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Snezhnost [94]
3 years ago
15

The difference of y and 9 is at least 24 .

Mathematics
1 answer:
polet [3.4K]3 years ago
7 0
Show a photo so we know what your talking Abuja
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Pls answer this with a picture showing the correct box plot
faust18 [17]

Answer/Step-by-step explanation:

To find out the mistake of the student, let's find the min, max, median, Q1 and Q3, which make up the 5 important values that are represented in a box plot.

Given, {2, 3, 5, 6, 10, 14, 15},

Minimum value = 2

Median = middle data point = 6

Q1 = 3 (the middle value of the lower part of the data set before the median)

Q3 = 14 (middle value of the upper part of the data set after the median)

Maximum value = 15

If we examine the diagram the student created, you will observe that he plotted the median wrongly. The median, which is represented by the vertical line that divides the box, ought to be at 6 NOT 10.

See the attachment below for the correct box plot.

3 0
3 years ago
prove that sin theta cos theta = cot theta is not a trigonometric identity by producing a counterexample
Zolol [24]

Answer:

To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.

Begin with the right hand side:

R.H.S = cot θ = \frac{cos \ \theta}{sin \ \theta}

L.H.S = sin θ cos θ

so, sin θ cos θ ≠ \frac{cos \ \theta}{sin \ \theta}

So, the equation is not a trigonometric identity.

=========================================================

<u>Anther solution:</u>

To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.

Assume θ with a value and substitute with it.

Let θ = 45°

So, L.H.S = sin θ cos θ = sin 45° cos 45° = (1/√2) * (1/√2) = 1/2

R.H.S = cot θ = cot 45 = 1

So, L.H.S ≠ R.H.S

So, sin θ cos θ = cot θ is not a trigonometric identity.

5 0
3 years ago
Solve the equation <br><br> 3x-5&lt; 8x-1/2<br><br> x+7≤ 4x+1
vladimir1956 [14]
This is all I got. Hope this helps:

6 0
1 year ago
Read 2 more answers
A. Incorrect; the student should add 8 + 2 to calculate the x-coordinate and should add 5 + 9 to calculate the y-coordinate. B.
likoan [24]

Answer:

A

Step-by-step explanation:

I think it’s a based on what you given

3 0
3 years ago
. Use the quadratic formula to solve each quadratic real equation. Round
Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}, with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}

\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5

So B has two solutions of 5 and -1.5.

Now to C!

\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}

\frac{44}{8} = 5.5

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0
3 years ago
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