The first answer of part A is 32
So 4y means you have to multiply 4 by y, which is 20 since 4x5=20
And then you add 12, which is 32.
The second answer to part A is 40
With the second question of part a, the expression is basically saying 4 times whatever 5+3 is. So 5+3= 8, and 4(8) is 40
For part B, they are equivalent because let’s pretend y=2. 12+4y= 20. And then 4(y+3) would equal 20 because 4(2)=8 and 4(3)=12 and 8+12=20. This might not be the answer that your teacher is looking for, but it’s still a right answer so technically they can’t say it’s wrong unless you have a super unfair teacher
Answer:
A
Step-by-step explanation:
1.9 about 2, and 4.4 is about 4.
2x4=8
So it is safe to say between 4 and 10.
Answer:
no solution
not sure but you can use math
way to solve
We are given these three people age below:
- Jim's age
- Carla's age
- Tomy's age
We define the age of Jim as any variable, because the problem doesn't give any specific age. I will define Jim's age as x.
Next, Carla is 5 years older than Jim. That means if Carla is 5 years older, her age would be x+5.
Then Tomy is 6 years older than Carla. That means the age would be 6+(x+5).
The sum of their three ages is 31 years old. That means we add all these ages and equal to 31.
Combine like terms and solve for x.
Then we substitute the value of x in ages to find these three people ages.
- Jim's age = x = 5
- Carla's age = x+5 = 5+5 = 10
- Tomy's age = 6+(x+5) = 6+(5+5) = 6+10 = 16.
Answer
- Jim's age = 5
- Carla's age = 10
- Tomy's age = 16
Answer:
95% confidence interval for the difference between the average mass of eggs in small and large nest is between a lower limit of 0.81 and an upper limit of 2.39.
Step-by-step explanation:
Confidence interval is given by mean +/- margin of error (E)
Eggs from small nest
Sample size (n1) = 60
Mean = 37.2
Sample variance = 24.7
Eggs from large nest
Sample size (n2) = 159
Mean = 35.6
Sample variance = 39
Pooled variance = [(60-1)24.7 + (159-1)39] ÷ (60+159-2) = 7619.3 ÷ 217 = 35.11
Standard deviation = sqrt(pooled variance) = sqrt(35.11) = 5.93
Difference in mean = 37.2 - 35.6 = 1.6
Degree of freedom = n1+n2 - 2 = 60+159-2 = 217
Confidence level = 95%
Critical value (t) corresponding to 217 degrees of freedom and 95% confidence level is 1.97132
E = t×sd/√(n1+n2) = 1.97132×5.93/√219 = 0.79
Lower limit = mean - E = 1.6 - 0.79 = 0.81
Upper limit = mean + E = 1.6 + 0.79 = 2.39
95% confidence interval for the difference in average mass is (0.81, 2.39)
