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Arlecino [84]
3 years ago
12

An animal shelter spends $3.00 per day to care for each bird and $8.50 per day to care for each cat. Kayla noticed that the shel

ter spent $123.00 caring for birds and cats on Tuesday. Kayla found a record showing that there were a total of 30 birds and cats on Tuesday. How many birds were at the shelter on Tuesday?
Mathematics
2 answers:
Lady bird [3.3K]3 years ago
8 0
$123 = ($3b + $8.50c)

b + c = 30

b = 24 birds ($72)
c = 6 cats ($51)

51+72=123

There were 24 birds at the shelter on Tuesday.
Irina-Kira [14]3 years ago
5 0
Thirty birds were at the shelter on Tuesday
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0.98 = 98% probability that the average midterm score of these students is at most 75 points.

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Problems of normal distributions can be solved using the z-score formula.

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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.

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The average midterm score of students in a certain course is 70 points.

This means that \mu = 70

29 students are randomly selected and the standard deviation of their scores is found to be 13.15 points.

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This is the pvalue of Z when X = 75. So

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Z = 2.05

Z = 2.05 has a pvalue of 0.98.

0.98 = 98% probability that the average midterm score of these students is at most 75 points.

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