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zalisa [80]
1 year ago
5

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.Refer to

Exhibit 6-3. What is the minimum weight of the middle 95% of the players?
Mathematics
1 answer:
Dafna11 [192]1 year ago
3 0

Given that

the weight of football players is distributed with a mean of 200 pounds and a standard deviation of 25 pounds.

And we need to find What is the minimum weight of the middle 95% of the players?

Explanation -

Using the Empirical Rule, 95% of the distribution will fall within 2 times of the standard deviation from the mean.

Two standard deviations = 2 x 25 pounds = 50 pounds

So the minimum weight = 200 pounds - 50 pounds = 150 pounds

Hence the final answer is 150 pounds.

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1. Add 6 to both sides.
<span><span><span><span>4h</span>−6</span>+6</span>=<span>22+6</span></span>
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</span>2.Divide both sides by 4.
<span><span><span>4h/</span>4</span>=<span>28/4
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</span></span>−8s+1=33
1: Subtract 1 from both sides.
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−8s=32
2: Divide both sides by -8.
−8s/−8=32/−8
<span>s=-4
</span><span><span><span>

−<span>4w</span></span>−4</span>=8
</span>Step 1: Add 4 to both sides.
<span><span><span><span>−<span>4w</span></span>−4</span>+4</span>=<span>8+4
</span></span><span><span>−<span>4w</span></span>=12
</span>2: Divide both sides by -4.
<span><span><span>−<span>4w/</span></span><span>−4</span></span>=<span>12/<span>−4
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<span>

Step 1: Simplify both sides of the equation.</span>
<span><span><span><span>1/7</span>b</span>+5</span>=<span>−2
</span></span> 2: Subtract 5 from both sides.
<span><span><span><span><span>1/7</span>b</span>+5</span>−5</span>=<span><span>−2</span>−5
</span></span><span><span><span>17</span>b</span>=<span>−7
</span></span>3: Multiply both sides by 7.
<span><span>7*<span>(<span><span>1/7</span>b</span>)</span></span>=<span><span>(7)</span>*<span>(<span>−7</span>)
</span></span></span><span>b=<span>−49</span></span>
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