Patrice Patriot has dimes and uarters in a piggy bank. She has a total of 20 coins for atotal of $4.25. How many dimes does she have?

Answer:

its 5000

Step-by-step explanation:

for it to be 50000 like the other person said it would be asking what is 0.5 not 0.05 like it says in the question

Answer: N >/= 7 bits

Minimum of 7 bits

Step-by-step explanation:

The minimum binary bits needed to represent 65 can be derived by converting 65 to binary numbers and counting the number of binary digits.

See conversation in the attachment.

65 = 1000001₂

65 = 7 bits :( 0 to 2^7 -1)

The number of binary digits is 7

N >/= 7 bits

To represent the unsigned decimal integer 65 in binary format, we need a minimum of 7 binary bits.

This question is related to the conversion of decimal numbers to binary format. To present an unsigned decimal 65 in binary, we first need to find the highest power of 2 that is less than or equal to 65, which is 2^6=64.

Then we continue to find the next highest power of 2 for the remainder of the value until we reach zero. This process would require a total of 7 bits. Therefore, to represent 65 as an unsigned binary integer, we need a minimum of 7 bits.

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Answer:

a) [tex]\frac{28}{349}[/tex] of the prize winners were women.

b) [tex]\frac{321}{349}[/tex] of the prize winners were men.

Step-by-step explanation:

Given:

Prize awarded to women = 56

Prize awarded to men = 642

we need to find;

a. fraction of the prize winners were​ women

b. fraction were​ men

Solution:

First we will find the Total prize awarded.

Total Prize awarded is equal to sum of Prize awarded to women and Prize awarded to men.

framing in equation form we get;

Total Prize awarded = [tex]56+642 = 698[/tex]

Now we need to find the fraction of prize winners were women.

to find the fraction of prize winners were women we will divided Prize awarded to women from Total Prize awarded we get;

fraction of the prize winners were​ women = [tex]\frac{56}{698} = \frac{2\times28}{2\times 349}=\frac{28}{349}[/tex]

a) Hence [tex]\frac{28}{349}[/tex] of the prize winners were women.

Now we can say that;

to find the fraction of prize winners were men we will divided Prize awarded to men from Total Prize awarded we get;

fraction of the prize winners were​ men = [tex]\frac{642}{698}= \frac{2\times321}{2\times 349} = \frac{321}{349}[/tex]

b) Hence [tex]\frac{321}{349}[/tex] of the prize winners were men.

Answer:

Step-by-step explanation:

x2 = 36?

solution

x^2-(36)=0

Factoring:  x2-36

Check : 36 is the square of 6

Check :  x2  is the square of  x1  

Factorization is :       (x + 6)  •  (x - 6)

(x + 6) • (x - 6)  = 0

x+6 = 0

x = -6

x-6 = 0

add 6 to both side

x = 6

x = -6

The solutions of an equation are the true values of the equation.

The expression that represents the solution(s) of [tex]\mathbf{x^2 = 36}[/tex] is [tex]\mathbf{x = \pm6}[/tex]

The equation is given as:

[tex]\mathbf{x^2 = 36}[/tex]

Take square roots of both sides

[tex]\mathbf{x = \pm\sqrt{36}}[/tex]

Express the square root of 36 as 6.

So, we have

[tex]\mathbf{x = \pm6}[/tex]

So, the expression that represents the solution(s) of [tex]\mathbf{x^2 = 36}[/tex] is [tex]\mathbf{x = \pm6}[/tex]

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Answer:

(0.084,0.396)

Step-by-step explanation:

The 99% confidence interval for the proportion of customers who use debit card monthly can be constructed as

[tex]p-z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} } <P<p+z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} }[/tex]

[tex]p=\frac{x}{n}[/tex]

[tex]p=\frac{12}{50}[/tex]

[tex]p=0.24[/tex]

[tex]q=1-p=1-0.24=0.76[/tex]

[tex]\frac{\alpha }{2} =\frac{\0.01 }{2}=0.005[/tex]

[tex]p-z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} } <P<p+z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} }[/tex]

[tex]0.24-z_{0.005} \sqrt{\frac{0.24*0.76}{50} } <P<0.24+z_{0.005} \sqrt{\frac{0.24*0.76}{50} }[/tex]

[tex]0.24-2.58(0.0604)<P< 0.24+2.58(0.0604)[/tex]

[tex]0.24-0.155832<P<0.24+0.155832[/tex]

By rounding to three decimal places we get,

[tex]0.084<P<0.396[/tex]

The 99% confidence interval for the proportion of customers who use debit card monthly is (0.084,0.396).

