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Igoryamba
3 years ago
14

State the degree: 11m^3n^2p

Mathematics
2 answers:
Vadim26 [7]3 years ago
7 0
<h3>Answer:  6</h3>

Explanation:

Using the rule that x = x^1, we can rewrite the p as p^1

So 11m^3n^2p is the same as 11m^3n^2p^1

The exponents are: 3, 2, 1

Those exponents add up to 3+2+1 = 6

The degree of a monomial like this is simply equal to the sum of the exponents.

Licemer1 [7]3 years ago
7 0

Answer:

6

Step-by-step explanation:

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In all, 1000 students took a statistics exam worth 150 points. The first quartile (or Q1) for all 1000 scores is 90 points. Abou
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Answer:

The number of students who scored more than 90 points is 750.

Step-by-step explanation:

Quartiles are statistical measures that the divide the data into four groups.

The first quartile (Q₁) indicates that 25% of the observation are less than or equal to Q₁.

The second quartile (Q₂) indicates that 50% of the observation are less than or equal to Q₂.

The third quartile (Q₃) indicates that 75% of the observation are less than or equal to Q₃.

It is provided that the first quartile is at 90 points.

That is, P (X ≤ 90) = 0.25.

The probability that a student scores more than 90 points is:

P (X > 90) = 1 - P (X ≤ 90)

                = 1 - 0.25

                = 0.75

The number of students who scored more than 90 points is: 1000 × 0.75 = 750.

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if you see on the screenshot if you draw a line between the 2 it mesures 9.4

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<h2>1)</h2>

(x - 4) {}^{2}  - 28 = 8 \\ (x - 4) {}^{2}  = 8 + 28 \\ (x - 4) {}^{2}  = 36

This must be true for some value of x, since we have a quantity squared yielding a positive number, and since the equation is of second degree,there must exist 2 real roots.

\sqrt{(x - 4) {}^{2} }  =  ± \sqrt{36}  \\x - 4 = ±6 \\ x _{1}- 4 = 6 \:  \:  \:  \:  \:  \:  \: \:  \:  \:  x _{2}- 4 =  - 6 \\ x_{1} = 6 + 4 \:  \:  \:  \:  \:  \:  \:  \:  \:  \: x_{2} =  - 6 + 4 \\ x_{1} = 10 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \: \:  \: x_{2} =  - 2

<h2>2)</h2>

Well he started off correct to the point of completing the square.

(x - 3)  {}^{2}  = 16 \\ x - 3 = ±4 \\  \: x_{1}  - 3 = 4 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: x_{2}  - 3 =  - 4 \\ x_{1}  = 7 \:  \:  \:  \:  \:  \:  \:  \:  \:  \: x_{2}  =  - 1

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1 year ago
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