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liubo4ka [24]
3 years ago
8

Kim orders catering from Midtown Dinerfor $35. She spends $5 on a large order of potato salad and the rest on turkey sandwich. E

ach sandwich is 2.50. How many sandwiches Kim buy?
Mathematics
1 answer:
aivan3 [116]3 years ago
3 0

Answer:

Kim had bought 12 sandwiches.

Step-by-step explanation:

Given:

Amount spend at Midtown Dine = $35

Amount spend on large order of potato salad = $5

Cost of each sandwich = $2.5

We need to find the number of sandwich did Kim bought.

Solution:

Let the number of sandwich Kim bought be 'x'.

Now we know that;

Amount spend at Midtown Dine is equal to sum of Amount spend on large order of potato salad and number of sandwich Kim bought multiplied by Cost of each sandwich.

framing in equation form we get;

5+2.5x=35

Subtracting both side by 5 we get;

5+2.5x-5=35-5\\\\2.5x=30

Dividing both side by 2.5 we get;

\frac{2.5x}{2.5}=\frac{30}{2.5}\\\\x=12

Hence Kim had bought 12 sandwiches.

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Suppose that one-way commute times in a particular city are normally distributed with a mean of 15.43 minutes and a standard dev
vovikov84 [41]

Answer:

Yes, a commute time between 10 and 11.8 minutes would be unusual.

Step-by-step explanation:

A probability is said to be unusual if it is lower than 5% of higher than 95%.

We use the normal probability distribution to solve this question.

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.

In this problem, we have that:

\mu = 15.43, \sigma = 2.142

Would it be unusual for a commute time to be between 10 and 11.8 minutes?

The first step to solve this problem is finding the probability that the commute time is between 10 and 11.8 minutes. This is the pvalue of Z when X = 11.8 subtracted by the pvalue of Z when X = 10. So

X = 11.8

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

Z = \frac{11.8 - 15.43}{2.142}

Z = -1.69

Z = -1.69 has a pvalue of 0.0455

X = 10

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

Z = \frac{10 - 15.43}{2.142}

Z = -2.54

Z = -2.54 has a pvalue of 0.0055

So there is a 0.0455 - 0.0055 = 0.04 = 4% probability that the commute time is between 10 and 11.8 minutes.

This probability is lower than 4%, which means that yes, it would be unusual for a commute time to be between 10 and 11.8 minutes.

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2 years ago
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