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Genrish500 [490]
3 years ago
12

Andre makes a three-digit number.

Mathematics
1 answer:
s2008m [1.1K]3 years ago
8 0

Answer:  115,151,115,133,313,331

Step-by-step explanation:

The Andre's  number can consist from 1+1+5 or 3+3+1. There are no any other sets of 3 odd digit to get 7.

Lets prove this statement.

Lets 1 of the digit is bigger than 5. However the digit is odd so it can be 7 only. However in this case the residual 2 digits are 0 . This is not possible so the gigits are odd however 0 is even.

Lets check the case when the biggest digit in the set is smaller than 3.

So it can be 1 only.

So the residual 2 digits can be 1 only. The sum of 1+1+1<7 .

SO we've  prooven that the only sets of the digits are 1;1;5 or 3;3;1

The set   1;1;5 can give 3 numbers:

115,151,115

The set   1;3;3 can give 3 numbers as well:

133,313,331

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Answer:

Step-by-step explanation:

Hello!

To study if the average dollars spend per shopper is different than the expected population average of $84.5 there was a sample of 120 shoppers taken and their spending habit registered.

Then the study variable is:

X: Amount of money spent by a shopper ($)

This population standard deviation is σ= $16

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X~N(μ;σ²)

I've also calculated the sample mean and standard deviation:

X[bar]= $81.01

S= $19.62

a.

If the objective is to test that the true mean spending for the population is significantly different than the expected average of 84.5, the parameter of interest is the population mean μ and the hypothesis test will be two-tailed.

H₀: μ = 84.5

H₁: μ ≠ 84.5

α:0.05

The statistic is a standard normal for one sample:

Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)

Z_{H_0}= \frac{81.01-84.5}{\frac{16}{\sqrt{120} } } = -2.389

p-value two tailed  0.016894

b.

The decision rule using the p-value method is:

If p-value ≤ α ⇒ Reject the null hypothesis.

If p-value > α ⇒ Do not reject the null hypothesis.

Since the p-value is less than the significance level, the decision is to reject the null hypothesis.

c.

Using a level of significance of 5% the null hypothesis was rejected. So there is enough statistical evidence to support the researcher's claim that the true mean spending for the population is significantly different than the expected average of $84.5.

I hope it helps!

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