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Tomtit [17]
2 years ago
9

1. how do you use a two-way frequency table to find possible associations and trends in data?

Mathematics
1 answer:
maw [93]2 years ago
8 0
I think by <span> Using a row conditional relative frequency</span>
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A store manager wishes to reduce the price on her fresh ground coffee by mixing two grades. If she has 35 pounds of coffee which
kari74 [83]

35 lbs, your welcome even though you probably don't need the answer anymore

8 0
2 years ago
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Find the quotient.
Strike441 [17]

Answer:

Step-by-step explanation:

Find the quotient.

(4.86 × 109)

(2.43 × 103)

What is the exponent of the power in the quotient?

6

What is the coefficient in the quotient?

2  

Which operation produced the exponent of the power in the quotient?

Subtracting the original exponents

6 0
2 years ago
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What is the first term of the fibonacci like sequence whose second term is 4 and whose fith term is 22?
bagirrra123 [75]

Answer: -2

Step-by-step explanation:

The information in the question can be used to form an equation which goes thus:

2nd term = a + d = 4 ...... i

5th term = a + 4d = 22 ....... ii

From equation i

a = 4 - d ........ iii

Put equation iii into ii

a + 4d = 22

(4 - d) + 4d = 22

4 - d + 4d = 22

4 + 3d = 22

3d = 22 - 4

3d = 18

d = 18/3

d = 6

Commons difference is 6

Since a + d = 4

a + 6 = 4

a = 4 - 6

a = -2

The first term is -2

3 0
2 years ago
A human resources manager keeps a record of how many years each employee at a large company has been working in their current ro
REY [17]

Answer:

0.3085 = 30.85% probability that the mean years of experience from the sample of 4 is greater than 3.5 years.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 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 proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Distribution of years of experience:

Mean 3, so \mu = 3

Standard deviation 2, so \sigma = 2

Sample of 4:

n = 4, s = \frac{2}{\sqrt{4}} = 1

What is the probability that the mean years of experience from the sample of 4 is greater than 3.5?

1 subtracted by the pvalue of Z when X = 3.5. So

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

By the Central Limit Theorem

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

Z = \frac{3.5 - 3}{1}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915

1 - 0.6915 = 0.3085

0.3085 = 30.85% probability that the mean years of experience from the sample of 4 is greater than 3.5 years.

3 0
2 years ago
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You and a friend play a game where you each toss a balanced coin. If the upper faces on the coins are both tails, you win $1; if
oksian1 [2.3K]

Answer:  The mean and variance of Y is $0.25 and $6.19 respectively.

Step-by-step explanation:

Given : You and a friend play a game where you each toss a balanced coin.

sample space for tossing two coins : {TT, HT, TH, HH}

Let Y denotes the  winnings on a single play of the game.

You win $1; if the faces are both heads

then P(Y=1)=P(TT)=\dfrac{1}{4}=0.25

You win $6; if the faces are both heads

then P(Y=6)=P(HH)=\dfrac{1}{4}=0.25

You loose $3; if the faces do not match.

then P(Y=1)=P(TH, HT)=\dfrac{2}{4}=0.50

The expected value to win : E(Y)=\sum_{i=1}^{i=3} y_ip(y_1)

=1\times0.25+6\times0.25+(-3)\times0.50=0.25

Hence, the mean of Y : E(Y)= $0.25

E(Y^2)=\sum_{i=1}^{i=3} y_i^2p(y_i)\\\\=1^2\times0.25+6^1\times0.25+(-3)^2\times0.5\\\\=0.25+1.5+4.5=6.25

Variance = E[Y^2]-E(Y)^2

=6.25-(0.25)^2=6.25-0.0625=6.1875\approx6.19

Hence, variance of Y = $ 6.19

6 0
2 years ago
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