Answer:
For 1) you must do that yourself from least to greatest.
2) most snow was 3 and 3/4
3) least snow was 1 and 2/4
4) it snowed 3 and 1/4 3 days long.
5) it could have been 4 and 2/4 with another 1 and 2/4 since that makes three or maybe just 4 with 1 since that also makes three there are a lot of possibilities considering it only has to be subtracted and equal three.
Step-by-step explanation:
Hope this helped.
A brainliest is always appreciated.
Answer:
See below in bold.
Step-by-step explanation:
For the fair coin Prob(head) = 1/2 and Prob(Tail) = 1/2.
For the biased coin it is Prob(head) = 2/3 and Prob(Tail) = 1/3.
a) Prob(2 heads) = 1/2 * 2/3 = 1/3.
b) Prob(2 tails) = 1/2 * 1/3 = 1/6.
c) Prob(1 head ) = Prob(H T or T H) = 1/2 * 1/3 + 1/2 * 2/3) = 1/6+1/3 = 1/2.
d) Prob (at least one head) = prob (HH or TH or HT) = 1/3 + 1/2 =<em> </em>5/6.
The person HAD homework 75/100 or 75%
the remaining days he did NOT have homework , which is 25/100 or 25%
Answer:
answer is D
Step-by-step explanation:
because to even it out, you would need ore ones to have rolled to make an average
Answer:
Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .
Step-by-step explanation:
The sample proportion is p2= 7/27= 0.259
and q2= 0.74
The sample size = n= 27
The population proportion = p1= 0.4
q1= 0.6
We formulate the null and alternate hypotheses that the new program is effective
H0: p2> p1 vs Ha: p2 ≤ p1
The test statistic is
z= p2- p1/√ p1q1/n
z= 0.259-0.4/ √0.4*0.6/27
z= -0.141/0.09428
z= -1.496
The significance level ∝ is 0.05
The critical region for one tailed test is z ≤ ± 1.645
Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .
