Refer to the diagram shown below.
Let x = m∠ADB.
Because m∠BDC is 32° greater than m∠ADB, therefore
m∠BDC = x + 32°
Each angle of a rectangle is 90°, therefore
x + (x+32) = 90
2x + 32 = 90
2x = 58
x = 29°
x+32 = 61°
Answer:
m∠BDC = 61°
m∠ADB = 29°
4/5-2/3= 12/15-10/15 (common denominator is 15)=2/15 pounds of seeds left
To find the average assessment, add the 5 grades together, then divide by the quantity of grades added together (in this case 5).
=56 + 42 + 47 + 59 + 48
=252
=252 ÷ 5
=50.4 average
ANSWER: Yes, he passes his course with an average grade of 50.4.
Hope this helps! :)
Answer:
Step-by-step explanation:
Hello!
X₁: speed of a motorcycle at a certain intersection.
n₁= 135
X[bar]₁= 33.99 km/h
S₁= 4.02 km/h
X₂: speed of a car at a certain intersection.
n₂= 42 cars
X[bar]₂= 26.56 km/h
S₂= 2.45 km/h
Assuming
X₁~N(μ₁; σ₁²)
X₂~N(μ₂; σ₂²)
and σ₁² = σ₂²
<em>A 90% confidence interval for the difference between the mean speeds, in kilometers per hour, of motorcycles and cars at this intersection is ________.</em>
The parameter of interest is μ₁-μ₂
(X[bar]₁-X[bar]₂)±
* 
[(33.99-26.56) ± 1.654 *(
)]
[6.345; 8.514]= [6.35; 8.51]km/h
<em>Construct the 98% confidence interval for the difference μ₁-μ₂ when X[bar]₁= 475.12, S₁= 43.48, X[bar]₂= 321.34, S₂= 21.60, n₁= 12, n₂= 15</em>
[(475.12-321.34) ± 2.485 *(
)]
[121.96; 185.60]
I hope this helps!
