Show that (2, 1) is a solution of the system of equations.
x + 3y = 5,
y = –x + 3

1. Substitute (2, 1) into x + 3y = 5 to get .____
Options: 1 + 32 = 5, 1 + 3(2) = 5, 2 + 31 = 5, 2 + 3(1) = 5

2. Simplify the equation to get ____
Options: 5 = 5 is true, 7 = 7 is true, 7 = 5 is false, 33 = 5 is false

3. Substitute (2, 1) into y = –x + 3 to get ____
Options: 1 = -(2) + 3, 1 = -(2 + 3), 2 = -(1) + 3, 2 = -(1 + 3)

4. Simplify the equation to get .____
Options: 1 = 1 is true, 1 = -1 is false, 2 = 2 is true, 2 = -2 is false

Answer:

30 square inch

Step-by-step explanation:

[tex]area \: of \: base = 6 \times 5 = 30 \: {inch}^{2} \\ [/tex]

Answer:

2/5

Step-by-step explanation:

The total number of candies are 2+1+2 = 5 candies

P (peppermint) = number of peppermints/total

                        =2/5

Answer:

2/5

Step-by-step explanation:

The probability is 2/5.There are five in all and two peppermint.Put it as a fraction and you get 2/5.

Answer:

a) 0.214 or 21.4%

b) P=0.011

c) The realtor should sample at least 551 homes.

Step-by-step explanation:

The current thinking is that housing prices follow an approximately normal model with mean $238,000 and standard deviation $5,041.

a) We need to know the proportion of housing prices in Athens that are less than $234,000. We can calculate this from the z-score for the population distribution.

[tex]z=\dfrac{X-\mu}{\sigma}=\dfrac{234,000-238,000}{5,041}=\dfrac{-4,000}{5.041}=-0.793\\\\\\ P(x<234,000)=P(z<-0.793)=0.214[/tex]

The proportion of housing prices in Athens that are less than $234,000 is 0.214.

b) Now, a sample is taken. The size of the sample is n=134.

We have to calculate the probability that the average selling price is greater than $239,000.

In this case, we have to use the standard error of the sampling distribution to calculate the z-score:

[tex]z=\dfrac{\bar x-\mu}{\sigma/\sqrt{n}}=\dfrac{239,000-238,000}{5,041/\sqrt{134}}=\dfrac{1,000}{435.476}= 2.296 \\\\\\P(\bar x>239,000)=P(z>2.296)=0.011[/tex]

The probability that the average selling price is greater than $239,000 is 0.011.

c) We have another sample taken from a distribution with the same parameters.

We have to calculate the sample size so that the margin of error for a 98% confidence interval is $500.

The expression for the margin of error of the confidence interval is:

[tex]E=z\cdot \sigma/\sqrt{n}[/tex]

We can isolate n from the margin of error equation as:

[tex]E=z\cdot \sigma/\sqrt{n}\\\\\sqrt{n}=\dfrac{z\cdot \sigma}{E}\\\\n=(\dfrac{z\cdot \sigma}{E})^2[/tex]

We have to look for the critical value of z for a 98% CI. This value is z=2.327.

Now we can calculate the minimum value for n to achieve the desired precision for the interval:

[tex]n=(\dfrac{z\cdot \sigma}{E})^2\\\\\\n=(\dfrac{2.327*5,041}{500})^2= 23.461 ^2=550.410\approx551[/tex]

The realtor should sample at least 551 homes.

Answer:

a) 0.214 or 21.4%

b) P=0.011

c) The realtor should sample at least 551 homes

Step-by-step explanation:

Answer:

0.7

Step-by-step explanation:

0.3+0.7=1.0=100%

Final answer:

The probability of not winning the game that Martin is playing is 0.7 or 70%, which is obtained by subtracting the probability of winning (0.3) from 1.

