Answer:
There were 84 students in 6th grade, 68 in 7th grade, and 90 in 8th grade.
Step-by-step explanation:
Given that Elk middle school has 264 students, and there are 84 students in the eighth grade, and there are 22 fewer seventh-grade students than sixth-grade students, to determine how many students there are in each grade, the following calculation must be performed:
(264 - 84) / 2) - 22 = X
(180/2) - 22 = X
90 - 22 = X
68 = X
6th grade = 84
7th grade = 68
8th grade = 90
Therefore, there were 84 students in 6th grade, 68 in 7th grade, and 90 in 8th grade.
Answer:
-9.1 degrees at midnight
Step-by-step explanation:
15.1 - 24.2 = -9.1 degrees
Question:
Isabella pitches for her high school softball team. Of the 7 pitches she throws, 5 are strikes. She uses a standard deck of cards to model her pitches, with hearts, clubs, and spades representing strikes. Which statement about her model is true?
Answer:
B. Her model can be improved the most by removing the queens and kings of hearts, clubs, and spades.
Explanation:
I took the test and passed.
<span>(a) This is a binomial
experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer
than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
</span>
SE is hypotenuse so with having <span>coordinates of S and E we can say midpoint is (6+0/2,8+0/2)=(3,4)
i really don't know what are those options </span>
</h3><h3 /><h3>Decimal Form:</h3><h3>4678899.08686137</h3><h3>Mixed Number Form:</h3><h3>4678899 1912705/22020202</h3><h3>or</h3><h3>
</h3><h2>____________________________________</h2><h2>I know this is probably just a troll but your welcome!</h2><h2>Hope you have a good day, Loves~ <3</h2>
