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Nat2105 [25]
3 years ago
8

The scores on a quiz are normally distributed. The mean of the quiz is 93 and the standard deviation is 4.2. By using the Empiri

cal rule, what scores fall 1 standard deviation from the mean?
89 and 101

84.6 and 101.4

89.2 and 96.8

88.8 and 97.2
Mathematics
1 answer:
xz_007 [3.2K]3 years ago
6 0

Answer:

any score that lies between 88.8 and 97.2 is within one std. dev. of the mean

Step-by-step explanation:

One std. dev. above the mean would be 93 + 4.2, or 97.2.  One std. dev. below the mean would be 93 - 4.2, or 88.8.

So:  any score that lies between 88.8 and 97.2 is within one std. dev. of the mean.

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Simplify: 40x7- (-12x7)<br> Enter the original expression if it cannot be<br> simplified.
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Answer:

52x^7

Step-by-step explanation:

40x^7 -(-12x^7) =

= 40x^7 + 12x^7 =

= 52x^7

Hope this helps!

If not, I am sorry.

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Mom bought 8 apples we ate 1/4 of them. How many did we eat?
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Answer:

you ate 2

Step-by-step explanation:

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What is the approximate volume of a cone with a height of 11 ft and radius of 3 ft?
nadya68 [22]
Substitute the values of the radius <span>(<span><span>r=3</span>),</span></span><span> the </span>height <span>(<span><span>h=11</span>),</span></span><span> and an approximation of </span>Pi <span>(<span>3.14)</span></span><span> into the </span>formula<span> to find the </span>volume<span> of the </span>cone<span>.
</span><span>V≈+<span>1/3</span>⋅3.14⋅<span>3^2</span>⋅11

</span><span>Simplify.
</span>Write 3.14<span> as a </span>fraction<span> with </span>denominator <span>1.</span><span> 
</span><span>V≈+<span>1/3</span>⋅<span>3.14/1</span></span> <span>

</span>Combine <span>1/3</span><span> and </span><span>3.14/1</span><span> to get </span><span><span>3.14/3</span>.</span><span> 
</span><span>V≈+<span>3.14/3</span></span><span> 
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Replace back in to larger expression<span>. 
</span><span>V≈+<span>3.14/3</span>⋅<span>3^2</span>⋅11</span><span> 
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Divide 3.14<span> by </span>3<span> to get </span><span>1.04666667.
</span><span>V≈+1.04666667⋅<span>3^2</span>⋅11

</span>Raise 3<span> to the </span>power<span> of </span>2<span> to get </span><span>9.
</span><span>V≈+1.04666667⋅9⋅11

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</span><span>V≈+103.62

</span>Remove trailing zeros<span> from </span><span>103.62.
</span><span>V≈+103.62</span><span>f<span>t^3

Volume:
</span></span>V≈+103.62ft^3
8 0
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