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Masteriza [31]
2 years ago
14

You are the senior class president and are selling items for a school fundraiser. You have cell phone cases and t-shirts that ha

ve the school logo on
them for sale. Each case costs $12, and each t-shirt costs $7 After selling a total of 60 items, you have made a total of $450. How many cases and t-
shirts were sold?
Mathematics
1 answer:
andrew11 [14]2 years ago
3 0

The $450 of total amount made and the total number of items sold of 60

gives the items sold as 54 T-shirts and 6 cases.

Response:

  • 54 T-shirts sold and 6 cases where sold

<h3>Which method can be used to find the number of T-shirts and cases sold?</h3>

The given parameters are;

The cost of each case = $12

Cost of each T-shirt = $7

Number of items sold = 60

The amount made = $450

Required;

The number of shirt and T-shirt sold

Solution:

Let <em>x</em> represent the number of T-shirt sold and let <em>y</em> represent the number

of cases sold, we have the following simultaneous equation;

  • 7·x + 12·y = 450
  • x + y = 60

Which gives;

y = 60 - x

7·x + 12·(60 - x) = 450

7·x + 12 × 60 - 12·x = 450

5·x = 12 × 60 - 450 = 270

x = \dfrac{270}{5} = 54

x = 54

  • The number of T-shirts sold, x = <u>54</u>
  • The number of cases sold, y = 60 - 54 =<u> 6</u>

Learn more about simultaneous equations here:

brainly.com/question/904961

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P(X&lt; ) 1-P(X&gt; ) A softball pitcher has a 0.626 probability of throwing a strike for each curve ball pitch. If the softball
Mashutka [201]

Answer:

19.49% probability that no more than 16 of them are strikes

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 30, p = 0.626

So

\mu = E(X) = np = 30*0.626 = 18.78

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{30*0.626*0.374} = 2.65

What is the probability that no more than 16 of them are strikes?

Using continuity correction, this is P(X \leq 16 + 0.5) = P(X \leq 16.5), which is the pvalue of Z when X = 16.5. So

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

Z = \frac{16.5 - 18.78}{2.65}

Z = -0.86

Z = -0.86 has a pvalue of 0.1949

19.49% probability that no more than 16 of them are strikes

7 0
3 years ago
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