1. The county fair charges $1.50 per ticket for the rides. Jermaine bought 20 tickets for the rides and spent a

Mathematics

1 answer:

makvit [3.9K]3 years ago

4 0

Cost per ticket for the rides = $1.50.

a) Let us assume the number of ride tickets = x.

Total cost of x number of rides = y.

Total amount spent = $48.75.

Amount spent on 20 tickets = 20 × 1.50 = $30

<em>Fair admission charge = Total amount spent - Amount spent on 20 tickets</em>

Let us apply slope-intercept form y=mx+b to get the equation.

Total cost of x number of tickets at the rate $1.50 each ticket = 1.50x.

To<em>tal cost of fair = Total cost of x number of tickets + Fair admission charge.</em>

You might be interested in

And the last question is:

Leya [2.2K]

It should be " b " That should be right but u may want to wait for others to respond to this to make sure

6 0

3 years ago

Read 2 more answers

The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability

amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

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

Also, important to remember that the standard deviation is the square root of the variance.

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 $s = \frac{\sigma}{\sqrt{n}}$