Answer:
B. weight of a bag of apples
Step-by-step explanation:
A continuous random variable is a variable that is measured not counted. It can be any number between integers in decimal form.
So from the options given above, only the weight of bag of apples is a continuous variable as it will be measured.
While the others cannot be expressed in the form of decimals.
So the correct answer is:
B. weight of a bag of apples ..
We can write a system of equations:
1x + 10y = 182
x + y = 56
Where 'x' is the number of $1 bills, and 'y' is the number of $10 bills.
To find this we can solve using substitution.
Re-arrange the 2nd equation:
x + y = 56
Subtract 'y' to both sides:
x = -y + 56
Now we can plug in '-y + 56' for 'x' in the first equation.
1x + 10y = 182
1(-y + 56) + 10y = 182
-y + 56 + 10y = 182
Subtract 56 to both sides:
-y + 10y = 126
Combine like terms:
9y = 126
Divide 9 to both sides:
y = 14
Now we can plug this into any of the two equations to find the 'x' value.
x + y = 56
x + 14 = 56
Subtract 14 to both sides:
x = 42
So our final answer is (42, 14).
This means that the motel clerk had 42 $1 bills, and 14 $10 bills.
Answer:
The answer to
1. 22/5
2. 3
Step-by-step explanation:
Answer:
-7
Step-by-step explanation:
XZ ≅ EG and YZ ≅ FG is enough to make triangles to be congruent by HL. Option b is correct.
Two triangles ΔXYZ and ΔEFG, are given with Y and F are right angles.
Condition to be determined that proves triangles to be congruent by HL.
<h3>What is HL of triangle?</h3>
HL implies the hypotenuse and leg pair of the right-angle triangle.
Here, two right-angle triangles ΔXYZ and ΔEFG are congruent by HL only if their hypotenuse and one leg are equal, i.e. XZ ≅ EG and YZ ≅ FG respectively.
Thus, XZ ≅ EG and YZ ≅ FG are enough to make triangles congruent by HL.
Learn more about HL here:
brainly.com/question/3914939
#SPJ1
In ΔXYZ and ΔEFG, angles Y and F are right angles. Which set of congruence criteria would be enough to establish that the two triangles are congruent by HL?
A.
XZ ≅ EG and ∠X ≅ ∠E
B.
XZ ≅ EG and YZ ≅ FG
C.
XZ ≅ FG and ∠X ≅ ∠E
D.
XY ≅ EF and YZ ≅ FG
