The answer is D.
First, separate the compound equality into 2
1st -3 (x-4) greater than/ equal to 21
2nd -3( x-4) < 30
Second, solve the inequality for x
1st x less than/ equal to -3
2nd x > -6
Third, find the intersection
D. -6 < x (less than/equal) to -3
I hope this helped, and if you don’t mind, giving me brainiest!
One number<span> is eight </span>less than<span> five </span>times<span> another. </span>If the sum<span> of the </span>two numbers<span> is 28, find the </span><span>numbers . Please levae this i need points i have 0000
</span>
Answer:
a) The probability is 0.04
b) The probability is 0.36
c) The pprobability is 0,25
d) The probability is 0.09
Step-by-step explanation:
Lets calculate areas:
the target has a radius of 10 inces, hence the target area has a area on 10²*π = 100π square inches.
a) A circle of 2 inches of radius has an area of 2²π = 4π square inches, hence the probability of hitting that area is 4π/100π = 1/25 = 0.04
b) If the dart s within 2 inches of the rim, then it is not at distance 8 inches from the center (that is the complementary event). The probability for the dart to be at 8 inches of the center is 8²π/100π = 64/100 = 16/25 = 0.64, thus, the probability that the dart is at distance 2 or less from the rim is 1-0.64 = 0.36.
c) The first quadrant has an area exactly 4 times smaller than the area of the target (each quadrant has equal area), thus the probability for the dart to fall there is 1/4 = 0.25
d) If the dart is within 2 inches from the rim (which has probability 0.36 as we previously computed), then it will be equally likely for the dart to be in either of the 4 quadrants (the area that is within 2 inches from the rim forms a ring and it has equal area restricted on each quadrant). Therefore, the probability for the dice to be in the first qudrant and within 2 inches from the rim is 0.36*1/4 = 0.09.
I have a feeling, these were the answer choices you were trying to make: <span>A ∪ B = {7}
B ⊆ B
B ⊂ A Also, the second answer is the only one that's true
</span>
1) Yes, the relationship in the table is proportional. If, when you've been walking for 10 minutes, you are 1.5 miles away from home, and when you've been walking for 20 minutes, you are 1 mile away from home, and when you've been talking 30 minutes, you are 0.5 miles away from home, then we can see that there is a proportion that happens here. For every 10 minutes you walk, you get 0.5 miles closer to your home.
2) We know that you've been walking 10 minutes already at the start of this problem, and we know that you walk at a steady pace of 0.5 miles every 10 minutes, so we just need to add 0.5 miles to our starting point to get the distance from the school to home, which makes it 2 miles away.
3) An equation representing the distance between the distance from school and time walking could be something like this:
t = 20d
Where t is the amount of time it takes to get home (in this case, t = 40 minutes) and d is the distance you can walk in 10 minutes (in this case, 0.5 miles)
The equation is lame, but that's the best I could do :\
Hope that helped =)
