For this case, the first thing we must do is define variables.
We have then:
n: number of cans that each student must bring
We know that:
The teacher will bring 5 cans
There are 20 students in the class
At least 105 cans must be brought, but no more than 205 cans
Therefore the inequation of the problem is given by:
Answer:
105 <u><</u> 20n + 5 <u><</u> 205
the possible numbers n of cans that each student should bring in is:
105 <u><</u> 20n + 5 <u><</u> 205
$2.6, multiply 13 by .2 and you get 2.6
4
You can just put this into a calculator:
1/2^2 = 0.25
3×5-3 = 12
1^3 = 1
0.25 × 12 + 1 = 4
Answer:
<u>The approximate total weight of the grapefruits, using the clustering estimation technique is B. 35 ounces.</u>
Step-by-step explanation:
We notice that the weights of the grapefruits given are slightly down or above 7, then we will use <em>7 as our cluster</em> for the estimation, as follows:
Weights
7.47 ⇒ 7
7.23 ⇒ 7
6.46 ⇒ 7
7.48 ⇒ 7
6.81 ⇒ 7
<u>Now we can add up 7 + 7 + 7 + 7 + 7 for the weights of the grapefruits and the approximate total weight is B. 35 ounces.</u>
Answer:
The p-value is 0.0229
Step-by-step explanation:
With 
we have an upper-tail alternative. Because the p-value is defined as the probability of getting a value at least as extreme as the value observed. The observed value is given by the test statistic z = 1.997 which comes from a standard normal distribution. Therefore, we compute the p-value in the following way P(Z > 1.997) = 0.0229, i.e., the p-value is 0.0229
