Answer: To find the volume of a cube, use the formula V=lwh, or Volume equals length times width times height.
Answer:
s = 71 mph, and s+9 = f = 80 mph
Step-by-step explanation:
The distances the two busses travel add up to 604 mi.
Letting f be the faster speed and s the slower. Then f = s + 9 (mph).
Then (s + 9)(mph)(4 hr) + (s)(mph)(4 hr) = 604 mi. Solve this for s:
4s + 36 + 4s = 604 mi, or 8s = 568. Finally, s = 71 mph, and s+9 = f = 80 mph.
Answer:

Step-by-step explanation:



or

The valid conclusions for the manager based on the considered test is given by: Option
<h3>When do we perform one sample z-test?</h3>
One sample z-test is performed if the sample size is large enough (n > 30) and we want to know if the sample comes from the specific population.
For this case, we're specified that:
- Population mean =
= $150 - Population standard deviation =
= $30.20 - Sample mean =
= $160 - Sample size = n = 40 > 30
- Level of significance =
= 2.5% = 0.025 - We want to determine if the average customer spends more in his store than the national average.
Forming hypotheses:
- Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get:

- Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically

where
is the hypothesized population mean of the money his customer spends in his store.
The z-test statistic we get is:

The test is single tailed, (right tailed).
The critical value of z at level of significance 0.025 is 1.96
Since we've got 2.904 > 1.96, so we reject the null hypothesis.
(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).
Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.
Learn more about one-sample z-test here:
brainly.com/question/21477856
Answer:
After finding the prime factorization of $2010=2\cdot3\cdot5\cdot67$, divide $5300$ by $67$ and add $5300$ divided by $67^2$ in order to find the total number of multiples of $67$ between $2$ and $5300$. $\lfloor\frac{5300}{67}\rfloor+\lfloor\frac{5300}{67^2}\rfloor=80$ Since $71$,$73$, and $79$ are prime numbers greater than $67$ and less than or equal to $80$, subtract $3$ from $80$ to get the answer $80-3=\boxed{77}\Rightarrow\boxed{D}$.
Step-by-step explanation:
hope this helps