Answer:
We reject H₀
we accept Hₐ seeds in the packet would germinate smaller than 93%
Step-by-step explanation:
Test of proportions
One tail-test (left side)
93 % = 0.93
p₀ = 0,93
1.- Hypothesis
<h3>
H₀ ⇒ null hypothesis p₀ = 0.93</h3><h3>
Hₐ ⇒ Alternative hypothesis p = 0.875</h3><h3>
2.-Confidence interval 95 %</h3><h3>
α = 0,05 </h3><h3>
and </h3><h3>
z(c) = - 1.64</h3><h3>
3.- Compute z(s)</h3><h3>
z(s) = (p - p₀)/√(p₀*q₀)/n z(s) = (0.875-0.93)/√0.93*0.07)200</h3><h3>
z(s) = - 0,055/ √0.0003255</h3><h3>
z(s) = - 0.055/ 0.018</h3><h3>
z(s) = - 3,06</h3><h3>
4.-Compere z(c) and z(s)</h3><h3>
z(s) < z(c) -3.06 < -1.64</h3><h3>
z(s) is in rejection region, we reject H₀</h3>
Answer:
Cathedrals
Step-by-step explanation: Trust my brain i studied this to
Answer:
The probability that a randomly selected student didn’t take economics but did take statistics is 13%.
Step-by-step explanation:
Let the event that a student offers Economics be E.
The event that a student does NOT offer Economics is E'.
Let the event that a student offers Statistics be S.
The event that a student does NOT offer Statistics be S'.
P(E) = 13.5% = 0.135
P(S) = 24.7% = 0.247
P(E n S) = 11.7% = 0.117
Find the probability that a randomly selected student didn’t take economics but did take statistics
This probability = P(E' n S)
Since E and E' are mutually exclusive events,
P(S) = P(E' n S) + P(E n S)
P(E' n S) = P(S) - P(E n S)
P(E' n S) = 0.247 - 0.117 = 0.13 = 13%
Hope this Helps!!!
Answer:
Step-by-step explanation:
The zeroes are -3,2, and -7. You have to solve each individual factor for zero.
