All categories

3 years ago

Can someone find the median? I got 177 divided by 2 but not sure

Mathematics

1 answer:

Lostsunrise [7]3 years ago

8 0

Greetings!

The Median refers to the middle of the set of number! So line up the numbers from largest to smallest or vice versa:
97,94,89,88,88,83

The middle number(s) is/are:
89 and 88.

Now we can use the same technique to find the mode, using the two number:
$=88+89$

$=177$

Now divide the sum by the number of numbers (2):
$= \frac{177}{2}$

$=88.5$

The Median is:
$\left[\begin{array}{ccc}88.5\end{array}\right]$

Hope this helps!
-Benjamin

You might be interested in

The ratio of boys to girls in the sixth grade is 2:3. If there are 24 boys, how many girls are there?

Yuri [45]

36 girls because 24/2=12 and 12•3=36

4 0

3 years ago

Read 2 more answers

Complete the identity.<br> 1) sec^4 x + sec^2 x tan^2 x - 2 tan^4 x = ?

Alecsey [184]

Answer:

See Explanation

Step-by-step explanation:

<em>Question like this are better answered if there are list of options; However, I'll simplify as far as the expression can be simplified</em>

Given

$sec^4 x + sec^2 x tan^2 x - 2 tan^4 x$

Required

Simplify

$(sec^2 x)^2 + sec^2 x tan^2 x - 2 (tan^2 x)^2$

Represent $sec^2x$ with a

Represent $tan^2x$ with b

The expression becomes

$a^2 + ab- 2 b^2$

Factorize

$a^2 + 2ab -ab- 2 b^2$

$a(a + 2b) -b(a+ 2 b)$

$(a -b) (a+ 2 b)$

Recall that

$a = sec^2x$

$b = tan^2x$

The expression $(a -b) (a+ 2 b)$ becomes

$(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

..............................................................................................................................

In trigonometry

$sec^2x =1 +tan^2x$

Subtract $tan^2x$ from both sides

$sec^2x - tan^2x =1 +tan^2x - tan^2x$

$sec^2x - tan^2x =1$

..............................................................................................................................

Substitute 1 for $sec^2x - tan^2x$ in $(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

$(1) (sec^2x+ 2 tan^2x)$

Open Bracket

$sec^2x+ 2 tan^2x$ ------------------This is an equivalence

$(secx)^2+ 2 (tanx)^2$

Solving further;

................................................................................................................................

In trigonometry

$secx = \frac{1}{cosx}$

$tanx = \frac{sinx}{cosx}$

Substitute the expressions for secx and tanx

................................................................................................................................

$(secx)^2+ 2 (tanx)^2$ becomes

$(\frac{1}{cosx})^2+ 2 (\frac{sinx}{cosx})^2$

Open bracket

$\frac{1}{cos^2x}+ 2 (\frac{sin^2x}{cos^2x})$

$\frac{1}{cos^2x}+ \frac{2sin^2x}{cos^2x}$

Add Fraction

$\frac{1 + 2sin^2x}{cos^2x}$ ------------------------ This is another equivalence

................................................................................................................................

In trigonometry

$sin^2x + cos^2x= 1$

Make $sin^2x$ the subject of formula

$sin^2x= 1 - cos^2x$

................................................................................................................................

Substitute the expressions for $1 - cos^2x$ for $sin^2x$

$\frac{1 + 2(1 - cos^2x)}{cos^2x}$

Open bracket

$\frac{1 + 2 - 2cos^2x}{cos^2x}$

$\frac{3 - 2cos^2x}{cos^2x}$ ---------------------- This is another equivalence

8 0

3 years ago

I need help with this

Sloan [31]

Answer:

slope = - $\frac{4}{3}$ , point on line = (4, - 3 )

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b ) a point on the line

y + 3 = - $\frac{4}{3}$ (x - 4) ← is in point- slope form

with slope m = - $\frac{4}{3}$ and point on line (a, b ) = (4, - 3 )

3 0

2 years ago

4 (negative one-half x + 7) + 5 x = 15

Ronch [10]

Answer:

Step 3 is incorrect

Step-by-step explanation:

4(-1/2x + 7) + 5x = 15

-2x + 28 + 5x = 15

3x + 28 = 15

3x + 28 - 28 = 15 - 28

3x = -13

4 0

3 years ago

X+9=13352643-2x answer get brainliest

Ymorist [56]

Answer:

4450878

Step-by-step explanation:

7 0

3 years ago