Answer:
SSS (Side side side)
Step-by-step explanation:
Answer:
Total hours that Jenny ran = 3.63 hours.
Step-by-step explanation:
Jenny ran on Tuesday for = 2/5 hours or 0.4 hours.
Time consumed to run on Thursday = 11/6 hours or 1.83 hours.
Time consumed to run on Saturday = 21/ 15 hours or 1.4 hours.
Here, the total hours can be calculated by just adding all the running hours. So the running hours of Tuesday, Thursday, and Saturday will be added to find the total hours.
Total hours that Jenny ran = 0.4 + 1.83 + 1.4 = 3.63 hours.
Answer: It will take 1.875 hours
Step-by-step explanation:
Given that time t required to do a job varies inversely as the number of people working.
we have it expressed as
t∝1/N
where t is time required to do a job
N is the number of people working.
Introducing our constant of proportionality, we have tht
t=k1/N
t=k/N
Now since it takes 5 hr for 3 cooks to prepare the food , K becomes
t=k/N
5=K/3
K=5X3=15
Given that there are 8 cooks, time required to prepare dinner will be
t=k/N
t=15/8
t=1.875 hours
<span>The fraction of baked goods not sold is 6/10. The numerator is the number of left over baked goods (10-4 -6) and the denominator is the total loaves of bread (10). An equivalent fraction is 12/20. This was obtained by multiplying the original fraction by two (6 x2 = 12 and 10 x2 =20).</span>
Answer:
The maximum number of red beans that can be put in one of the stacks is 5.
Step-by-step explanation:
First, Every pile is made with 7 beans, but they need to have at least one bean of each color, so we need to include in the pile minimum one yellow bean and one green bean. If we make a stack with this condition, the stack is going to have: 1 yellow bean, 1 green bean, and 5 red beans.
Additionally, it is necessary to know if there are enough red beans for the other piles. In this case, we have 10 red beans so if we have 5 red beans in one stack, the other two stacks have enough red beans to fulfill with the condition to have at least one red bean.
So, the maximum number of red beans that can be put in one of the stacks is 5.
