Answer:
the first field (rate 3/4) has 32 square yards and the second field (rate 2/3) has 24 square yards.
Step-by-step explanation:
With the statement we can make a system of 2x2 equations, where:
"x" is the area of the first field
"y" is the area of the second field
However,
x + y = 56 => x = 56 - y
3/4 * x + 2/3 * y = 40
replacing we have:
3/4 * (56 - y) + 2/3 * y = 40
42 - 3/4 * y + 2/3 * y = 40
-0.0833 * y = 40 - 42
y = -2 / -0.0833
y = 24
now for x:
x = 56 - 24
x = 32
This means that the first field (rate 3/4) has 32 square yards and the second field (rate 2/3) has 24 square yards.
Answer:
7.1 cm
Step-by-step explanation:
Let s represent the side length of the square. The Pythagorean theorem tells you the relationship to the diagonal length is ...
(10 cm)² = s² +s² . . . . the sum of the squares of the legs is the square of the hypotenuse
100 cm² = 2s² . . . . . . simplify
50 cm² = s² . . . . . . . . divide by 2
√(50) cm = s ≈ 7.1 cm . . . . . take the square root
The length of a side of the square to the nearest tenth is 7.1 cm.
First you substitute into the equation is/of=%/100. By that, I mean, 119/x=70/100. Cross multiply to get 119(100)=70x. Solve for x=170
Answer:
15.3feet
Step-by-step explanation:
Given the scale model of a train as;
2cm = 3feet
We are to find the height of real train of height 10.2cm, we can write;
10.2cm = x
Divide both expressions
2/10.2 = 3/x
2x = 3×10.2
2x = 30.6
x= 15.3
Hence the height if the real train is 15.3feet
The relative frequency for the class with lower class limit 31 is 8.57%.
<h3>How to calculate the frequency?</h3>
It should be noted that the lowrr class frequency is 31 in this case.
Therefore, the relative frequency for the class with lower class limit 31 will be:
= 3/(6+4+4+9+3+9) × 100
= 3/35 × 100
= 8.57%
Therefore, the relative frequency for the class with lower class limit 31 is 8.57%.
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