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3 years ago

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The high temperatures at a park for 10 days are 40, 42, 41, 41, 43, 43, 44, 45, 44, and 45 degrees Fahrenheit. What is the inter

quartile range of the data?

2 answers:

Ksenya-84 [330]3 years ago

5 0

Answer:

The interquartile range of the data is 3.

Step-by-step explanation:

Here, the given data is,

40, 42, 41, 41, 43, 43, 44, 45, 44, 45,

Step 1 : The ascending order of the data,

40, 41, 41, 42, 43, 43, 44, 44, 45, 45,

Step 2 : Median of the data,

40, 41, 41, 42, 43, 43, 44, 44, 45, 45

Step 3 : Make a mark in the center of the data,

40, 41, 41, 42, 43, | 43, 44, 44, 45, 45

Step 4 : Make lower half and upper half of the data,

(40, 41, 41, 42, 43) | (43, 44, 44, 45, 45)

Step 5 : Median of lower half,

$Q_1$ = 41,

Median of upper half,

$Q_3$ = 44

Step 6 : Thus, the interquartile range of the data = $Q_3-Q_1$

= 44 - 41

= 3

aleksklad [387]3 years ago

4 0

Range in a distribution is the difference between the smallest and the largest value in a distribution. Interquartile range is the difference between the upper quartile and the lower quartile in a distribution.
In an ungrouped data , the first step is to arrange the values in ascending order;
40,41,41,42,43,43,44,44,45,45
The lower quartile is 41 while the upper quartile is 44
Thus the interquartile range is 3 (44-41)

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2 years ago

Consider the parabola given by the equation: f(x) = 4x² - 6x - 8 Find the following for this parabola: A) The vertex: Preview B)

jeyben [28]

Answer:

The vertex: $(\frac{3}{4},-\frac{41}{4} )$

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Step-by-step explanation:

To find the vertex of the parabola $4x^2-6x-8$ you need to:

1. Find the coefficients <em>a</em>, <em>b</em>, and <em>c </em>of the parabola equation

<em> $a=4, b=-6, \:and \:c=-8$ </em>

2. You can apply this formula to find x-coordinate of the vertex

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3. To find the y-coordinate of the vertex you use the parabola equation and x-coordinate of the vertex ( $f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c$ )

$f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c\\f(\frac{3}{4})=4\cdot (\frac{3}{4})^2-6\cdot (\frac{3}{4})-8\\y=\frac{-41}{4}$

To find the vertical intercept you need to evaluate x = 0 into the parabola equation

$f(x)=4x^2-6x-8\\f(0)=4(0)^2-6\cdot 0-0\\f(0)=-8$

To find the coordinates of the two intercepts of the parabola you need to solve the parabola by completing the square

$\mathrm{Add\:}8\mathrm{\:to\:both\:sides}$

$x^2-6x-8+8=0+8$

$\mathrm{Simplify}$

$4x^2-6x=8$

$\mathrm{Divide\:both\:sides\:by\:}4$

$\frac{4x^2-6x}{4}=\frac{8}{4}\\x^2-\frac{3x}{2}=2$

$\mathrm{Write\:equation\:in\:the\:form:\:\:}x^2+2ax+a^2=\left(x+a\right)^2$

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=2+\left(-\frac{3}{4}\right)^2\\x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{41}{16}$

$\left(x-\frac{3}{4}\right)^2=\frac{41}{16}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

$x_1=\frac{\sqrt{41}+3}{4},\:x_2=\frac{-\sqrt{41}+3}{4}$

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3 years ago

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Greeley [361]

Answer:

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Step-by-step explanation:

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Now we determine the ratio of V2 : V1 as shown below:

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2 years ago

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OLEGan [10]

Answer: B, C, And F

Step-by-step explanation:

My answer is correct because, for the negative numbber whatever is close to 1 like -1 is greater then the other negative number.

# $AnimePower$

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2 years ago