Answer:
odd: 4/7
not 3: 6/7
4 or 5: 2/7
spinner spun 2: there both 1/7 i think im
only in 7th grade ;-;
Step-by-step explanation: well there are 7 spaces so the probabilty
of it landing on an odd space on the wheel is how many odd numbers there are over 7 . and there is only 1 three so the chance that it will not land on 3 is 6/7. there is a 2/7 chance it will land on 4 or 5. and im not to sure abt the last one but like i said im only in the 7th grade i am taking p classes tho so the rest should be right. :D
U would multiply the number on the side and the number on the bottom. hope that helps :)
Answer:
x equals 30
Step-by-step explanation:
An equilateral triangle adds up to a total of 180 degrees so divided 60 into two and you get 30
Question 1: <span>
The answer is D. which it ended up being <span>
0.9979</span>
Question 2: </span>
The expression P(z > -0.87) represents the area under the standard normal curve above a given value of z. What is P(z > -0.87)? Express your answer as a decimal to the nearest ten thousandThe expression P(z > -0.87) represents the area under the standard normal curve above a given value of z. What is P(z > -0.87)? Express your answer as a decimal to the nearest ten thousandth (four decimal places). So being that rounding it off would mean your answer would be = ?
Question 3: <span>
Assume that the test scores from a college admissions test are normally distributed, with a mean of 450 and a standard deviation of 100. a. What percentage of the people taking the test score between 400 and 500?b. Suppose someone receives a score of 630. What percentage of the people taking the test score better? What percentage score worse?c. A university will not admit a student who does not score in the upper 25% of those taking the test regardless of other criteria. What score is necessary to be considered for admission? </span>
z = 600-450 /100 = .5 NORMSDIST(0.5) = .691462<span><span>
z = 400-450 /100 = -.5 NORMSDIST(-0.5) = .30854
P( -.5 < z <.5) = .691462 - .30854 = .3829 Or 38.29%
Receiving score of 630:
z = 630-450 /100 = 1.8 NORMSDIST(1.8) = .9641
96.41% score less and 3.59 % score better
upper 25%
z = NORMSINV(0.75)= .6745
.6745 *100 + 450 = 517 Would need score >517 to be considered for admissions
</span><span>
Question 4: </span>
The z-score for 45cm is found as follows:</span>
Reference to a normal distribution table, gives the cumulative probability as 0.0099.<span>
Therefore about 1% of newborn girls will be 45cm or shorter.</span>
Answer:
B . . . . . .. . ....... .......