To find a 99% confidence interval for the proportion of customers who use their debit card monthly, you can use the formula for the confidence interval for a proportion. The formula is:

[tex]\[ \text{Confidence Interval} = \hat{p} \pm Z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \][/tex]

Where:

- [tex]\(\hat{p}\)[/tex] is the sample proportion (in this case, 12 out of 50 customers).

- Z is the critical value for a 99% confidence level (you can find this value in a standard normal distribution table or use a calculator, and it's approximately 2.576 for a 99% confidence level).

- n is the sample size (50 customers).

Now, plug in the values:

- [tex]\(\hat{p}\)[/tex] = [tex]\frac{12}{50} = 0.24\) (the sample proportion)[/tex].

- [tex]\(Z = 2.576\) (for a 99% confidence level)[/tex].

- [tex]\(n = 50\) (the sample size)[/tex].

Calculate the standard error:

[tex]\[ \text{Standard Error} = \sqrt{\frac{0.24 \cdot (1-0.24)}{50}} \][/tex]

Now, calculate the margin of error:

[tex]\[ \text{Margin of Error} = Z \cdot \text{Standard Error} = 2.576 \cdot \sqrt{\frac{0.24 \cdot (1-0.24)}{50}} \][/tex]

Finally, calculate the confidence interval:

[tex]\[ \text{Confidence Interval} = 0.24 \pm \text{Margin of Error} \][/tex]

Calculate the upper and lower bounds:

Lower Bound: [tex]\(0.24 - \text{Margin of Error}\)[/tex]

Upper Bound: [tex]\(0.24 + \text{Margin of Error}\)[/tex]

Now, calculate these values:

[tex]\[ \text{Margin of Error} = 2.576 \cdot \sqrt{\frac{0.24 \cdot (1-0.24)}{50}} \][/tex]

[tex]\[ \text{Margin of Error} \approx 0.0944 \][/tex]

Lower Bound: [tex]\(0.24 - 0.0944 \approx 0.1456\)[/tex]

Upper Bound: [tex]\(0.24 + 0.0944 \approx 0.3344\)[/tex]

So, the 99% confidence interval for the proportion of customers who use their debit card monthly is approximately 0.1456 to 0.3344.

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Answer:

If a 1% level of significance is used to test a null hypothesis, there is a probability of ____ less than 1%______ of rejecting the null hypothesis when it is true

Step-by-step explanation:

Given that a hypothesis testing is done.

Level of significance used is 1%

i.e. alpha = 1%

When we do hypothesis test, we find out test statistic Z or t suitable for the test and find p value

If p value is < 1% we reject null hypothesis otherwise we accept null hypothesis.

So p value can be atmost 1% only for accepting null hypothesis.

So the answer is 1%

If a 1% level of significance is used to test a null hypothesis, there is a probability of ____less than 1%______ of rejecting the null hypothesis when it is true.

Answer:

Option a) The confidence interval increases as the standard deviation increases.

Step-by-step explanation:

We have to find the true statements.

Confidence Interval:

[tex]\mu \pm Test_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

a. The confidence interval increases as the standard deviation increases.

As the standard deviation increases, the margin of error increases, thus, the width of the confidence interval increases.

Thus, the given statement is true.

b. The confidence interval increases as the standard deviation decreases.

The given statement is false. The explanation is similar to above part.

c. The confidence interval increases as the sample size increases.

As the sample size increases, the margin of error decreases, thus, the confidence interval width decreases.

Thus, the given statement is false.

d. The t-value for a confidence interval of 95% is smaller than for a confidence interval of 90%

The t-value depend on the level of significance as well as degree of freedom. But for a particular degree of freedom  t-value for a confidence interval of 95% is greater than for a confidence interval of 90%.

Thus, the given statement is false.

In this exercise we have to use our knowledge of statistics to identify the correct alternative that best matches, so we have:

Letter a.

So with the knowledge in statistics topics we can say that:

a. True, as the standard deviation increases, the margin of error increases, thus, the width of the confidence interval increases.

b. False, the confidence interval increases as the standard deviation decreases.

c. False, as the sample size increases, the margin of error decreases, thus, the confidence interval width decreases.

d. False, the t-value depend on the level of significance as well as degree of freedom. But for a particular degree of freedom  t-value for a confidence interval of 95% is greater than for a confidence interval of 90%.