Explanation:

If Martin is playing a game where the probability of winning is 0.3, then the probability of not winning can be calculated by subtracting the probability of winning from 1. This is because the sum of the probabilities of all possible outcomes must equal 1. Since the probability of winning is 0.3, we calculate the probability of not winning as follows:

Therefore, the probability of not winning is 0.7 or 70%.

Answer:

it is 61.3

Step-by-step explanation:

Answer:

Approximately 20,579 units.

Answer:

y =20

Step-by-step explanation:

5(y+4)=6y

Distribute

5y +20 = 6y

Subtract 5y from each side

5y-5y+20=6y-5y

20 =y

Answer:

solution

5y+20=6y

5y-6y=20

-y=20

the answer is 89

Step-by-step explanation:

it does not matter if the number is negative the absolute value is the number inside the lines

Answer:

The absolute value of this one is 89. Because for example: |-3|=3 because any number is in that sign || the number will turn to positive. For example, If it is |-3| it will turn to 3

Answer:

To find why the crimes were committed

It's a parallelogram, opposite sides congruent.

6x - 7 = 2x + 9

4x = 16

x = 4

12 = y + 3

9 = y

Answer: x=4, y=9

Answer:

x = 4 and y = 9

Step-by-step explanation:

This is a parallelogram, which we can tell because of the arrows. Basically, opposite sides are parallel. By definition, then, opposite sides of this polygon are equal: LM = ON and LO = MN. That means we can set the various expressions equal to each other:

LM = ON  ⇒  6x - 7 = 2x + 9  ⇒  4x = 16  ⇒  x = 4

LO = MN  ⇒  12 = y + 3  ⇒  y = 9

Thus, x = 4 and y = 9.

Hope this helps!

Answer:

8b+7p=20

p=2b

Step-by-step explanation:

You're welcome. Thou shall complete thou work without any trouble.

Answer:

a. [tex]\mu=\bar x =14.6[/tex]

b. The 95% CI for the population mean is (14.22, 14.98).

c. B. "The statement is incorrect. A correct statement would be​"One can be​ 95% confident that the true mean number of people named per person will fall in the interval computed in part b"

d. D. It does not impact the validity of the interpretation because the sampling space of the sample mean is approximately normal according to the Central Limit Theorem.

Step-by-step explanation:

a) The sample mean provides a point estimation of the population mean.

In this case, the estimation of the mean is:

[tex]\mu=\bar x =14.6[/tex]

b) With the information of the sample we can estimate the

As the sample size n=2824 is big enough, we can aproximate the t-statistic with a z-statistic.

For a 95% CI, the z-value is z=1.96.

The sample standard deviation is s=10.3.

The margin of error of the confidence is then calculated as:

[tex]E=z\cdot s/\sqrt{n}=1.96*10.3/\sqrt{2824}=20.188/53.141=0.38[/tex]

The lower and upper limits of the CI are:

[tex]LL=\bar x-z\cdot s/\sqrt{n}=14.6-0.38=14.22\\\\UL=\bar x+z\cdot s/\sqrt{n}=14.6+0.38=14.98[/tex]

The 95% CI for the population mean is (14.22, 14.98).

c. "95% of the​ time, the true mean number of people named per person will fall in the interval computed in part b"

The right answer is:

B. "The statement is incorrect. A correct statement would be​"One can be​ 95% confident that the true mean number of people named per person will fall in the interval computed in part b"

The confidence interval gives bounds within there is certain degree of confidence that the true population mean will fall within.

It does not infer nothing about the sample means or the sampling distribution. It only takes information from a sample to estimate a interval for the population mean with certain degree of confidence.

d. It is unlikely that the personal network sizes of adults are normally distributed. In​ fact, it is likely that the distribution is highly skewed. If​ so, what​ impact, if​ any, does this have on the validity of inferences derived from the confidence​ interval?

The answer is:

D. It does not impact the validity of the interpretation because the sampling space of the sample mean is approximately normal according to the Central Limit Theorem.

The reliability of a confidence interval depends more on the sample size, not on the distribution of the population. As the sample size increases, the absolute value of the skewness and kurtosis of the sampling distribution decreases. This sample size relationship is expressed in the central limit theorem.

