A Farmer Weighs Two Calves After They Are Born. One Calf Weighs 38 7/10 Kilograms. Another Calf Weighs 37 1/10 Kilog Rams. How Much More Does The First Calf
The first calf weighs 38 7/10 kilograms.
The second calf weighs 37 1/10 kilograms.
To find out how much more the first calf weighs, we need to subtract the weight of the second calf from the weight of the first calf.
Calculating 38 7/10 - 37 1/10, we get 1 6/10.
So, the first calf weighs 1 6/10 kilograms more than the second calf.
4/7+1/8
To add 4/7 and 1/8, we need to find a common denominator.
The common denominator is 56 since it is divisible by both 7 and 8.
Now, we can rewrite the fractions with the common denominator:
4/7 = (4 * 8)/(7 * 8) = 32/56
1/8 = (1 * 7)/(8 * 7) = 7/56
Now that the fractions have the same denominator, we can add them:
32/56 + 7/56 = (32 + 7)/56 = 39/56
Therefore, 4/7 + 1/8 equals 39/56.
2/3-1/4
To subtract 1/4 from 2/3, we need to find a common denominator.
The common denominator is 12 since it is divisible by both 3 and 4.
Now, we can rewrite the fractions with the common denominator:
2/3 = (2 * 4)/(3 * 4) = 8/12
1/4 = (1 * 3)/(4 * 3) = 3/12
Now that the fractions have the same denominator, we can subtract them:
8/12 - 3/12 = (8 - 3)/12 = 5/12
Therefore, 2/3 - 1/4 equals 5/12.
A Scientist Measured The Lengths Of Catfish In A Lake. Below Are The Lengths Of 11 Catfish In Inches. 6, 8, 10, 11, 12, 15, 17, 18, 19, 20, 22 Which Box Plot Represents The Data?
Based on the given lengths of catfish in inches:
6, 8, 10, 11, 12, 15, 17, 18, 19, 20, 22
The box plot that represents this data should have the following characteristics:
- The minimum value (smallest length) should be 6.
- The maximum value (largest length) should be 22.
- The median (middle value) should be close to 15, as it is the average of 12 and 17.
- The lower quartile should be close to 10, as it is the average of 8 and 11.
- The upper quartile should be close to 19, as it is the average of 18 and 20.
- There should be no outliers indicated by any individual points outside of the whiskers (lines connecting the box to the minimum and maximum values).
Based on these characteristics, the correct box plot should have a box stretching from approximately 10 to 19, with a median line close to 15, and whiskers connecting to minimum and maximum values of approximately 6 and 22, respectively.
Quadrilateral ABCD Is A Kite With An Area Of 14cm2 . The Length Of Overline ED Is 5 Cm. What Is The Length Of Overline BE ? A 7 Cm B 2 Cm C 4 Cm D 3 Cm
In a kite, the diagonals are perpendicular and bisect each other.
Let's denote the length of overline BE as "x".
Since overline ED is a diagonal of the kite and it bisects overline BC, we can divide the area of the kite by 2 to find the area of triangle EBC.
Area of triangle EBC = (1/2) * base * height
14 cm² = (1/2) * x * 5 cm
Simplifying the equation:
14 cm² = (5/2) * x cm
28 cm² = 5x cm
28/5 cm = x cm
Therefore, the length of overline BE is approximately 5.6 cm.
None of the given answer choices match this calculation.
To find out how much more the first calf weighs compared to the second calf, we need to subtract the weight of the second calf from the weight of the first calf.
The weight of the first calf is 38 7/10 kilograms, and the weight of the second calf is 37 1/10 kilograms.
To subtract these two weights, we can convert the mixed fractions to improper fractions and then subtract:
Weight of first calf: 38 7/10 kilograms = (38 * 10 + 7) / 10 = 387 / 10
Weight of second calf: 37 1/10 kilograms = (37 * 10 + 1) / 10 = 371 / 10
Now we can subtract the weight of the second calf from the weight of the first calf:
Weight of the first calf - Weight of the second calf = (387 / 10) - (371 / 10)
To subtract fractions, we need a common denominator, which in this case is 10:
Weight of the first calf - Weight of the second calf = (387 - 371) / 10 = 16 / 10
Simplifying the fraction, we get:
16 / 10 = 8 / 5
Therefore, the first calf weighs 8/5 kilograms more than the second calf.