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Answer:

The probability that we have a fair coin is 0.2717

Step-by-step explanation:

Let Ci be the event that the coin i is used for while we let i = 1,2,3 (representing the three coin)

We also let H be the event that the coin which is flipped lands (which is head)

Therefore, the problem arise:

That:  P(H/C1) = 1/2.............fair coin

          p(H/C2) = 2/3..........Chance of head

          p(H/C3) = 2/3............Chance of tail

Now, noting that the coin was picked at random,

We have, : p(Ci) = P(C1) = P(C2) = P(C3) = 1/3

We can then say that or calculate thus:

P( C2 | H ) = P (H ∩ C2) ÷ P(H)

Where P (H ∩ C2) means the probability of event intersection

= P(H ∩ C2) ÷ P(H ∩ C1) + P(H ∩ C2) + P(H ∩ C3)

= P(H | C2) P(C2) ÷ P(H | C1) P(C1) + P(H | C2) P(C2) + P(H | C3) P(C3)

= (1 / 2) (1 / 3) ÷  (1/2)(1/3) + (2/3)(1/3) + (2/3) (1/3)

0.5 × 0.3 ÷ (0.5 × 0.3) + (0.67 × 0.3) + (0.67 × 0.3)

0.15 ÷ 0.15 + 0.201 + 0.201

=0.15 / 0.552

= 0.2717

Answer:72

Step-by-step explanation:

Answer:

Option C.

Step-by-step explanation:

The given equations are

[tex]3A-B=5[/tex]

[tex]2A+3B=-4[/tex]

We need  isolate one of the variables, such that it appears by itself on one side of the equation.

Isolating a variable in two equations is easiest when one of them has a coefficient 1.

In equation 1, coefficient of B is 1. So, we can easily isolate one of the variables.

The equation is

[tex]3A-B=5[/tex]

Subtract 3A from both sides.

[tex]-B=5-3A[/tex]

Multiply both sides by -1.

[tex]B=3A-5[/tex]

Therefore, the correct option is C.

1 a) Pairs of Independent Events are A and B, A and D

b) Triples of Independent Events are A, B, C

2 a) Both HHHHHHHHH and HHHHHHHHT are equally likely.

b) Probability of each event:

HHHHHHHHH: 1/512

HHHHHHHHT: 1/512

c) The person is suffering from gambler's fallacy

d) The person is suffering from misconception of randomness.

1) Events

A: Heads on the first toss.

B: Tails on the second toss.

C: Heads on the third toss.

D: All three outcomes the same (HHH or TTT).

E: Exactly one head turns up.

a) To check if two events are independent, we see if [tex]P(A \cap B) = P(A) \times P(B)[/tex]

Calculating Individual Probabilities

P(A) = 1/2 (Heads on the first toss)

P(B) = 1/2 (Tails on the second toss)

P(C) = 1/2 (Heads on the third toss)

P(D) = P(HHH) + P(TTT) = 1/8 + 1/8 = 1/4

P(E) = 3/8 (Exactly one head: H, T, T; T, H, T; T, T, H)

Checking Independence

i. A and B

P(A ∩ B): Probability of A and B happening together (H on first and T on second):

There are 4 outcomes: HTH, HTT, HHH, HHT. Only HTT has T on second.

P(A ∩ B) = 1/4.

P(A) × P(B) = 1/2 × 1/2 = 1/4.

They are independent

ii. A and D

P(A ∩ D): Probability of A and D (first is H and all are the same):

Only HHH satisfies this.

P(A ∩ D) = 1/8.

P(A) × P(D) = 1/2 × 1/4 = 1/8.

They are independent

iii. A and E

P(A ∩ E): Probability of A and E (first is H and only one head):

Possible outcomes: H, T, T (1 possibility).

P(A ∩ E) = 1/8.

P(A) × P(E) = 1/2 × 3/8 = 3/16.

They are not independent.

iv. D and E

P(D ∩ E): Probability of D and E (all the same and exactly one head):

Impossible since D can only be HHH or TTT, and E requires one head.

P(D ∩ E) = 0.

P(D) × P(E) = 1/4 × 3/8 = 3/32.

They are not independent

b) To check if three events are independent, we check if P(A ∩ B ∩ C) = P(A) × P(B) × P(C)

i. A, B, C

P(A ∩ B ∩ C): Probability of H on first, T on second, H on third:

Outcome: HTH.

P(A ∩ B ∩ C) = 1/8.

P(A) × P(B) × P(C) = 1/2 × 1/2 × 1/2 = 1/8.

They are independent

ii. A, B, D

P(A ∩ B ∩ D): Cannot have T on second and all the same

P(A ∩ B ∩ D) = 0.

P(A) × P(B) × P(D) = 1/2 × 1/2 × 1/4 = 1/16.

They are not independent

iii. C, D, E

P(C ∩ D ∩ E): Cannot have all the same (D) and exactly one head (E).

