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Neporo4naja [7]
3 years ago
10

The two dot plots show the results of rolling a die 15 times on two occasions. Comparing the variability of the dot plots, which

inference, if any, can you make about the result?

Mathematics
2 answers:
Gnesinka [82]3 years ago
5 0
I think it's "The greater variability of the second 15 rolls is more like the expected result".
allsm [11]3 years ago
5 0

Answer: C. The greater variability of the second 15 rolls is more like the expected result.

Step-by-step explanation:

From the given picture we can see two dot plots displaying the results of rolling a die 15 times on two occasions.

When we compare both the dot plots , we conclude that the first one is less variable compare to the second one or the second one more variable than the first one (as second one has dots everywhere on all 6 outcomes but in first there has only 4 outcomes).

Hence, the inference from the dot plots we concluded that :-

The greater variability of the second 15 rolls is more like the expected result.

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If f(x)=(-x) cubed, then f(-3)?
velikii [3]

Answer:

f(-3) = 27

Step-by-step explanation:

f(x) = (-x)^3

We want the value when x =-3

Substitute this in

f(-3) = (--3)^3

f(-3) = 3^3

f(-3) = 27

7 0
3 years ago
In the expansion of (1/ax +2ax^2)^5 the coefficient of x is five. Find the value of the constant a.
DedPeter [7]

Answer:

80x⁴

Step-by-step explanation:

(\frac{1}{ax} + 2ax^2)^5 = 5C_0(\frac{1}{ax})^5(2ax^2)^0 + 5C_1(\frac{1}{ax})^4(2ax^2)^1 + 5C_2(\frac{1}{ax})^3 (2ax^2)^2

                           + 5C_3 (\frac{1}{ax})^2(2ax^2)^3 + 5C_4(\frac{1}{ax})^1(2ax^2)^4 + 5C_5(\frac{1}{ax})^0(2ax^2)^5

5C_0(\frac{1}{ax})^5(2ax^2)^0  =1 \times (\frac{1}{ax})^5 \times 1 = \frac{1}{a^5x^5}\\\\5C_1(\frac{1}{ax})^4(2ax^2)^1  = 5 \times (\frac{1}{ax})^4 \times (2ax^2)^1 = 10 ax^2 \times \frac{1}{a^4x^4} = \frac{10}{a^3x^2}\\\\5C_2 (\frac{1}{ax})^3 (2ax^2)^2= 10 \times (\frac{1}{ax})^3 \times (2ax^2)^2 = 10 \times \frac{1}{a^3x^3} \times 4a^2x^4 = \frac{40x}{a}\\\\5C_3 (\frac{1}{ax})^2 (2ax^2)^3 = 10 \times (\frac{1}{ax})^2 \times (2ax^2)^3 = 10 \times \frac{1}{a^2x^2} \times 8a^3 x^6 = 80ax^4\\\\

5C_4(\frac{1}{ax})^1(2ax^2)^4 = 5 \times \frac{1}{ax} \times 16a^4x^8 = 80a^3x^7\\\\5C_5(\frac{1}{ax})^0(2ax^2)^5 = 1 \times 1 \times 32a^5x^{10}

The fourth term of the expansion has the constant a,

the coefficient of a is 80x⁴

6 0
3 years ago
Estimate 14.182 + 3.56 by first rounding each number to the nearest tenth.
Licemer1 [7]

Answer:

17.8

Step-by-step explanation:

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5*2 = 100 - 100 = 0

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What she said hope that helps
7 0
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