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vampirchik [111]
2 years ago
13

Please hurry ASAP One ride on a city bus costs $1.50. Martina has $18 on her bus pass. Write and solve an equation to find how m

any rides she can take without loading more money on her bus pass,
r= ?
She can ride the bus ? times before she needs to load more money on her pass.
Mathematics
2 answers:
bogdanovich [222]2 years ago
6 0

Answer:

She can ride the bus 12 times before she needs to load more money on  her pass.

Step-by-step explanation

1. You would take 18 and divide it by 1.5 (1.50) to get 12.

seraphim [82]2 years ago
4 0

Answer:

it,s 12

Step-by-step explanation:

you keep on minus 1.50 from 18 e.g. 17.50, 16, 15.50, 14, 13.50, 12

another e.g. 1.50[_18

                       = 12

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I don't really understand but, the Y coordinate is always on the right side so therefore y= -2, and 1

(xYz)

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Edward can run 1/2 mile in 300 seconds what is edwards unit rate
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Rate/ speed = distance / time;
= 1/2mile / 300 seconds;
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Can any one figure this out?
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A sample size 25 is picked up at random from a population which is normally
Margarita [4]

Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

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

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

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

Mean of 100 and variance of 36.

This means that \mu = 100, \sigma = \sqrt{36} = 6

Sample of 25:

This means that n = 25, s = \frac{6}{\sqrt{25}} = 1.2

(a) P(X<99)

This is the pvalue of Z when X = 99. So

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

By the Central Limit Theorem

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

Z = \frac{99 - 100}{1.2}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

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

Z = \frac{100 - 100}{1.2}

Z = 0

Z = 0 has a pvalue of 0.5

X = 98

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

Z = \frac{98 - 100}{1.2}

Z = -1.67

Z = -1.67 has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

So

P(98 < X < 100) = 0.4525

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