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Sergeeva-Olga [200]
2 years ago
10

An owner of a small store knows that in the last week 52 customers paid with cash, 42 paid with a debit card, and 153 paid with

a credit card. Based on the number of customers from last week, which fraction is closest to the probability that the next customer will pay with cash?
A. 1/5

B.1/4

C.1/3

D.1/2
Mathematics
2 answers:
Pie2 years ago
8 0

Answer:

around  1/4

Step-by-step explanation:

Alexxandr [17]2 years ago
6 0
The answer is B ...............
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ISSS THISSSSS CORRECTTT??????please help
LenaWriter [7]

Answer:

I think that is correct, sorry is i get it wrong

7 0
2 years ago
30 tens - 3 tens 3 tenths
lubasha [3.4K]
300-30.3=269.7 because you barrow from 3 in 300 because you can't round into 0 and that equalled 7 and than you do 9-0=9 and there's 009.7 than you do 9-3=6 and that's 069.7 and you do 2-0=2 and there's 269.7
4 0
3 years ago
A bacteria culture begins with 15 bacteria which double in amount at the end of every hour. How many bacteria are grown during t
Eva8 [605]
The answer is 30720

A = P * rⁿ
A - the final population
r - the growth rate
P - the initial number of bacteria
n - the number of hours

We know:
A = ?
r = 2 (it doubles)
P = 15
n = 12

After 11 hours
A(11) = 15 * 2¹²
A(11) = 15 * 2048
A(11) = 30720

After 12 hours
A(12) = 15 * 2¹²
A(12) = 15 * 4096
A(12) = 61440

<span>How many bacteria are grown during the 12th hour?
A(12) - A(11) = 61440 - 30720 = 30720</span>
8 0
3 years ago
Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample
fomenos

Answer:

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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.

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}}

37% of the company's orders come from first-time customers.

This means that p = 0.37

A random sample of 225 orders will be used to estimate the proportion of first-time-customers.

This means that n = 225

Mean and standard deviation:

\mu = p = 0.37

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37*0.63}{225}} = 0.0322

What is the probability that the sample proportion is between 0.26 and 0.38?

This is the pvalue of Z when X = 0.38 subtracted by the pvalue of Z when X = 0.26.

X = 0.38

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

By the Central Limit Theorem

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

Z = \frac{0.38 - 0.37}{0.0322}

Z = 0.31

Z = 0.31 has a pvalue of 0.6217

X = 0.26

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

Z = \frac{0.26 - 0.37}{0.0322}

Z = -3.42

Z = -3.42 has a pvalue of 0.0003

0.6217 - 0.0003 = 0.6214

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

5 0
2 years ago
Hey can you please me asap!
crimeas [40]

Answer:

It's D

Step-by-step explanation:


6 0
3 years ago
Read 2 more answers
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