Answer:
55%
Step-by-step explanation:
$598.99–$385.99
=$213
$213/$385.99 × 100%
=55%
Since the sum of the probabilities of all possible outcomes must be 100%, we can deduce the following:
- Cooking in under 20 minutes: 10%
- Cooking between 20 and 30 minutes: 85%
- Cooking in more than 30 minutes: 5%
In fact, the probabilities of cooking in less than 20 or more than 30 sum up to 15%, which means that the remaining outcome (i.e. cooking time between 20 and 30) must complete this probability to 15, and in fact 15+85=100.
That being said, all three answers are simply a combination of these three scenarios: let C be the cooking time, for aesthetic reasons:
![P(C\geq 20) = P(20 \leq C \leq 30)+P(C\geq30) = 85\%+5\%=90\%](https://tex.z-dn.net/?f=P%28C%5Cgeq%2020%29%20%3D%20P%2820%20%5Cleq%20C%20%5Cleq%2030%29%2BP%28C%5Cgeq30%29%20%3D%2085%5C%25%2B5%5C%25%3D90%5C%25)
![P(20 \leq C \leq 30) = 85\%](https://tex.z-dn.net/?f=P%2820%20%5Cleq%20C%20%5Cleq%2030%29%20%3D%2085%5C%25)
![P(C\leq 30) = P(C\leq20)+P(20 \leq C \leq 30) = 10\%+85\%=95\%](https://tex.z-dn.net/?f=P%28C%5Cleq%2030%29%20%3D%20P%28C%5Cleq20%29%2BP%2820%20%5Cleq%20C%20%5Cleq%2030%29%20%3D%2010%5C%25%2B85%5C%25%3D95%5C%25)
Answer:
The answer is below
Step-by-step explanation:
Pythagoras theorem states that for a right angled triangle, the square of the hypotenuse side is equal to the sum of the square of the remaining sides. The hypotenuse is the longest side (that is side opposite to the 90° angle).
In right angle triangle ABD:
AB² = AD² + BD² (1)
In right angle triangle ACD:
AC² = AD² + CD² (2)
Also:
AC² + AB² = BC² (3)
But BC = BD + CD
AC² + AB² = (BD + CD)² (4)
Adding equation 1 and 2 gives:
AB² + AC² = (AD² + BD²) + (AD² + CD²)
AB² + AC² = 2AD² + BD² + CD²
substituting AC² + AB² = (BD + CD)²:
(BD + CD)² = 2AD² + BD² + CD²
BD² + 2(BD)(CD)+ CD² = 2AD² + BD² + CD²
2AD² = 2(BD)(CD)
AD² = BD * CD
There are no figures shown, post the figure in comments and I will answer in comments.
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)±
* ![Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }](https://tex.z-dn.net/?f=Sa%5Csqrt%7B%5Cfrac%7B1%7D%7Bn_1%7D%20%2B%5Cfrac%7B1%7D%7Bn_2%7D%20%7D)
![t_{n_1+n_2-2;1-\alpha /2}= t_{175; 0.95}= 1.654](https://tex.z-dn.net/?f=t_%7Bn_1%2Bn_2-2%3B1-%5Calpha%20%2F2%7D%3D%20t_%7B175%3B%200.95%7D%3D%201.654)
![Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{134*16.1604+41*6.0025}{135+42-2} } = 3.71](https://tex.z-dn.net/?f=Sa%3D%20%5Csqrt%7B%5Cfrac%7B%28n_1-1%29S_1%5E2%2B%28n_2-1%29S_2%5E2%7D%7Bn_1%2Bn_2-2%7D%20%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B134%2A16.1604%2B41%2A6.0025%7D%7B135%2B42-2%7D%20%7D%20%3D%203.71)
[(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>
![t_{n_1+n_2-2;1-\alpha /2}= t_{25; 0.99}= 2.485](https://tex.z-dn.net/?f=t_%7Bn_1%2Bn_2-2%3B1-%5Calpha%20%2F2%7D%3D%20t_%7B25%3B%200.99%7D%3D%202.485)
![Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{11*(43.48)^2+14*(21.60)^2}{12+15-2} } = 33.06](https://tex.z-dn.net/?f=Sa%3D%20%5Csqrt%7B%5Cfrac%7B%28n_1-1%29S_1%5E2%2B%28n_2-1%29S_2%5E2%7D%7Bn_1%2Bn_2-2%7D%20%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B11%2A%2843.48%29%5E2%2B14%2A%2821.60%29%5E2%7D%7B12%2B15-2%7D%20%7D%20%3D%2033.06)
[(475.12-321.34) ± 2.485 *(
)]
[121.96; 185.60]
I hope this helps!