Answer:
a) 48.21 %
b) 45.99 %
c) 20.88 %
d) 42.07 %
e) 50 %
Note: these values represent differences between z values and the mean
Step-by-step explanation:
The test to carry out is:
Null hypothesis H₀ is μ₀ = 30
The alternative hypothesis m ≠ 30
In which we already have the value of z for each case therefore we look directly the probability in z table and carefully take into account that we had been asked for differences from the mean (0.5)
a) z = 2.1 correspond to 0.9821 but mean value is ubicated at 0.5 then we subtract 0.9821 - 0.5 and get 0.4821 or 48.21 %
b) z = -1.75 P(m) = 0.0401 That implies the probability of m being from that point p to the end of the tail, the difference between this point and the mean so 0.5 - 0.0401 = 0.4599 or 45.99 %
c) z = -.55 P(m) = 0.2912 and this value for same reason as before is 0.5 - 0.2912 = 0.2088 or 20.88 %
d) z = 1.41 P(m) = 0.9207 0.9207 -0.5 0.4207 or 42.07 %
e) z = -5.3 P(m) = 0 meaning there is not such value in z table is too small to compute and difference to mean value will be 0.5
d) z= 1.41 P(m) =
Answer:
cos h=x/h=30/g=60/x=rad3/2...y=1/2/z=1
Step-by-step explanation:
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Answer:
a) p + m = 10 and 
b) The mixing amount is 5 lb peanut and 5 lb mixture.
Step-by-step explanation:
Given that p pounds of peanuts and m pounds of the 80% almond and 20% peanut mixture are used to make 10 pounds of 60% peanuts and 40% of almonds mixture.
Now, we can write that p + m = 10 ........ (1)
Now, m pounds of 80% almond and 20% peanut mixture contain 0.2m pounds of peanuts and 0.8m pounds of almonds.
Now, from the condition given it can be written that
.......... (2)
⇒
⇒ 2p + 0.4m = 2.4m
⇒ 2p = 2m
⇒ p = m
Now from equation (1) we get p = m = 5 pounds.
a) Therefore, equations (1) and (2) are the system of equations that models the situation.
b) The mixing amount is 5 lb peanut and 5 lb mixture. (Answer)