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zlopas [31]
2 years ago
13

The length of an animal is 24.5 inches that is normally distributed. The standard deviation is 1.3 inches. What would be the sec

tions that are one and two standard deviations from the mean? Select all that apply
A) 21.9
B) 22.5
C) 23.2
D) 23.5
E) 25.5
F) 25.8
G) 27.1
Mathematics
1 answer:
Alborosie2 years ago
6 0
G thanks i am the best lol
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PLSS HELPP MMEEE FASSTT
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Please please please help and solve this with steps, much help needed thank you :) 20 points for this!
leonid [27]

Answer:

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

\frac{dy}{dx}y=x^3+2x-5x+3

\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}

  • \mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)

1. \mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:

y'\:y=x^3+2x-5x+3

2. \mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}

yy'\:=x^3-3x+3

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3. \mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1

4. \mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}

5. \mathrm{Simplify}

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

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oee [108]

Answer:

A=3

Step-by-step explanation:

2353032/99 = 23768

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3 years ago
I can't figure out how to do this problem. Could somebody help me?
Verizon [17]

start by plugging in the value you already have, in this case, 5 for y.

-4x-3(5)=17

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however, since a is a+1, a would be 7 (making x 8)

also plz tell me if i read this wrong and it’s actually 1,5 rather than a+1 and 5 cause if i did i’m so sorry lol.

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