<span>Y is directly proportional to x^2. It could be represented by the expression:
y </span>α x^2
We can make it into an equality by inserting the proportionality constant, k.
y = kx^2
k would be constant for any value of y with a corresponding value of x. We solve the problem by this concept as follows:
y1/(x1)^2 = y2/(x2)^2
10/(x1)^2 = y2/(x1/2)^2
10/4 = y2
Therefore, when the value of x is halved, y is equal to 10/4.
Easy, you just replace m with 2, and when a letter is placed next to a number then it means you multiply them, therefore
6*2-1=12-1 = 11
Answer:
16c + 38d + 48f
Step-by-step explanation:
12c + 36d + 48f + 4c + 2d + 12f
= 16c + 38d + 60f
Answer:
81.85.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation , the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Suppose the mean height for men is 70 inches with a standard deviation of 2 inches.
This means that
What percentage of men are between 68 and 74 inches tall?
The proportion is the p-value of Z when X = 74 subtracted by the p-value of Z when X = 68.
X = 74
has a p-value of 0.9772.
X = 68
has a p-value of 0.1587.
0.9772 - 0.1587 = 0.8185
0.8185*100% = 81.85%.
Thus the percentage is 81.85%, and the answer is 81.85.
Answer:
Solution given:
A triangle PQR is right angled at R, with hypotenuse{h}PQ=80cm
and
base[b]PR=60cm.
perpendicular [P]= QR
<u>by</u><u> </u><u>using</u><u> </u><u>Pythagoras</u><u> </u><u>law</u>
<u>h²</u><u>=</u><u>p²</u><u>+</u><u>b²</u>
80²=QR²+60²
QR²=80²-60²
QR=
QR=20=52.9=53cm
<u>QR</u><u>=</u><u>5</u><u>3</u><u>c</u><u>m</u><u>.</u>