3/3=1 and 7/8 is < 1 so 3/3 is bigger
Reduce a 24 cm by 36 cm photo to 3/4 original size.
The most logical way to do this is to keep the width-to-height ratio the same: It is 24/36, or 2/3. The original photo has an area of (24 cm)(36 cm) = 864 cm^2.
Let's reduce that to 3/4 size: Mult. 864 cm^2 by (3/4). Result: 648 cm^2.
We need to find new L and new W such that W/L = 2/3 and WL = 648 cm^2.
From the first equation we get W = 2L/3. Thus, WL = 648 cm^2 = (2L/3)(L).
Solve this last equation for L^2, and then for L:
2L^2/3 = 648, or (2/3)L^2 = 648. Thus, L^2 = (3/2)(648 cm^2) = 972 cm^2.
Taking the sqrt of both sides, L = + 31.18 cm. Then W must be 2/3 of that, or W = 20.78 cm.
Check: is LW = (3/4) of the original 864 cm^2? YES.
Answer: D
Step-by-step explanation:
all possible rational zeros are the factors of the last term divided by the coefficient of the first term
so it's (±1, ±3, ±9) / (±1, ±2)
(±1, ±3, ±9) / ±1 = ±1, ±3, ±9
(±1, ±3, ±9) / ±2 = ±1/2, ±3/2, ±9/2
--> ±1, ±3, ±9, ±1/2, ±3/2, ±9/2
Z = 17.
2(4z -8) = 120
8z -16 = 120
z -2 = 15
z = 17
Answer: the observer should consider to eliminate or to retake the third measure.
Explanation:
The four measures taken are 124.53, 124.55, 142.51 and 124.52.
As it can be easily seen, the third measure is much different from the other three. This means that something went wrong during the observation: it can be either the measure taken wrong or that the number was written wrong (if you switch the 2 and the 4 you get a number similar to the other ones).
If the third measure is not considered, an estimate of the mean would place it around 124.5, while if the outlier (the detatched number) is considered an estimate of the mean would increase to about 129.
Therefore, in order to obtain a more reliable mean, the observer should consider to eliminate or to retake the third measure.
