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ValentinkaMS [17]

3 years ago

13

What is the difference between an equation and a function

1 answer:

natta225 [31]3 years ago

5 0

Answer:

Equation has an equal sign

Function has a variable

Step-by-step explanation:

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How do you solve this by grouping x^3-3x^2+3x-9=0

frozen [14]

Answer:

x = 3

x = ± √3 = ± 1.7321

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How do you solve number 13 and 14?

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Solve each equation using the quadratic formula. Find the solutions.<br><br> x^2 - 6x + 7 = 0

LuckyWell [14K]

X^2-6x+7=0

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4 years ago

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The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas

Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is $= \frac{(n+1)}{2}$ .

$\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5$ position orders

$10^{th} \ and \ 11^{th}$ place an average of observations so, the

$Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430$

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

$Q_1$ often falls throughout the ordered BELOW average is $= \frac{(n+1)}{2}$ place.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered position

Average $5^{th} \ and \ 6^{th}$ location findings

$Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5$

In $= \frac{(n+1)}{2}$ the ordered place, Q3 always falls ABOVE the median.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered At median ABOVE

Consequently $Q_3$ falls between $5^{th} \ and \ 6^{th}$ ABOVE the median position

$15^{th}\ and \ 16^{th}$ place average of observations

$\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2} = 502$

$\to IQR = Q_3 - Q_1= 502-358.5= 143.5$

$\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25$

4 0

3 years ago

Ierofanga [76]

Answer:

A

Step-by-step explanation:

Sum the product of the components in the first row of A with the corresponding components of the first column in B

Repeat this with the components in the second row of A with the corresponding components of the second column in B, that is

AB

= $\left[\begin{array}{ccc}2&1\\3&4\\\end{array}\right]$ $\left[\begin{array}{ccc}3&1\\5&2\\\end{array}\right]$

= $\left[\begin{array}{ccc}2(3)+1(5)&2(1)+1(2)\\3(3)+4(5)&3(1)+4(2)\\\end{array}\right]$

= $\left[\begin{array}{ccc}6+5&2+2\\9+20&3+8\\\end{array}\right]$

= $\left[\begin{array}{ccc}11&4\\29&11\\\end{array}\right]$ → A

6 0

3 years ago