The point estimate for μ is 14.6. The confidence interval will provide the range where the true mean falls with 95% confidence. The Central Limit Theorem suggests that the deviation from the normal distribution will not significantly affect the answers.

a- The point estimate for μ is x=14.6. This is calculated as the mean of all measured values.

b- An interval estimate can be calculated with the formula: x ± Z*(s/√n) where Z is the Z-value from a Z-table corresponding to desired confidence level, here, 0.95. The result would give you the range in which the true mean, μ, falls with 95% confidence.

c- The correct answer is B: The statement is incorrect. A correct statement would be "One can be 95% confident that the true mean number of people named per person will fall in the interval computed in part b."

d- If the personal network sizes of adults are not normally distributed and the distribution is highly skewed, it will have an impact on the validity of inferences derived from the confidence interval. The correct answer is D: It does not impact the validity of the interpretation as the sampling space of the sample mean will still be approximately normal due to the Central Limit Theorem.

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The Answer is : ABC = 40

Answer:

To Prove: [tex]9e^x[/tex] is equal to the sum of its Maclaurin series.

Step-by-step explanation:

If [tex]f(x) = 9e^x[/tex], then [tex]f ^{(n + 1)(x)} =9e^x[/tex] for all n. If d is any positive number and   |x| ≤ d, then [tex]|f^{(n + 1)(x)}| = 9e^x\leq 9e^d.[/tex]

So Taylor's Inequality, with a = 0 and M = [tex]9e^d[/tex], says that [tex]|R_n(x)| \leq \dfrac{9e^d}{(n+1)!} |x|^{n + 1} \:for\: |x| \leq d.[/tex]

Notice that the same constant [tex]M = 9e^d[/tex] works for every value of n.

But, since [tex]lim_{n\to\infty}\dfrac{x^n}{n!} =0 $ for every real number x$[/tex],

We have [tex]lim_{n\to\infty} \dfrac{9e^d}{(n+1)!} |x|^{n + 1} =9e^d lim_{n\to\infty} \dfrac{|x|^{n + 1}}{(n+1)!} =0[/tex]

It follows from the Squeeze Theorem that [tex]lim_{n\to\infty} |R_n(x)|=0[/tex] and therefore [tex]lim_{n\to\infty} R_n(x)=0[/tex] for all values of x.

[tex]THEOREM\\If f(x)=T_n(x)+R_n(x), $where $T_n $is the nth degree Taylor Polynomial of f at a and $ lim_{n\to\infty} R_n(x)=0 \: for \: |x-a|<R, $then f is equal to the sum of its Taylor series on $ |x-a|<R[/tex]

By this theorem above, [tex]9e^x[/tex] is equal to the sum of its Maclaurin series, that is,

[tex]9e^x=\sum_{n=0}^{\infty}\frac{9x^n}{n!}[/tex]  for all x.

Answer:

a. length  = 0.7211 ft

b. width  = 0.7211 ft

c. height = 140.3846 ft

Step-by-step explanation:

This is an optimiztion with restriction problem.

We have to minimize the cost, with the restriction of the volume being equal to 72 ft3.

As the cost for the sides is constant, we know that length and width are equal.

Then, we can express the volume as:

[tex]V=x\cdot y\cdot z=x^2z=73[/tex]

being x: length and z: height

We can express the height in function of the length as:

[tex]x^2z=73\\\\z=73x^{-2}[/tex]

Then, the cost of the box can be expressed as:

[tex]C=0.62(x^2)+4*0.15(xz)+0.18(x^2)=(0.62+0.18)x^2+0.60xz\\\\C=0.8x^2+0.60x*x^{-2}=0.8x^2+0.6x^{-1}[/tex]

To optimize C, we derive and equal to zero

[tex]\dfrac{dC}{dx}=\dfrac{d}{dx}[0.8x^2+0.6x^{-1}]=1.6x-0.6x^{-2}=0\\\\\\1.6x=0.6x^{-2}\\\\x^{1+2}=0.6/1.6=0.375\\\\x=\sqrt[3]{0.375} =0.7211[/tex]

The height z is then

[tex]z=73x^{-2}=\dfrac{73}{0.7211^2}=\dfrac{73}{0.52}=140.3846[/tex]

Answer:

The sample size is smaller than 30, so we need to assume that the underlying population is normally distributed.

