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3 years ago

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According to a recent study, the carapace length for adult males of a certain species of tarantula are normally distributed with

mean of u = 18.55 mm
and a standard deviation of o = 1.51 mm. Complete parts (a) through (d) below.
Click here to view page 1 of the standard normal distribution table.
Click here to view page 2 of the standard normal distribution table.
a. Find the percentage of the tarantulas that have a carapace length between 17 mm and 18 mm.

1 answer:

seraphim [82]3 years ago

3 0

Idk tbh sorry bye ok hbye

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Lagrangian:

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Derivatives of the Lagrangian set equal to zero:

$L_x=2x+\lambda=0\implies x=-\dfrac\lambda2$
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$-\dfrac\lambda2-2(2+\lambda)+3\left(5-\dfrac{3\lambda}2\right)=6$
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The cost of one pound of peanuts is 60% of the cost of one pound of cashews. If one pound of cashews costs $2, what is the cost

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Based on the given problem above, this would appear that,
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3 years ago

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Luba_88 [7]

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2 years ago

What are the roots of f(x) = x2 – 48? –48 and 48 –24 and 24 Negative 8 StartRoot 3 EndRoot and 8 StartRoot 3 EndRoot Negative 4

Ivanshal [37]

Answer:

$x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Step-by-step explanation:

$x=4\sqrt{3},\:x=-4\sqrt{3}$

Considering the expression

$x^2\:-\:48$

Solving

$x^2-48=0$

$x^2-48+48=0+48$

$x^2=48$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{48},\:x=-\sqrt{48}$

Solving

$x=\sqrt{48}$

= $\sqrt{2^4\cdot \:3}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

= $\sqrt{3}\sqrt{2^4}$

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$=2^2\sqrt{3}$

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So,

$x=4\sqrt{3}$

Similarly,

$x=-\sqrt{48}=-4\sqrt{3}$

Therefore, $x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Keywords: roots, expression

Learn more about roots from brainly.com/question/3731376

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zhannawk [14.2K]

Answer:

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