Shoe sizes for men in the united states have a distribution that is bell shaped where = 10 and = 1.5. using the empirical rule, what percent of men have a shoe size between 7 and 13?

Step-by-step explanation:

Pictured is a normal distribution curve.  The mean is in the middle and each line going from the middle is either adding or subtracting the standard deviation.  Our mean is 10, so 10 goes in the middle, and the line to the right of 10 is 10+1.5=11.5.  The next line to the right of that is 11.5+1.5=13.  The line to the right of that is 13+1.5=14.5.

To the left of the mean we have 10-1.5=8.5.  To the left of that we have 8.5-1.5=7.  To the left of that we have 7-1.5=5.5.  You can see in the image what the percentage is within each separation.  From a size 7 to a size 13 we have the percentages 13.5+34+34+13.5=95%

In the greater realm of things, this statistic tells a shoe store manager that since 95% of men polled wear a shoe size between 7 and 13, it would be cost efficient for him to keep an abundance of these sizes on hand.  The greatest majority of men polled (68%) wear from a size 8.5 to a size 11.5.  

mark brainliest

step-by-step explanation:

2 tens + 6 ones

1 ten + 16 ones

0 tens + 26 ones

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.