The  sampling distribution of x overbar will be approximately normally distributed with mean 63 and standard deviation 5.69.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem

The sample size is smaller than 30, so we need to assume that the underlying population is normally distributed.

If it is:

[tex]\mu = 63, \sigma = 18, n = 10, s = \frac{18}{\sqrt{10}} = 5.69[/tex]

The  sampling distribution of x overbar will be approximately normally distributed with mean 63 and standard deviation 5.69.

Answer:

a) 30.50% probability that  at least 150 stay on the line for more than one minute.

b) 0% probability that more than 200 stay on the line for more than one minute.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 800, p = 0.18[/tex]

So

[tex]\mu = E(X) = np = 800*0.18 = 144[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{800*0.18*0.82} = 10.87[/tex]

a. at least 150 stay on the line for more than one minute.

Using continuity correction, [tex]P(X \geq 150 - 0.5) = P(X \geq 149.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 149.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{149.4 - 144}{10.87}[/tex]

[tex]Z = 0.51[/tex]

[tex]Z = 0.51[/tex] has a pvalue of 0.6950

1 - 0.6950 = 0.3050

30.50% probability that  at least 150 stay on the line for more than one minute.

b. more than 200 stay on the line.

Using continuity correction, [tex]P(X \geq 200 + 0.5) = P(X \geq 200.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 200.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{200.5 - 144}{10.87}[/tex]

[tex]Z = 5.2[/tex]

[tex]Z = 5.2[/tex] has a pvalue of 1

1 - 1 = 0

0% probability that more than 200 stay on the line for more than one minute.

The P-value for this hypothesis test is 0.0228.

To find the P-value for this hypothesis test, we need to calculate the proportion of alumni who favored firing the coach in the sample. Out of 100 alumni, 64 were in favor. So, the sample proportion is 64/100 = 0.64.

Now, we need to calculate the test statistic, which follows a normal distribution. The formula for the test statistic is: z = (p' - p) / sqrt(p * (1-p) / n), where p' is the sample proportion, p is the claimed proportion under the null hypothesis, and n is the sample size.

Plugging in the values, we get: z = (0.64 - 0.50) / sqrt(0.50 * (1-0.50) / 100) = 2.00

The P-value is the probability of observing a test statistic as extreme as 2.00, assuming the null hypothesis is true. We can look up this probability in a standard normal distribution table or use a statistical software. In this case, the P-value is 0.0228.

Answer:

$834.8 dollars that week

Step-by-step explanation:

All you have to do is multiply $20.87 by 10 hours to get your answer:)

By multiplying Harper's hourly wage ($20.87) by 40 hours, we determined that Harper will earn $834.80 in a 40-hour workweek.

To calculate how much Harper will earn in a 40-hour work week, you simply need to multiply his hourly wage by the number of hours he works. In this case, that's $20.87 times 40. Using direct multiplication:

$20.87 x 40 = $834.80

So, Harper will earn $834.80 in a 40-hour workweek.

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The day when they would meet first time after joining the gym together will be 12.

The analysis of mathematical representations is algebra, and the handling of those symbols is logic.

Kirk goes to the gym every 3 days.

Deshawn goes to the gym every 4 days.

If they join the gym on the same day.

Then the day when they would meet first time after joining the gym together will be

LCM of 4, 3 will be 12.

Then the day will be 12.

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Final answer:

Kirk and Deshawn will be at the gym together on the 12th day since they joined.

Explanation:

Gym memberships for Kirk and Deshawn occur every 3 days and 4 days respectively. To find the first day they'll be at the gym together, we need to find the lowest common multiple of 3 and 4.

