Answer:
Step-by-step explanation:
We want to determine a 95% confidence interval for the mean salary of all graduates from the English department.
Number of sample, n = 400
Mean, u = $25,000
Standard deviation, s = $2,500
For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean ± z × standard deviation/√n
It becomes
25000 ± 1.96 × 2500/√400
= 25000 ± 1.96 × 125
= 25000 ± 245
The lower end of the confidence interval is 25000 - 245 =24755
The upper end of the confidence interval is 25000 + 245 = 25245
Therefore, with 95% confidence interval, the mean salary of all graduates from the English department is between $24755 and $25245
You will have infinite solutions
500 let me know if it’s right i’m not the best at math
We know that the angles of a triangle sum to 180°. For ΔABC, this means we have:
(4x-10)+(5x+10)+(7x+20)=180
Combining like terms,
16x+20=180
Subtracting 20 from both sides:
16x=160
Dividing both sides by 16:
x=10
This means ∠A=4*10-10=40-10=30°; ∠B=5*10+10=50+10=60°; and ∠C=7*10+20=70+20=90.
For ΔA'B'C', we have
(2x+10)+(8x-20)+(10x-10)=180
Combining like terms,
20x-20=180
Adding 20 to both sides:
20x=200
Dividing both sides by 20:
x=10
This gives us ∠A'=2*10+10=20+10=30°; ∠B'=8*10-20=80-20=60°; and ∠C'=10*10-10=100-10=90°.
Since the angle are all congruent, ΔABC~ΔA'B'C' by AAA.
Answer:
Well, I don't think I can remember EVERY skill I was taught through the years but in Kindergarten we used Number Sense so we could accurately... also there's estimation and of course PEMDAS. I've also learned how to round up so that the solution is at least around those numbers if the skill is to be applied.
Step-by-step explanation:
I'm not sure if this'll help but go ahead and have it... just try and change it a bit so they don't get ya for plagirism
