Answer:
it moves four places to the right, becoming 4,500
2 ( x - 4) is the answer if that is shown out of the selections you can make.
Answer:
- The set with x ≤ 0 and y ≤ 0 is not a vector space.
- The 10th axiom does not hold
Step-by-step explanation:
Let u, v and w be vectors in the vector space V, and let c and d be scalars. A set S is defined as a vector space if it satisfies the following conditions:
- 1.
- 2. v + w = w + v
- 3. (u + v) + w = u + (v + w)
- 4. v + 0 = v = 0 + v
- 5. v + (−v) = 0
- 6. 1v = v
- 8. c(v + w) = cv + cw
- 9. (c + d)v = cv + dv
- 10.
Given the set of all vectors x, y in with x ≤ 0 and y ≤ 0
If the scalar c is such that .
Therefore, the 10th axiom is not satisfied and thus the set with x ≤ 0 and y ≤ 0 is not a vector space.
Answer:
9,000 bottle caps.
Step-by-step explanation:
The thousands place is represented by the first 8 in this number. When we round, we round up because 880 is closer to 1,000 than to 0. This means that the number of bottle caps is closer to 9,000 than it is to 8,000. We need to gain 120 bottle caps to reach 9,000. On the other hand we need to lose 880 to reach 8,000. 120 is a smaller number than 880.
Answer:
Step-by-step explanation:
Hello!
The variable of interest is
X: the amount a 10-11-year-old spends on a trip to the mall.
Assuming that the variable has a normal distribution, you have to construct a 98% CI for the average of the amount spent by the 10-11 year-olds on one trip to the mall.
For this you have to use a Student-t for one sample:
X[bar] ± *
n= 6
$18.31, $25.09, $26.96, $26.54, $21.84, $21.46
∑X= 140.20
∑X²= 3333.49
X[bar]= ∑X/n= 140.20/6= 23.37
S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/5[3333.49-(140.2)²/6]= 11.50
S= 3.39
X[bar] ± *
[23.37 ± 3.365 * ]
[18.66;29.98]
With a confidence level of 98%, you'd expect that the interval $[18.66;29.98] will include the population mean of the money spent by 10-11 year-olds in one trip to the mall.
I hope you have a SUPER day!