Plane UVWX and plane QRST
Answer:
ok is very easy
Step-by-step explanation:
so 1/5 of an hour is 12 mins
and 2 is 2 hrs so that 120 mins
now add 120+12=132
132 x -30 degrees
The answer to that is 132 x -30 =
-3960 sorry if its wrong and if you get a second try put -69.1150384 rad
I think sorry if you get it wrong have a good day
To write 8/33 as a decimal you have to divide numerator by the denominator of the fraction.
<span>We divide now 8 by 33 what we write down as 8/33 and we get 0.24242424242424 </span>
<span>And finally we have: </span>
8/33 as a decimal<span> equals </span><span>0.24242424242424</span>
The pattern is subtracting 9
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>
