Answer:
16 spoonfuls
Step-by-step explanation:
Round off the 1 1/8 oz spoon capacity to 1 oz. Then divide 16 oz by 1 oz, obtaining the estimate 16 spoonfuls.
Is this for real? Lol anyways I’m guessing B
Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by:
.
The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:


Z = 1.71
Z = 1.71 has a p-value of 0.9564.
1 - 0.9564 = 0.0436.
0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.
More can be learned about the normal distribution at brainly.com/question/24663213
#SPJ1
Step-by-step explanation:
First term(a) = 1
Common difference(d)=2-1=1
No.of term(n)=50
Sn=n÷2(n+1)=50÷2(50+1)=1275