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3 years ago

The mean preparation fee H&R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use t

astraxan [27]

Answer:

a)0.6192

b)0.7422

c)0.8904

d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

Step-by-step explanation:

Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then

z(p)= $\frac{ME*\sqrt{N}}{s }$ where

Me is the margin of error from the mean

s is the standard deviation of the population

N is the sample size

z(p)= $\frac{8*\sqrt{30}}{50 }$ ≈ 0.8764

by looking z-table corresponding p value is 1-0.3808=0.6192

z(p)= $\frac{8*\sqrt{50}}{50 }$ ≈ 1.1314

by looking z-table corresponding p value is 1-0.2578=0.7422

z(p)= $\frac{8*\sqrt{100}}{50 }$ ≈ 1.6

by looking z-table corresponding p value is 1-0.1096=0.8904

Minimum required sample size for 0.95 probability is

N≥ $(\frac{z*s}{ME} )^2$ where

N is the sample size

z is the corresponding z-score in 95% probability (1.96)

s is the standard deviation (50)

ME is the margin of error (8)

then N≥ $(\frac{1.96*50}{8} )^2$ ≈150.6

Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

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3 years ago

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Softa [21]

$y=\dfrac{e^{2x}}{4+3e^x}$
$\implies y'=\dfrac{2e^{2x}(4+3e^x)-3e^xe^{2x}}{(4+3e^x)^2}$
$\implies y'=\dfrac{8e^{2x}+3e^{3x}}{(4+3e^x)^2}$

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$8e^{2x}+3e^{3x}=e^{2x}(8+3e^x)=0$

$e^{2x}>0$ for all $x$ , so we can divide through:

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