Answer:
A) 68.33%
B) (234, 298)
Step-by-step explanation:
We have that the mean is 266 days (m) and the standard deviation is 16 days (sd), so we are asked:
A. P (250 x < 282)
P ((x1 - m) / sd < x < (x2 - m) / sd)
P ((250 - 266) / 16 < x < (282 - 266) / 16)
P (- 1 < z < 1)
P (z < 1) - P (-1 < z)
If we look in the normal distribution table we have to:
P (-1 < z) = 0.1587
P (z < 1) = 0.8413
replacing
0.8413 - 0.1587 = 0.6833
The percentage of pregnancies last between 250 and 282 days is 68.33%
B. We apply the experimental formula of 68-95-99.7
For middle 95% it is:
(m - 2 * sd, m + 2 * sd)
Thus,
m - 2 * sd <x <m + 2 * sd
we replace
266 - 2 * 16 <x <266 + 2 * 16
234 <x <298
That is, the interval would be (234, 298)
Let x be the unknown number. The equation is going to be 7/28=25/x, now we cross multiply. 7*x=25*28, 7x=700, divide both sides by 7 and you should get 100, So, x=100 and 7/28=25/100
-8.1
Step-by-step explanation:

<em>Times </em><em>all </em><em>by </em><em>9</em><em> </em><em>to </em><em>get </em><em>rid </em><em>of </em><em>fraction</em>
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<em>Take </em><em>2</em><em>1</em><em>.</em><em>6</em><em> </em><em>away</em><em> </em><em>from</em><em> </em><em>both</em><em> </em><em>sides</em>
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<em>Divide</em><em> </em><em>by </em><em>-</em><em>4</em>
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