Answer: By cross multiplication.
Step-by-step explanation: Given product is 3 × 292.
We know that after simple multiplication, we get 3 × 292 = 876.
Now, to check division with multiplication, either we need to divide 876 by 3 to get the answer 292,
or
we need to divide 876 by 292 to get the answer 3.
We will do that as follows -
Thus, doing cross-multiplication, we arrive at our conclusion.
Answer:
x = 25, y = 9
Step-by-step explanation:
Since the triangles are congruent then corresponding angles and sides are congruent, thus
∠ O = ∠ P , substitute values
6y - 14 = 40 ( add 14 to both sides )
6y = 54 ( divide both sides by 6 )
y = 9
and
NG = IT , that is
x - 2y = 7 ( substitute y = 9 into the equation )
x - 2(9) = 7
x - 18 = 7 ( add 18 to both sides )
x = 25
Answer:
Step-by-step explanation:
Terms are separated from one another by + and/or - signs. In this case neither appears, so the given expression constitutes a single term.
Answer:
A=10 cm
B=5 cm
Area of square = 25 cm
Step-by-step explanation:
Since that is an equilateral triangle based off the fact that all the sides are the same variable (a) then that means 30 divided by 3 which equals 10. So now we know that a = 10. Lets input that information for the rectangle. Since there are 2 a's that equals 20 and the perimeter is 30 so b has to be 5 since 5 times 2 is 10 plus 20 equals 30. so now we know that a=10 and b=5 so our squares sides are all 5. The area of the square is 25
Answer:
(−0.103371 ; 0.063371) ;
No ;
( -0.0463642, 0.0063642)
Step-by-step explanation:
Shift 1:
Sample size, n1 = 30
Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm
Shift 2:
Sample size, n2 = 25
Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17
Mean difference ; μ1 - μ2
Zcritical at 95% confidence interval = 1.96
Using the relation :
(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)
(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)
Lower boundary :
-0.02 - 0.0833710 = −0.103371
Upper boundary :
-0.02 + 0.0833710 = 0.063371
(−0.103371 ; 0.063371)
B.)
We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.
C.)
For sample size :
n1 = 300 ; n2 = 250
(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)
Lower boundary :
-0.02 - 0.0263642 = −0.0463642
Upper boundary :
-0.02 + 0.0263642 = 0.0063642
( -0.0463642, 0.0063642)
