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LenaWriter [7]
3 years ago
13

A total of 279 tickets were sold for the school play. they were either adult or student tickets. the number of student tickets s

old was two times the number of adult tickets sold. how many adult tickets were sold?
Mathematics
2 answers:
barxatty [35]3 years ago
8 0
S=2a

a+s=279, using a from above in this equation gives you:

a+2a=279  combine like terms on left side

3a=279  divide both sides by 3

a=93

So 93 adult tickets were sold.
Mice21 [21]3 years ago
6 0
A + s = 279
s = 2a

a + 2a = 279
3a = 279
a = 279/3
a = 93 <=== adult

s = 2a
s = 2(93)
s = 186 ....students
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Which expressions have a sum of -12 right answer only!!
kenny6666 [7]

Answer:

B and D

Step-by-step explanation:

2+-14 = -12

-4+(- 8) = -12

because its a negative number it wont go up it will go down

7 0
3 years ago
Read 2 more answers
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

6 0
3 years ago
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Furkat [3]

Answer:

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Hope this helps

3 0
3 years ago
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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serious [3.7K]
Step by step explaination:
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When exponent is -2, it equals: 1/25 or .04
6 0
3 years ago
Read 2 more answers
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