LCM(3, 4) = 12. Therefore, Kirk and Deshawn will be at the gym together on the 12th day since they joined the gym.

Answer:

x = 4

Step-by-step explanation:

14x - 15 + 139 = 180

(Alternate & Supplementary angles)

14x = 56

x = 4

Answer:

BC = 9

Step-by-step explanation:

Assuming this is a straight line

AB + BC = AC

2x-6 + x+2 = 17

Combine like terms

3x -4 = 17

Add 4 to each side

3x-4+4 = 17+4

3x = 21

Divide each side by 3

3x/3 =21/3

x =7

We want to find BC

BC =x+2

     =7+2

     =9

Answer:

The probability that none of the four cans selected contains an incorrect mix of paint is P=0.2545.

Step-by-step explanation:

We have 12 cans, out of which 3 are defective (incorrect mix of paint).

The man will choose 4 cans to paint his mother's house living room.

Let x = the number of the paint cans selected that are defective.

The variable x is known to follow a hypergeometric distribution.

The probability of getting k=0 defectives in a selected sample of K=4 cans, where there are n=3 defectives in the population of N=12 cans is:

[tex]P(X=k)=\dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}\\\\\\\\ P(X=0)=\dfrac{\binom{4}{0}\binom{12-4}{3-0}}{\binom{12}{3}}=\dfrac{\binom{4}{0}\binom{8}{3}}{\binom{12}{3}}=\dfrfac{1*56}{220}=\dfrac{56}{220}=0.2545[/tex]

The probability that none of the four cans selected contains an incorrect mix of paint is P=0.2545.

Final answer:

The probability that none of the four randomly selected cans are defective is approximately 0.2545, or 25.45%, which is determined using the hypergeometric distribution.

Explanation:

The student is faced with a scenario where a man has twelve 1-gallon paint cans, out of which three contain an incorrect mix of paint. The man randomly selects four of these cans to paint with, and the question is to find the probability that none of the four selected cans are defective, which follows the hypergeometric distribution.

The relevant parameters for the hypergeometric distribution in this scenario are: the total number of cans (N=12), the number of defective cans (K=3), the number of cans selected (n=4), and the number of defective cans selected that we are interested in (x=0). To compute the probability, we use the hypergeometric probability formula:

P(X = x) = [(C(K, x) * C(N-K, n-x)) / C(N, n)]

Substituting the given values, we have:

P(X = 0) = [(C(3, 0) * C(12-3, 4-0)) / C(12, 4)]
= [(1 * C(9, 4)) / C(12, 4)]
= (1 * 126) / 495
≈ 0.2545

This means the probability that none of the four randomly selected cans are defective is approximately 0.2545, or 25.45%.

Given:

Given that two similar cylinder have surface areas 24π cm² and 54π cm².

The volume of the smaller cylinder is 16π cm³

We need to determine the volume of the larger cylinder.

Volume of the larger cylinder:

The ratio of the two similar cylinders having surface area of 24π cm² and 54π cm², we have;

[tex]\frac{24 \pi}{54 \ pi}=\frac{4}{9}[/tex]

       [tex]=\frac{2^2}{3^2}[/tex]

Thus, the ratio of the surface area of the two cylinders is [tex]\frac{2^2}{3^2}[/tex]

The volume of the larger cylinder is given by

[tex]\frac{2^2}{3^2}\times \frac{2}{3}=\frac{16 \pi }{x}[/tex]

where x represents the volume of the larger cylinder.

Simplifying, we get;

[tex]\frac{2^3}{3^3}=\frac{16 \pi }{x}[/tex]

[tex]\frac{8}{27}=\frac{16 \pi }{x}[/tex]

Cross multiplying, we get;

[tex]8x=16 \pi \times 27[/tex]

[tex]8x=432 \pi[/tex]

 [tex]x=54 \pi \ cm^3[/tex]

Thus, the volume of the larger cylinder is 54π cm³

Answer:

54π cm³

Step-by-step explanation:

The question asks to perform a hypothesis test about the mean pH level in a river. Given a sample size of 29, a sample mean of 6.7, a sample standard deviation of 0.35, and a significance level of α = 0.05, the provided reference suggests that there is insufficient evidence to reject the company's claim of a mean pH of 6.8, due to the calculated p-value being greater than α.

