Answer:
32 and 56?
Step-by-step explanation:
32 and 56 are related to 8.
32=8x4
56=8x7
The answer is option B, if this helped mark me the brainiest!!
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:
True
Step-by-step explanation:
In order to calculate this, we simply need to replace the variable c with the value of -8 and solve the inequality.
-9c >= 63
-9 * -8 >= 63
72 >= 63
Since both values are negative the product will be positive. This product is 72 which is in fact greater than 63, thus making the statement True
Answer:
Step-by-step explanation:
<u>Linear Momentum
</u>
The momentum of a system of masses m1,m2,m3... with speeds v1,v2,v3,... is given by
When some interactions take place into the system of masses with no external forces interferring, the total momentum is not changed, which means that if the new speeds are v1', v2', v3'... then:
The problem relates the story of Batman (m1=100kg) who is initially assumed at rest and lands on a boat with mass m2=580 kg initially moving at 14 m/s to the positive reference. When the collision takes place, both masses join and a common speed 
is achieved by the common mass. Applying the conservation of momentum, we have
Since 
Solving for
