We know that all black beards cost the same money
And all red beards cost $25
And B represents number of black beards
And R represents number of red beards
We have been given the inequality 20B+25R > 350 to show the target for this month.
The above inequality shows that Merlin needs to make more than 350, hence he made 350 last month
Further, it shows that B has been multiplied by 20, hence the black beard's cost was 20
Answer:
3 or 3/1 (the 1 is the denominator)
Step-by-step explanation:
the shape got 3 times bigger. let me know
Answer:
Null hypothesis 
Alternative hypothesis 
Step-by-step explanation:
Consider the provided information.
The shower by an adult is 5 minutes.
The mean amount of time that college students spend in the shower is significantly different from 5 minutes. In a sample of 15 students, he found the average time was 4.29 minutes and the standard deviation was 0.75 minutes.
We need to determine the appropriate null and alternative hypotheses.
A null hypothesis is there is no statistical significance between the two variables.
An alternative hypothesis is there is a statistically significant relationship between two variables.
Therefore the required null and alternative hypothesis are:
Null hypothesis
Alternative hypothesis
<span>You can probably just work it out.
You need non-negative integer solutions to p+5n+10d+25q = 82.
If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80.
So this is the same as n + 2d + 5q ≤ 16
So now you simply have to "crank out" the cases.
Case q=0 [ n + 2d ≤ 16 ]
Case (q=0,d=0) → n = 0 through 16 [17 possibilities]
Case (q=0,d=1) → n = 0 through 14 [15 possibilities]
...
Case (q=0,d=7) → n = 0 through 2 [3 possibilities]
Case (q=0,d=8) → n = 0 [1 possibility]
Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81
Case q=1 [ n + 2d ≤ 11 ]
Case (q=1,d=0) → n = 0 through 11 [12]
Case (q=1,d=1) → n = 0 through 9 [10]
...
Case (q=1,d=5) → n = 0 through 1 [2]
Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42
Case q=2 [ n + 2 ≤ 6 ]
Case (q=2,d=0) → n = 0 through 6 [7]
Case (q=2,d=1) → n = 0 through 4 [5]
Case (q=2,d=2) → n = 0 through 2 [3]
Case (q=2,d=3) → n = 0 [1]
Total from case q=2: 1 + 3 + 5 + 7 = 16
Case q=3 [ n + 2d ≤ 1 ]
Here d must be 0, so there is only the case:
Case (q=3,d=0) → n = 0 through 1 [2]
So the case q=3 only has 2.
Grand total: 2 + 16 + 42 + 81 = 141 </span>
