600 ---> 250
600 - 250 = 350
350 / 5 = 70
Average decrease in value per year: $70
Answer:
Perimeter.
Step-by-step explanation:
Lee measures around the entire outside of the bulletin board and finds the distance is 32 units.
She is measuring the border of her bulletin board.
The sum of outer covering of any object is called its perimeter. Here, 32 unit shows the perimeter of the bulletin board.
240,000 is a whole number. Now do the power of ten lol. Use a calculator. Remember don't do it all at once. The powers mean multiplying separate for example: 10×10 equals 100. I hundred times10 equals 1000. That is powered by three because it is multiplying 1030 different times which makes it the power. Simple math these days ha ha
Answer:
Step-by-step explanation:
Let x represent the number of youth tickets purchased at the zoo.
Let y represent the number of adult tickets purchased at the zoo.
At the zoo, youth tickets cost $5 and adult tickets cost $9. A group spent a total of $90 on tickets. This means that
5x + 9y = 90
The domain of the relationship is the possible set of values of x and y that satisfies the equation. The domain of this relationship is
0 ≤ x ≤ 18
At x = 0, it means only adult tickets were purchased.
At x = 18, it means only youth tickets were purchased.
Answer:
We reject H₀. We support that the new average credit card debt is bigger than the previous average
Step-by-step explanation:
Five years earlier
μ = 8189
σ = 690
Sample size n = 32
Recent year debt
x = 8776
Sample size n = 32
a) Hypothesis Test:
Null Hypothesis H₀ x = μ = 8189
Alternative Hypothesis Hₐ x > μ
b) z(c) Alternative Hypothesis establishes that the test is a one tail-test to the right.
z(c) for significance level α = 0.05 is from z-table z(c) = 1,64
c) z(s) = ( x - μ ) / σ /√n
z(s) = ( 8776 - 8189 ) / 690 /√32
z(s) = 587 *5,66/ 690
z(s) = 4,81
d) Comparing z(c) and z(s)
z(s) > z(c) Then z(c) is in the rejection region and we reject H₀
e) We have evidence that at 95 % of confidence the new value for the debt in credit card is now bigger than the average
