How do i solve for this? I forgot

A countrys people consume 6.7 billion pounds of candy per year. Express this quantity in terms of pounds per month. Note that th

Vanyuwa [196]

The answer is 2 candies per person per month

(1) Find the amount of candy consumed per person per year:
6 700 000 000 ÷ 305 000 000 = 21.97 candies per person per year

(2) Find the amount of candy consumed per person per month:
1 year has 12 months.
21.97 candy per person per year ÷ 12 = 1.83 ≈ 2 candies per person per month

8 0

3 years ago

Write the point-slope form of the equation of the line that passes through the points (6, -9) and (7, 1). Include your work in y

lys-0071 [83]

First find the slope,
m=(y2-y1)/(x2-x1)=(1-(-9))/(7-6)=10
Next, use the point slope form to find the equation:
L : y-y1=m(x-x1) => y-1=10(x-7)
[in point slope form, point is (7,1), slope=m=10 ]

5 0

3 years ago

The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6.

Wittaler [7]

Answer:

a) There is a 10.75% probability of observing less than 60 hits in an hour.

b) The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) There is a 24% probability of observing between 80 and 90 hits an hour

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem, we have that

The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6, so $\mu = 75, \sigma = 8.6$ .