Answer: the value for the associated test statistic is 1.2653
Step-by-step explanation:
Given that;
sample size one n₁ = 10
mean one x"₁ = 6.4
standard deviation one S₁ = 1.1
sample size two n₁ = 11
mean two x"₂ = 5.6
standard deviation one S₁ = 1.7
H₀ : μ₁ = μ₂
H₁ : μ₁ ≠ μ₂
Pooled Variance
sp = √( { [(n₁ - 1) × s₁² + (n₂ - 1) × s₂²] / (n₁ + n₂ - 2)} × (1/n₁ + 1/n₂))
we substitute
= √( { [(10 - 1) × (1.1)² + (11 - 1) × (1.7)²] / (10 + 11 - 2)} × (1/10 + 1/11))
= √( { [(9) × 1.21 + (10) × 2.89] / (19) } × (0.1909))
= √({[ 39.79 ] / 19} × (0.1909))
= √( 2.0942 × 0.1909)
= √( 0.39978 )
= 0.63228
Now Test Statistics will be;
t = ( x"₁ - x"₂) / sp
we substitute
t = ( 6.4 - 5.6) / 0.63228
t = 0.8 / 0.63228
t = 1.2653
Therefore the value for the associated test statistic is 1.2653
Given that the perimeter of rhombus ABCD is 20 cm, the length of the sides will be:
length=20/4=5 cm
the ratio of the diagonals is 4:3, hence suppose the common factor on the diagonals is x such that:
AC=4x and BD=3x
using Pythagorean theorem, the length of one side of the rhombus will be:
c^2=a^2+b^2
substituting our values we get:
5²=(2x)²+(1.5x)²
25=4x²+2.25x²
25=6.25x²
x²=4
x=2
hence the length of the diagonals will be:
AC=4x=4×2=8 cm
BD=3x=3×2=6 cm
Hence the area of the rhombus wll be:
Area=1/2(AC×BD)
=1/2×8×6
=24 cm²
The answer is 2/3 times as heavy as the dog.
Since we are looking for the weight of the cat in relation to the dog's weight, we have to divide the cat's weight by the dog's weight.
5 1/2 = 5.5, 8 1/4 = 8.25.
5.5 / 8.25 = 2/3.
On the top problem: Angle A and Angle B are a right angle so it equals 90 degrees. If Angle B is 50 degrees that makes Angle A 40 degrees. Angle A and Angle C is a straight line and equals 180 degrees so if Angle A equals 40 degrees that means Angle C equals 140 degrees.
Answer:
(D)14
Step-by-step explanation:
17-2x=-11
17+11=2x
28=2x
28/2=2x/2
14=x
x=14
