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3 years ago

I need help plz math

Mathematics

2 answers:

My name is Ann [436]3 years ago

7 0

Answer:

D/40

Step-by-step explanation:

The answer is D because if you were to put them in the correct order it would be

21, 40, 52, 58, 72, 75, 96

Q1 Q2 Q3

Q1 is quartile 1 so the answer is 40

Hope this helps!

QveST [7]3 years ago

3 0

Answer:

Q1 = 40

Step-by-step explanation:

Set up numbers from least to greatest

21, 40, 52, 58, 72, 75, 96

First, find the median,

Median is: 58

Now Q1 is all numbers before the median (without counting the median)

21, 40, 52

The median here is 40

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3 years ago

Read 2 more answers

Which polynomial correctly combines the like term and express the given polynomial in standard form

ankoles [38]

Answer:

$8mn^5 - 2m^6 + 5m^2n^4 - m^3n^3 + n^6 - 4m^6 + 9m^2n^4 - mn^5 - 4m^3n^3 = -6m^6 - mn^5 - 5m^3n^3 + 14m^2n^4 + 8mn^5 + n^6$

Step-by-step explanation:

Given

$8mn^5 - 2m^6 + 5m^2n^4 - m^3n^3 + n^6 - 4m^6 + 9m^2n^4 - mn^5 - 4m^3n^3$

Required

Correctly combine the like terms

We have:

$8mn^5 - 2m^6 + 5m^2n^4 - m^3n^3 + n^6 - 4m^6 + 9m^2n^4 - mn^5 - 4m^3n^3$

Collect like terms (in decreasing degrees of m)

$- 2m^6 - 4m^6 - mn^5- m^3n^3 - 4m^3n^3+ 5m^2n^4 + 9m^2n^4 + 8mn^5 + n^6$

$-6m^6 - mn^5 - 5m^3n^3 + 14m^2n^4 + 8mn^5 + n^6$

So:

$8mn^5 - 2m^6 + 5m^2n^4 - m^3n^3 + n^6 - 4m^6 + 9m^2n^4 - mn^5 - 4m^3n^3 = -6m^6 - mn^5 - 5m^3n^3 + 14m^2n^4 + 8mn^5 + n^6$

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3 years ago

if k is a half-plane determined by QR, then for every real number 0 < x < or = to 180, there is exactly one ray, QQP, that

zloy xaker [14]

Let me start from the meaning of half plane.you must have learned that there are two axes X and Y axes which includes positive X axes and negative X axes and there is positive Y axes and negative y axes. So there are four quadrants which i have explained in the figure. I have explained the meaning of Half plane in the Diagram also. Half plane=Figure obtained by covering any two adjacent regions of the coordinate axes.So we get four half planes.

Now your question is k is a half plane determined by QR, such that 0<x≤180 and there is exactly one ray QRP that lies in k

So, $0< \angle PQR \leq 180^\cir{0}$

8 0

3 years ago

Find all values of k for which the function y=sin(kt) satisfies the differential equation y′′+16y=0. Separate your answers by co

VashaNatasha [74]

Answer: k = 4, k = -4 and k = 0.

Step-by-step explanation:

If we have y = sin(kt)

then:

y' = k*cos(kt)

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then, if we have the relation:

y'' - y = 0

we can replace it by the things we derivated previously and get:

-k^2*sin(kt) + 16*sin(kt) = 0

we can divide by sin in both sides (for t ≠0 and k ≠0 because we can not divide by zero)

-k^2 + 16 = 0

the solutions are k = 4 and k = -4.

Now, we have another solution, but it is a trivial one that actually does not give any information, but for the diff equation:

-k^2*sin(kt) + 16*sin(kt) = 0

if we take k = 0, we have:

-0 + 0 = 0.

So the solutions are k = 4, k = -4 and k = 0.

4 0

3 years ago