Ask question

Ask question

All categories

3 years ago

12

What is the median of the data set represented by the dot plot?

2 answers:

vladimir2022 [97]3 years ago

7 0

The answer is 5

hope this helps

Butoxors [25]3 years ago

6 0

So ur numbers are : 3,3,4,5,5,5,6,8,8

and the median (middle number) is 5 <==

because 2 dots above the 3 means u have 2 threes....and 1 dot above the 4 means there is 1 four..and so on

You might be interested in

Five less than two times a number is sixty-nine. A. 5 – 2t = 69 t = –32 B. 2t – 5 = 69 t = 37 C. 5 + 2t = 69 t = 32 D. none of t

Julli [10]

Let's translate the expression step by step:

"Five less than..." $\to$ we will have to subtract five from something

"two times a number" $\to 2t$ .

So, we have

"Five less than two times a number" $\to 2t-5$ .

This "is sixty-nine", where "is" means "is equal to". So, the equation is

$2t-5=69$

which is option B

5 0

3 years ago

I need help with the second once and can someone tell me if the first answer I did is ok or good

Ira Lisetskai [31]

Answer:

Sorry i dont know

Step-by-step explanation:

6 0

3 years ago

HURRY WILL MARK THE BRAINLIEST Al uses the box in the shape of a rectangular prism as shown in the diagram to for his coin coll

Digiron [165]

425 inches, Multiply all the numbers and you get the answer.

7 0

3 years ago

She used 25 equally sized stones to make a path that is 12 feet long. How long is each stone

MAVERICK [17]

Answer:

The length of each stone is

L = 12/25 = 0.48 ft

Hope this helps!

:)

8 0

3 years ago

The expression (secx + tanx)2 is the same as _____.

trapecia [35]

<u>Answer:</u>

The expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

<u>Solution:</u>

From question, given that $\bold{(\sec x+\tan x)^{2}}$

By using the trigonometric identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$ the above equation becomes,

$(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x$

We know that $\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}$

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

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

On simplication we get

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

By using the trigonometric identity $\cos ^{2} x=1-\sin ^{2} x$ ,the above equation becomes

$=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}$

By using the trigonometric identity $(a+b)^{2}=a^{2}+2ab+b^{2}$

we get $1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}$

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

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

By using the trigonometric identity $a^{2}-b^{2}=(a+b)(a-b)$ we get $1-\sin ^{2} x=(1+\sin x)(1-\sin x)$

$=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}$

$= \frac{1+\sin x}{1-\sin x}$

Hence the expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

8 0

3 years ago