Find the value of X.

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7 0

3 years ago

A company sells boxes of duck calls (d) for $35 and boxes of turkey calls (t) for $45. they make batches of duck calls that fill

SIZIF [17.4K]

Answer:

Option A is the correct choice.

Step-by-step explanation:

Let d be the number of boxes of duck calls and t be the number of boxes of turkey calls.

We have been given that a company sells boxes of duck calls for $35 and boxes of turkey calls (t) for $45, so the revenue earned from selling d boxes of duck and t boxes of turkey call will be 35d and 45t respectively.

Further, the company plan to make $300. We can represent this information as:

$35d+45t=300...(1)$

We are also told that they make batches of duck calls that fill 6 boxes and batches of turkey calls that fill 8 boxes. the company only has 42 boxes. We can represent this information as:

$6d+8t=42...(2)$

$6d=42-8t...(2)$

Therefore, our desired system of equation will be:

$35d+45t=300...(1)$

$6d=42-8t...(2)$

8 0

3 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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:

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

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 pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$