2 out of 50 are defective.
Divide 2 by 50: 2/50 = 0.04
Now multiply the total quantity by that:
2000 x 0.04 = 80
80 are likely to be defective.
You have to draw the numbers then you would have to divide them draw it
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)
Johnny is selling tickets to a school play. On the first day of ticket sales he sold 14 senior (S) citizen tickets and 4 child (C) tickets for a total of $200. On the second day of ticket sales he sold 7 senior (S) citizen tickets and 1 child (C) ticket for a total of $92. What is the price of one child ticket?
14S + 4C = 200
14S = 200 - 4C
S = (200 - 4C)/14
7S + 1C = 92
7S = 92 - C
S = (92 - C)/7
(200 - 4C)/14 = (92 - C)/7
7 x (200 - 4C) = 14 x (92 - C)
1400 - 28C = 1288 - 14C
1400 - 1288 = 28C - 14C
112 = 14C
C = 112/14 = 8
the price of one child ticket = $8
I think the answer would be x= 4/7 y = +-16/7