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Anastaziya [24]
3 years ago
11

PLEASE ANSWER ASAP

Mathematics
2 answers:
lakkis [162]3 years ago
8 0
Putting these words down so you can mark that guy brainliest.

Your welcome, that guy:)

That sounded really weird.

IGNORE
ivolga24 [154]3 years ago
6 0
The rate of change, or slope, is the change in y over change in x. To find the slope, pick two points. The slope will be the change in y-coordinates over the change in x-coordinates.
As an equation, this is
slope =  \frac{(y2-y1)}{(x2-x1)}

To find the slope for function 1, pick any two points (x,y) in the table.
Let's choose (-1, 3) and (-2, 5).
The slope is 
\frac{5-3}{-2+1} =  \frac{2}{-1} = -2.

Let's choose (-1, 0) and (0, 4) for the two points for function 2.
The slope is
\frac{4-0}{0+1} =  \frac{4}{1} = 4.

You want to compare the magnitude (without negatives)  of the slope when comparing the greater rate of change/slope. In other words, you take the absolute value of each slope.
The magnitude of function 1's slope is 2.
The magnitude of function 2's slope is 4.
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The ages of armadillos are normally distributed, with a mean of 14 years and a standard deviation of 1.2. Approximately what per
vovangra [49]

Answer:

Percentage of armadillos between 13 and 17 years = 79.052%f using Standard Normal Distribution Tables

Step-by-step explanation:

As we know from normal distribution: z(x) = (x - Mu)/SD

where x = targeted value; Mu = Mean of Normal Distribution; SD = Standard Deviation of Normal Distribution

Therefore using given data: Mu = 14, SD = 1.2 we have z(x) by using z(x) = (x - Mu)/SD as under:

Approach 1 using Standard Normal Distribution Table:

z for x=17: z(17) = (17-14)/1.2 gives us z(17) = 2.5

z for x=13: z(13) = (13-14)/1.2 gives us z(13) = -0.83

Afterwards using Normal Distribution Tables we find the probabilities as under:

P(17) using z(17) = 2.5 gives us P(17) = 99.379%

Similarly we have:

P(13) using z(13) = -0.83 gives us P(13) = 20.327%

Finally in order to find out the probability between 17 & 13 years we have:

Percentage of armadillos between 13 and 17 years = P(17) - P(13) = 99.379% - 20.327% = 79.052%

The standard normal distribution table is being attached for yours easiness.

Approach 2 using Excel or Google Sheets:

P(17) = norm.dist(17,14,1.2,1)

P(13) = norm.dist(13,14,1.2,1)

Percentage of armadillos between 13 and 17 years = { P(17) - P(13) } * 100

Download pdf
4 0
3 years ago
Read 2 more answers
The focus of a parabola is located at (0,–2). The directrix of the parabola is represented by y = 2. Which equation represents t
Archy [21]

Answer:

\boxed{x^2=-8y}

Step-by-step explanation:

(a,b)\ is \ the\ focus\\y=k\ is \ the\ directrice\\\\Formula\ to\ use:\ y=\dfrac{(x-a)^2}{2(b-k)} +\dfrac{b+k}{2} \\\\Here: \ a=0,\ b=-2, \ k=2\\\\y=\dfrac{(x-0)^2}{2(-2-2)} +\dfrac{-2+2}{2} \\\\so\ y=-\dfrac{x^2}{8} \\\\or\ x^2=-8y

3 0
3 years ago
An analyst must decide between two different forecasting techniques for weekly sales of roller blades: a linear trend equation a
Amiraneli [1.4K]

Answer:

Following are the responses to the given question:

Step-by-step explanation:

Mean Absolute Deviation MAD

MAD = \sum_{i=1}^{n} \frac{\left | actual_{i}-forecast_{i} \right |}{n}

Mean squared error  

MSE=\Sigma^{n}_{i=1} \frac{(Actual_i - Forecast_i)^2}{(n - 1)}

linear trend equation is Ft = 124 + 2t

MAD = \frac{23}{10} = 2.3\\\\MSE = \frac{91}{9}= 10.11\\\\

Naive method

MAD = \frac{36}{9} = 4\\\\MSE = \frac{202}{8}=25.25\\\\MAD (Naive) = 4\\\\MAD (Linear) =2.3\\\\MSE (Naive)=25.25\\\\MSE (Linear)=10.11\\\\

Please find the attached file.

7 0
3 years ago
What = variable
prisoha [69]
74.4 is the answer.
hope this helps.
6 0
3 years ago
Read 2 more answers
Assume that the weight of two year old babies have distribution that is approximately normal with a mean of 29 pounds and a stan
Lana71 [14]

Answer:

25.15 ponds is the weight of two year old baby corresponds to 10th percentile.      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 29 pounds

Standard Deviation, σ = 3 pounds

We are given that the distribution of weight of two year old babies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.10

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 29}{3})=0.10  

Calculation the value from standard normal z table, we have,  

P(z < -1.282) = 0.10

\displaystyle\dfrac{x - 29}{3} = -1.282\\x = 25.154 \approx 25.15

Thus, 25.15 ponds is the weight of two year old baby corresponds to 10th percentile.

5 0
3 years ago
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