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3 years ago

Correct answers only please!

Mathematics

2 answers:

polet [3.4K]3 years ago

7 0

Answer:

B. 3×10^13

Step-by-step explanation:

The first two digits are fixed at one of three pairs of values; the last digit is "fixed" as the check digit, leaving 16 - 3 = 13 free digits. These can take on any of 10^13 values.

The total number of possible card numbers is 3×10^13.

Brrunno [24]3 years ago

4 0

Answer:

The correct option is B.

Step-by-step explanation:

Total possible digits are 0,1,2,3,4,5,6,7,8,9.

Total number of digits in a credit card = 16

It is given that that the last digit of credit card is fixed.

Total number of possible ways for 16th digit = 1

U-Kan-Trust-Us Credit Card numbers begin with 61, 62, or 63.

Total number of possible ways for first two digits = 3

Remaining places = 16 - 1 - 2 = 13

In these 13 remaining places any of 10 digits can be occur.

Total number of possible ways for remaining 13 digits = $10^{13}$

The maximum number of credit cards that UKTU can issue is

$Total =3\times (10^{13})\times 1$

$Total =3(10^{13})$

Therefore the correct option is B.

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The difference of two numbers is 3. Their sum is 13. Find the numbers

tekilochka [14]

Subtract 3 from 13 you get 10, divide that by two, you get 5, so one number is 5 and then add the three to the other 5, you get 8
The numbers are 5 & 8

6 0

3 years ago

Test the claim that the mean GPA of night students is larger than 2 at the .025 significance level. The null and alternative hyp

exis [7]

Answer:

$H_0: \, \mu = 2$ .

$H_1:\, \mu > 2$ .

Test statistics: $z \approx 2.582$ .

Critical value: $z_{1 - 0.025} \approx 1.960$ .

Conclusion: reject the null hypothesis.

Step-by-step explanation:

The claim is that the mean $\mu$ is greater than $2$ . This claim should be reflected in the alternative hypothesis:

$H_1:\, \mu > 2$ .

The corresponding null hypothesis would be:

$H_0:\, \mu = 2$ .

In this setup, the null hypothesis $H_0:\, \mu = 2$ suggests that $\mu_0 = 2$ should be the true population mean of GPA.

However, the alternative hypothesis $H_1:\, \mu > 2$ does not agree; this hypothesis suggests that the real population mean should be greater than $\mu_0= 2$ .

One way to test this pair of hypotheses is to sample the population. Assume that the population mean is indeed $\mu_0 = 2$ (i.e., the null hypothesis is true.) How likely would the sample (sample mean $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ ) be observed in this hypothetical population?

Let $\sigma$ denote the population standard deviation.

Given the large sample size $n = 60$ , the population standard deviation should be approximately equal to that of the sample:

$\sigma \approx s = 0.06$ .

Also because of the large sample size, the central limit theorem implies that $Z= \displaystyle \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}}$ should be close to a standard normal random variable. Use a $Z$ -test.

Given the observation of $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ :

$\begin{aligned}z_\text{observed}&= \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} \\ &\approx \frac{\overline{X} - \mu_0}{s / \sqrt{n}} = \frac{2.02 - 2}{0.06 / \sqrt{60}} \approx 2.582\end{aligned}$ .

Because the alternative hypothesis suggests that the population mean is greater than $\mu_0 = 2$ , the null hypothesis should be rejected only if the sample mean is too big- not too small. Apply a one-sided right-tailed $z$ -test. The question requested a significant level of $0.025$ . Therefore, the critical value $z_{1 - 0.025}$ should ensure that $P( Z > z_{1 - 0.025}) = 0.025$ .

Look up an inverse $Z$ table. The $z_{1 - 0.025}$ that meets this requirement is $z_{1 - 0.025} \approx 1.960$ .

The $z$ -value observed from the sample is $z_\text{observed}\approx 2.582$ , which is greater than the critical value. In other words, the deviation of the sample from the mean in the null hypothesis is sufficient large, such that the null hypothesis needs to be rejected at this $0.025$ confidence level in favor of the alternative hypothesis.