The total amount of dividend will be:
Total=(Number of issued shares)*(dividend per share)
Number of issued shares=25000
dividend per share=$1.5
thus the total dividend will be:
1.5*25000
=$37500
Answer:
58.995 round to the nearest whole number is 59 miles per hour
Step-by-step explanation:
1 kilometer/hr = 0.62 miles per hour
Therefore 100 km/hr = 100 × 0.621
= 62 miles per hour
95 km/hr = 95 × 0.621
= 58.995 miles per hour
58.995 round to the nearest whole number is 59 miles per hour
(a)
The binomial distribution can be used because the current situation satisfies all of the following:
1. The probability of success (p=85%) is known and remains constant during the whole experiment
2. The number of trials (n=40) is known and constant.
3. Each trial is a bernoulli trial (success or failure only)
4. All trials are (assumed) independent of each other.
The probability of x successes is therefore
P(X=x)=C(n,x)(p^x)(1-p)^(n-x)
(b) P(X=35) means the probability of 35 successes out of 40 trials at p=0.85
and
P(X=35)=C(40,35)*0.85^35*0.15^5=658008*0.003386*0.00007594
=0.16918
(c) P(X>=35)=∑ P(X=i) for i=35 to 40
=0.16918+0.13315+0.08157+0.03649+0.01060+0.00150
=0.4325
(d) P(X<20)=∑ P(X=i) for i=0 to 19
=0.00000003513 (individual probabilities are very small).

Pooled Combined Variance 
Test Statistic:



Degrees of freedom = 


The table value is greater than the calculated value.
Thus we accept H0.