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Ivanshal [37]
2 years ago
11

while only 5% of babies have learned to walk by the age of 10 months, 75% are walking by 13 months. if the age at which babies d

evelop the ability to walk can be described by a normal model, find the parameters
Mathematics
1 answer:
7nadin3 [17]2 years ago
4 0

Answer:

mu=12.1299

\sigma=1.2987

Step-by-step explanation:

In the normal distribution curve, we will have 5% below 10 months [horizontal axis] and 75% below 13 months [horizontal axis].

We need to find z-score using z-table [normal table] that corresponds to

5% = 0.05

and

75% = 0.75

Zscore formula:

z=\frac{x-\mu}{\sigma}

Where mu is mean and sigma is standard deviation [these are the 2 parameters we are seeking]

So, 0.05 corresponds to z = -1.64, and

0.75 corresponds to z = 0.67

Now we put both of these into z-score formula and solve both equations for mu and sigma.

-1.64=\frac{10-\mu}{\sigma}

and

0.67=\frac{13-\mu}{\sigma}

The first equation becomes:

-1.64=\frac{10-\mu}{\sigma}\\-1.64\sigma+\mu=10\\mu=10+1.64\sigma\\

Now, simplifying 2nd equation and putting this in:

0.67=\frac{13-\mu}{\sigma}\\0.67\sigma=13-(10+1.64\sigma)\\0.67\sigma=13-10-1.64\sigma\\2.31\sigma=3\\\sigma=1.2987

Now finding mu:

\mu=10+1.64(1.2987)\\\mu=12.1299

These two MU(mean) and SIGMA(standard deviation) are the 2 parameters.

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In-s [12.5K]

Given:

|x-4(72)|=2

To find:

The highest and lowest scores Sam could have made in the tournament.

Solution:

We have,

|x-4(72)|=2

|x-288|=2

It can be written as

x-288=\pm 2

Add 288 on both sides.

x=288\pm 2

x=288-2 and x=288+2

x=286 and x=290

Therefore, the highest and lowest scores Sam could have made in the tournament are 290 and 286 respectively.

8 0
2 years ago
F(x)=3x 2 +9f, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 9 and g(x)=\dfrac{1}{3}x^2-9g(x)= 3 1 ​ x 2
34kurt

Answer:

f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252

g(f(x)) = 3x^4 + 18x^2 + 18

<em>f(x) and g(x) and not inverse functions</em>

Step-by-step explanation:

Given

f(x) = 3x^2 + 9

g(x) = \dfrac{1}{3}x^2 - 9

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

f(x) = 3x^2 + 9

f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9

f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9

Expand Brackets

f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9

f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9

f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9

f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252

Calculating g(f(x))

g(x) = \dfrac{1}{3}x^2 - 9

g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9

g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9

g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9

Expand Brackets

g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9

g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9

g(f(x)) = 3x^4 + 18x^2 + 18

Checking for inverse functions

f(x) = 3x^2 + 9

Represent f(x) with y

y = 3x^2 + 9

Swap positions of x and y

x = 3y^2 + 9

Subtract 9 from both sides

x - 9 = 3y^2 + 9 - 9

x - 9 = 3y^2

3y^2 = x - 9

Divide through by 3

\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}

y^2 = \frac{x}{3} - 3

Take square root of both sides

\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}

y = \sqrt{\frac{x}{3} - 3}

Represent y with g(x)

g(x) = \sqrt{\frac{x}{3} - 3}

Note that the resulting value of g(x) is not the same as g(x) = \dfrac{1}{3}x^2 - 9

<em>Hence, f(x) and g(x) and not inverse functions</em>

4 0
2 years ago
A contractor has submitted bids on three state jobs: an office building, a theater, and a parking garage. State rules do not all
german

Answer:

The following are the answer to this question:

Step-by-step explanation:

In the given question the numeric value is missing which is defined in the attached file please fine it.  

Calculating the probability of the distribution for x:

\to f(x) = 0.19\  for \ x=14\\\\\to  f(x) = 0.29 \ for\ x=7\\\\\to f(x) = 0.38\  for \ x=1\\\\\to f(x)=0.14 \ for \ x=0\\

The formula for calculating the mean value:

\bold{ E(X)= x \times f(x)}

          =14 \times 0.19+7 \times 0.29+1 \times 0.38+0\times 0.14\\\\=2.66 + 2.03+0.38+ 0\\\\=5.07

\bold{E(X^2) = x^2 \times f(x)}

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use formula for calculating the Variance:

\to \bold{\text{Variance}= E(X^2) -[E(X)]^2}

                  = 51.83 - (5.07)^2\\\\= 51.83 -  25.70\\\\=26.13

calculating the value of standard deivation:

Standard Deivation (SD) = \sqrt{Variance}

                                          = \sqrt{26.13} \\\\=5.111

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