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S_A_V [24]
3 years ago
15

What is 2x times x? I keep forgetting. Please help

Mathematics
2 answers:
Stels [109]3 years ago
5 0

Answer:

2x^2

Step-by-step explanation:

2x*1x= 2x^2

PIT_PIT [208]3 years ago
3 0
2x times x equals 2x^2
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From 5 a.m. to noon, the temperature rose 15 °F to a high of 10 °F.
kodGreya [7K]

Answer:

The temperature was 5 °F

7 0
3 years ago
Suppose that the data for analysis includes the attributeage. Theagevalues for the datatuples are (in increasing order) 13, 15,
Bas_tet [7]

Answer:

a) \bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96

Median = 25

b) Mode = 25, 35

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

c) Midrange = \frac{70+13}{3}=41.5

d) Q_1 = \frac{20+21}{2} =20.5

Q_3 =\frac{35+35}{2}=35

e) Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

f) Figura attached.

g) When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

Step-by-step explanation:

For this case w ehave the following dataset given:

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.

Part a

The mean is calculated with the following formula:

\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96

The median on this case since we have 27 observations and that represent an even number would be the 14 position in the dataset ordered and we got:

Median = 25

Part b

The mode is the most repeated value on the dataset on this case would be:

Mode = 25, 35

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

Part c

The midrange is defined as:

Midrange = \frac{Max+Min}{2}

And if we replace we got:

Midrange = \frac{70+13}{3}=41.5

Part d

For the first quartile we need to work with the first 14 observations

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25

And the Q1 would be the average between the position 7 and 8 from these values, and we got:

Q_1 = \frac{20+21}{2} =20.5

And for the third quartile Q3 we need to use the last 14 observations:

25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70

And the Q3 would be the average between the position 7 and 8 from these values, and we got:

Q_3 =\frac{35+35}{2}=35

Part e

The five number summary for this case are:

Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

Part f

For this case we can use the following R code:

> x<-c(13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70)

> boxplot(x,main="boxplot for the Data")

And the result is on the figure attached. We see that the dsitribution seems to be assymetric. Right skewed with the Median<Mean

Part g

When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

6 0
3 years ago
I will send you a pic of the question... &lt;3​
RSB [31]

Answer:

kk send the pictures

Step-by-step explanation:

  • kkkkk
6 0
3 years ago
Please please please
lapo4ka [179]
The answer is y= 3/4x - 4.
6 0
3 years ago
Write an equation of the line satisfying the given conditions. Through (2, -5), parallel to 5x 6y + 7
mojhsa [17]

We have to determine the equation of the line passing through the point (2,-5) and parallel to the line 5x = 6y+7

When two lines are parallel, then the slopes of the two lines are equal.

Equation of line with point (x_1, y_1) and slope 'm' is given by:

(y-y_1) = m(x-x_1)

Since, we have to determine the equation of a line with point (2,-5).

So, the equation of the line is : (y-(-5)) = m(x-2)

y+5 = m(x-2)

Since, the line is parallel to the line 5x = 6y+ 7

So, 6y +7 = 5x

6y = 5x - 7

y = \frac{5}{6}x - \frac{7}{6}

So, slope of the line  'm' is \frac{5}{6}.

Therefore, the equation of the line is:

(y+5) = \frac{5}{6}(x-2)

6y + 30 = 5x - 10

6y - 5x = -10-30

6y - 5x = -40

Therefore, 6y - 5x = -40 is the required equation of the line.

8 0
3 years ago
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