Some animals on farms eat hay to get energy. A cow can eat 24 pounds of hay each day. Write and evaluate an expression to find h
1 answer:
Answer:
<u>8064 pounds</u> of hay a group of 24 cows can eat in two weeks.
Step-by-step explanation:
Given:
Some animals on farms eat hay to get energy. A cow can eat 24 pounds of hay each day.
Now, to find how many pounds a group of 24 cows can eat in two weeks.
In 1 one day a cow can eat hay = 24 pounds.
Number of cows = 24.
Number of days in 2 weeks = 7 ×2 = 14.
Now, the expression of getting the pounds of hay a group of 24 cows can eat in two weeks is:
24 ( 24 × 14 ).
<u><em>So, to evaluate the expression:</em></u>
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Therefore, 8064 pounds of hay a group of 24 cows can eat in two weeks.
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Answer:
0.3557 = 35.57% probability that one selected subcomponent is longer than 118 cm.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Normally distributed with a mean of 116 cm and a standard deviation of 5.4 cm.
This means that
Find the probability that one selected subcomponent is longer than 118 cm.
This is 1 subtracted by the pvalue of Z when X = 118. So
has a pvalue of 0.6443
1 - 0.6443 = 0.3557
0.3557 = 35.57% probability that one selected subcomponent is longer than 118 cm.