I think is 1/27.6 because 2x equals 4 and times 3 equals 12. 5.2y equals 15.6. Then 15.6 plus 12 equals 27.6.
The linear parent function is f(x)=x
Answer:
Area of trapezium = 4.4132 R²
Step-by-step explanation:
Given, MNPK is a trapezoid
MN = PK and ∠NMK = 65°
OT = R.
⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).
Now, sum of interior angles in a quadrilateral of 4 sides = 360°.
⇒ x + x + 65° + 65° = 360°
⇒ x = 115°.
Here, NS is a tangent to the circle and ∠NSO = 90°
consider triangle NOS;
line joining O and N bisects the angle ∠MNP
⇒ ∠ONS = 
= 57.5°
Now, tan(57.5°) = 
⇒ 1.5697 = 
⇒ SN = 0.637 R
⇒ NP = 2×SN = 2× 0.637 R = 1.274 R
Now, draw a line parallel to ST from N to line MK
let the intersection point be Q.
⇒ NQ = 2R
Consider triangle NQM,
tan(∠NMQ) = 
⇒ tan65° =
⇒ QM = 
QM = 0.9326 R .
⇒ MT = MQ + QT
= 0.9326 R + 0.637 R (as QT = SN)
⇒ MT = 1.5696 R
⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R
Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).
⇒ A = (
) × (ST)
= (
) × 2 R
= 4.4132 R²
⇒ Area of trapezium = 4.4132 R²
Answer:
x = 1 y = -4
Step-by-step explanation:
- Plug the value of y into the other equation.
x + 3y = -11
x + 3(-4x) = -11
x - 12x = -11
-11x = -11
x = 1
- Now substitute the value of x into any equation.
y = -4x
y = -4(1)
y = -4
Answer:
Step-by-step explanation:
We would set up the hypothesis test.
For the null hypothesis,
µ = 68.8
For the alternative hypothesis,
µ ≠ 68.8
This is a two tailed test.
Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.
Since n = 26,
Degrees of freedom, df = n - 1 = 26 - 1 = 25
t = (x - µ)/(s/√n)
Where
x = sample mean = 72.8
µ = population mean = 68.8
s = samples standard deviation = 2.5
t = (72.8 - 68.8)/(2.5/√26) = 8.16
We would determine the p value using the t test calculator. It becomes
p < 0.00001
Assuming a significance level of 0.05, then
Since alpha, 0.05 > than the p value, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the average life span of the residence of Oregon differs from the world average.
