68.5. You take 62 and divide it by the percentage as a decimal, .905. 62 / .905 = 68.5.
Answer:
3
Step-by-step explanation:
Answer:
There is strong evidence that less than 87% of the orders are delivered in less than 10 minutes.
Decision rule: Reject the null hypothesis if (P-value < level of significance)
Test statistic z=-2.70
Decision: Reject the null hypothesis (0.003<0.010)
Step-by-step explanation:
In this question we have to test an hypothesis.
The null and alternative hypothesis are:

The significance level is assumed to be 0.01.
The sample of size n=80 gives a proportion of p=61/80=0.7625.
The standard deviation is:

The statistic z is then

The P-value is

The P-value (0.003) is smaller than the significance level (0.010), so the effect is significant. The null hypothesis is rejected.
There is strong evidence that less than 87% of the orders are delivered in less than 10 minutes.
Decision rule: Reject the null hypothesis if (P-value < level of significance)
Test statistic z=-2.70
Decision: Reject the null hypothesis (0.003<0.010)
I believe the answer is B, Oval.
Answer:
40$
Step-by-step explanation:
they have
20$
5%
40hours
i=prt
multiplication of both
=40×20×5/100
=40$
the amount required at the week is 40$