In this problem, we are testing the hypothesis that the mean pH level of water in a nearby river is 6.8. The company claims this as the true population mean. The hypothesis under test is called the Null hypothesis.

The level of significance is given as α = 0.05. We have a sample of size 29 with mean 6.7 and standard deviation 0.35.

In hypothesis testing, we calculate a test statistic and compare it with a critical value corresponding to the level of significance α. Here, we would be calculating a t-score because we have the sample standard deviation, not the population standard deviation and the sample size is less than 30. If the test statistic falls in the critical region, then we reject the null hypothesis.

Without specific calculations, the given reference suggests that the decision is to not reject the null hypothesis, citing p-value > α. In this case, the calculated p-value from testing statistics is higher than 0.05, meaning that the observed test statistic would be quite likely if the null hypothesis is true.

This results in the conclusion that there is insufficient evidence in the sampled data to reject the company's claim of a mean pH of 6.8.

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There is not enough evidence to reject the company's claim at the α=0.05 level of significance.

Given:

Population mean =6.8

Sample mean  =6.7

Sample standard deviation s=0.35

Sample size n=29

Level of significance α=0.05

We'll perform a one-sample t-test since the population standard deviation is unknown and the sample size is less than 30.

The hypotheses are:

Null hypothesis (o):

The mean pH level of the water in the river is 6.8 (μ=6.8).

Alternative hypothesis (H1):

The mean pH level of the water in the river is not equal to 6.8 (≠6.8)

We'll use the formula for the test statistic of a one-sample t-test:

t = (x-  ) / [tex]\frac{s}{\sqrt{n} }[/tex]

t= -0.1/ 0.0651

t≈−1.535

Now, we'll find the critical value for a two-tailed test at α=0.05 significance level with n−1=28 degrees of freedom. Using a t-distribution table or statistical software,

we find the critical values to be approximately ±2.048.

Since −1.535 falls within the range −2.048 to 2.048, we fail to reject the null hypothesis.

So, there is not enough evidence to reject the company's claim at the α=0.05 level of significance.

Answer:

The player should be required to pay $5 to make this a fair game.

Step-by-step explanation:

U ~ Uniform(0, 10)

E[U] = (0 + 10)/2

         = 5

X | U ~ Poisson(U)

E[X | U] = U

By law of total probability for expectations,

E[X] = E[E[X|U]] = E[U] = $5

Therefore the player should be required to pay $5 to make this a fair game.

The probability of first selecting a chocolate chip cookie and then selecting a sugar cookie from a box containing 24 cookies in total is 6/46 or approximately 0.1304.

The question refers to calculating the probability of selecting cookies of different flavors one after the other without replacement from a box. To begin with, we must find the probability of selecting a chocolate chip cookie followed by the probability of selecting a sugar cookie after one chocolate chip cookie has been removed.

Firstly, the total count of cookies is 12 chocolate chip + 6 peanut butter + 6 sugar cookies = 24 cookies. The probability (P) of selecting a chocolate chip cookie first is P(chocolate chip) = 12/24 = 1/2. After eating the chocolate chip cookie, there are 23 cookies left and the probability of then selecting a sugar cookie is P(sugar) = 6/23 since there are 6 sugar cookies left out of the remaining 23 cookies.

Since these events are sequential without replacement, we can find the combined probability of both events by multiplying the probabilities of each event. Thus, the combined probability is P(chocolate chip then sugar) = P(chocolate chip) *P(sugar) = (1/2) * (6/23) = 6/46.

The combined probability of first selecting a chocolate chip cookie and then selecting a sugar cookie is therefore 6/46 or about 0.1304.