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lora16 [44]
3 years ago
10

The lengths of one-year-old baby girls can be described using a normal distribution with a mean of 29 inches and a standard devi

ation of 1.2 inches. using the empirical '68-95-99.7' rule, what percent of one-year-old girls are shorter than 31.4 inches?
Mathematics
1 answer:
crimeas [40]3 years ago
6 0

The '68-95-99.7' rule says

-About 68% of values fall within one standard deviation of the mean.

-About 95% of the values fall within two standard deviations from the mean.

-Almost all of the values — about 99.7% — fall within three standard deviations from the mean.

Here 31.4=29+2*1.2 26.6=29-2*1.2

Therefore, the probability

P(26.6

The required probability

P(L

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Is this function linear or nonlinear? *
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Answer:

It's linear

Step-by-step explanation:

5 0
3 years ago
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Use the number line to add the fraction. + + 1 1 7 3 -1 10 9 10 1 -1 10 4 5 3 5 Niat 1 2 + 1 5 N GIN 3 10 0 -1 3 10 1 10 + 1 2 +
user100 [1]
This is hard to understand , do u have a photo
8 0
2 years ago
How do you do this and what are the answers
scoray [572]
I can answer 1 to 4...
The area of a circle can be found by \pi r^{2}. Reminder: \pi = 3.14 (usually used when finding out the area of a circle)

1. 4^{2} = 16
\pi *16 or 3.14 * 16 = 50.24
50.24/4=12.56

2. 6^{2} = 36
\pi *36 or 3.14 * 36 = 113.04
113.04/10=11.304

(Not really sure if this is correct) 3. 2^{2}=4
\pi *4 or 3.14 * 4
12.56/2.66666667=11.70999999

4. 6^{2}=36
\pi *36= 113.04
113.04/6=18.84
5 0
3 years ago
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

#SPJ4

3 0
2 years ago
Match the circle equations in general form with their corresponding equations in standard form. x2 y2 − 4x 12y − 20 = 0 (x − 6)2
frosja888 [35]

The equation given is matched with the respective vertex forms.

<h3>What is the standard Equation for a circle ?</h3>

The standard equation for a circle is given by

(x-h)² + (y-k)² = r²

The equation given in the question are

x² + y²  − 4x + 12y − 20 = 0

(x − 6)²  + (y − 4)²  = 56

x²  + y²  + 6x − 8y − 10 = 0

(x − 2)²  + (y + 6)²  = 60

3x²  + 3y²  + 12x + 18y − 15 = 0

(x + 2)²  + (y + 3)²  = 18

5x²  + 5y²  − 10x + 20y − 30 = 0

(x+3)²  + (y-4)²  = 35

2x²  + 2y²  − 24x − 16y − 8 = 0

(x-1)²  + (y+2)²   = 11

The first equation is

x² + y²  − 4x + 12y − 20 = 0

It can be written ad

x² -4x + 4 - 4 +y²+12y +36-36 -20 = 0

(x-2)² +(y+6)²= 60

The second equation is

x²  + y²  + 6x − 8y − 10 = 0

x² +6x +9-9 +y²  -8y +16-16 -10 = 0

(x+3)²  + (y-4)²  = 35

The third equation is

3x²  + 3y²  + 12x + 18y − 15 = 0

(x + 2)²  + (y + 3)²  = 18

The fourth equation is

5x²  + 5y²  − 10x + 20y − 30 = 0

(x-1)²  + (y+2)²   = 11

The fifth equation is

2x²  + 2y²  − 24x − 16y − 8 = 0

(x − 6)²  + (y − 4)²  = 56

To know more about standard equation of Circle

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7 0
2 years ago
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