Answer:
<h3>
y = 20</h3><h3 />
Step-by-step explanation:
line RPN = 180 = 4y - 10 + 90 + y
180 - 90 + 10= 5y
100 = 5y
y = 100/5
y = 20
First you must know that for remarkable angles: cos (0) = 1, cos (π) = - 1, cos (π / 2) = 0, cos (3π / 2) = 0, cos (2π) = 1. Then, by simple substitution in the given formula, you can find the solutions of x. Which for the interval [0, 2π) are: x = π, x = pi divided by two and x = three pi divided by two.Attached solution.
9(2w−y)=21w−9y
First you multiply the numbers in the parenthesis by 9
subtract. 21w-18w=3w
add. -9y+9y=0
divide.
Final answer is 0.
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:
0.136
Step-by-step explanation:
If you have a calculator with statistical functions, this problem is straightforward.
Use the normcdf function as follows:
normcdf(300,350,400,50) = 0.136
This result is the probability that a worker makes between $300 and $350 per week: 0.136
