Answer:
1) m<1 = 25
2) m<4 = 155
3) m< 5 = 25
4) m<7 = 25
5) 110
6) 30.5
7) 74
Step-by-step explanation:
1)
x + 155 = 180
x = 25
2)
<4 = 155 because of vertical angle thm
3)
<5 = 25 (congruent to <1) b/c of corresponding angles thm
4)
<7 = <5 (25) because of vertical angles thm
5)
5x + 15 = 8x - 42
-3x = -57
x = 19
5(x) + 15
5(19) + 15 = 110
6)
180 - 110 = 70
3x + x + 4 + 70 = 180
4x + 74 = 180
4x = 106
x = 26.5
x+ 4 = 30.5
7)
180 - 115 = 65
x + 4 + 2x + 65 = 180
3x + 69 = 180
3x = 111
x = 37
2x = 74
Answer:
The chance that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses is P=0.0488.
Step-by-step explanation:
To solve this problem we divide the tossing in two: the first 5 tosses and the last 5 tosses.
Both heads and tails have an individual probability p=0.5.
Then, both group of five tosses have the same binomial distribution: n=5, p=0.5.
The probability that k heads are in the sample is:

Then, the probability that exactly 2 heads are among the first five tosses can be calculated as:

For the last five tosses, the probability that are exactly 4 heads is:

Then, the probability that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses can be calculated multypling the probabilities of these two independent events:

No it does not. :) hope this helps
16*28=448
28 is 12 inches higher than 16 so it should work.
Step-by-step explanation:
P(t) = 12,000 (2)^(-t/15)
9,000 = 12,000 (2)^(-t/15)
0.75 = 2^(-t/15)
ln(0.75) = ln(2^(-t/15))
ln(0.75) = (-t/15) ln(2)
-15 ln(0.75) / ln(2) = t
t = 6.23