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3 years ago

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Citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal dist

ribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm. Find the 80th percentile for the diameters of mandarin oranges, round 4 decimal places

1 answer:

SSSSS [86.1K]3 years ago

8 0

Answer:

6.0520

Step-by-step explanation:

Given that:

Mean = 5.85

Standard deviation = 0.24

Let N denote the 80th percentile.

P(X < N) = 0.8

$P(Z < \dfrac{(N - 5.85)}{0.24}) = 0.8$

Using the Excel formula: = NORM.INV(0.8,5.85,0.24)

The 80th percentile for the diameters of mandarin oranges:

= 6.0520 (to 4 decimal place)

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xxMikexx [17]

Answer:

-1, 2, 6

Step-by-step explanation:

We have to solve the equation as follows: 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).

Now, we have, $\frac{1}{x-6} +\frac{x}{x-2} = \frac{4}{x^{2}-8x+12 }$

⇒ $\frac{(x-2)+x(x-6)}{(x-2)(x-6)} = \frac{4}{x^{2}-8x+12 }$

⇒ $\frac{x-2+x^{2}-6x }{(x-2)(x-6)} =\frac{4}{(x-2)(x-6)}$

⇒ $\frac{(x-2)(x-6)}{x^{2}-5x-2 }=\frac{(x-2)(x-6)}{4}$

⇒ $(x-2)(x-6)[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$

⇒ $(x-2)(x-6) =0$ or, $[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$

If, (x-2)(x-6) =0, then x=2 or x=6

If, $[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$ , then $x^{2} -5x-2=4$

and (x-6)(x+1) =0

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3 years ago

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In this we know all three zeros and one point from which the graph pass.
So we will let specific cubic polynomial function of the form
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So required equation is
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Solve the equation 2p/3-12=-2

-BARSIC- [3]

Hey there!

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Firstly, we are going add $12$ on each of the sides that we're working with. like ↓

$p - 12 + 12 \\ \\ = -2 + 12$

This gives us $\frac{2}{3}p = 10$ (if you are wondering how we got the out come of $10$ it is because I added $-2 + 12 = 10$

Now multiply $\frac{3}{2}$ on each of your sides

$\frac{3}{2} ( \frac{2}{3}p) \\ \\ = \frac{3}{2} (10)$

Cancel the first set and you will find the value of $p$

$p = 15$

Good luck on your assignment and enjoy your day

~ $LoveYourselfFirst:)$

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