P(C ∩ D ∩ E) = 0.

P(C) × P(D) × P(E) = 1/2 × 1/4 × 3/8 = 3/64.

They are not independent.

2a) Both events are equally likely since they are specific sequences of tosses.

b) Probability of each event:

Probability of HHHHHHHHH

9 heads in a row:

Probability = (1/2)^9 = 1/512.

Probability of HHHHHHHHT

8 heads followed by 1 tail:

Probability = (1/2)^9 = 1/512.

c) They are suffering from the gambler's fallacy, which is the incorrect belief that past events influence independent events.

d) They may be suffering from the misconception of randomness, believing that certain sequences are more likely due to patterns they perceive.

Answer:

Step-by-step explanation:

Given that price, p, and quantity, x, are inearly related

We are given two points on this line as (p,x) = (96.90, 121) and (40.20,310)

Using two point formula we find linear equation as

[tex]\frac{y-121}{310-121} =\frac{0-96.90}{40.20-96.90} \\(x-121)(-56.70)=189(p-96,.90)\\189p+56.70x =25174.80[/tex]

a)[tex]189p+56.70x =25174.80[/tex] is the linear equation

b) A(x)

Substitute to get

189(50.70)+56.70x = 25174.80

x=275 units

c) This is linear function hence no local maxima or minima

d) No maximia or minima

Answer:

At least two of the restaurant have different mean delivery time.

Step-by-step explanation:

The ANOVA known as Analysis of variance involves the hypothesis testing of means on the basis of variances. The null hypothesis in ANOVA is taken as the equality mean i.e. H0: All means are equal. Whereas alternative hypothesis consists of at least two means are not equal.

The problem states that a pizza lover is comparing average delivery time. The null and alternative hypothesis in this case would be

H0: All restaurant have equal mean delivery time.

H1: At least two restaurant have different mean delivery time.

In an ANOVA comparing four local pizza restaurants' delivery times, the alternative hypothesis is that at least two restaurants have different mean delivery times.

When performing an ANOVA with data comparing the average delivery times for four local pizza restaurants, the alternative hypothesis is that at least two of the restaurants have different mean delivery times. This is because ANOVA tests the null hypothesis that all group means are equal against the alternative that at least one group mean is different. Specifically, the null hypothesis for a one-way ANOVA test with four groups is μ1 = μ2 = μ3 = μ4, where each μ represents the mean delivery time of a different restaurant. The alternative hypothesis, which the student is inquiring about, is Ha: μi ≠ μj for some i ≠ j.

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Answer:

a) There is a 63.02% probability that no one has done an one time fling.

b) There is a 36.98% probability that at least one person has done a one time fling

c) There is a 99.15% pprobability that no more than two people have done a one time fling.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have done an one time fling, or they have not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

[tex]n = 9, p = 0.05[/tex]

A. no one has done a one time fling

This is P(X = 0).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{9,0}.(0.05)^{0}.(0.95)^{9} = 0.6302[/tex]

There is a 63.02% probability that no one has done an one time fling.

B. at least one person has done a one time fling

Either no one has done a one time fling, or at least one person has. The sum of the probabilities of these events is decimal 1.

So

[tex]P(X = 0) + P(X \geq 1) = 1[/tex]

[tex]P(X \geq 1) = 1 - `(X = 0) = 1 - 0.6302 = 0.3698[/tex]

There is a 36.98% probability that at least one person has done a one time fling

c. no more than two people have done a one time fling

This is

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

[tex]P(X = 1) = C_{9,1}.(0.05)^{1}.(0.95)^{8} = 0.2985[/tex]

[tex]P(X = 2) = C_{9,2}.(0.05)^{2}.(0.95)^{7} = 0.0628[/tex]

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.6302 + 0.2985 + 0.0628 = 0.9915[/tex]

There is a 99.15% pprobability that no more than two people have done a one time fling.

In a group of 9 adults, the probability that no one has done a 'one-time fling' is 0.630, the probability that at least one person has done it is 0.370, and the probability that no more than two people have done it is approximately 0.853.

The question here involves probability given certain conditions. The probability of an event happening can be found by dividing the number of times the event occurs by the total number of occurrences. Here we have the probability of the 'one-time fling' happening as 5% or 0.05, the complement of which (it not happening) is 1 - 0.05 = 0.95.

A. The probability of none out of nine friends having done a one-fling time is (0.95)^9 = 0.630. So, there’s a 0.630 chance that none of them have done a one-time fling.

B. The probability of at least one person having done a one-time fling would be the complement of no one having done it, which equals 1 - (the probability of no one having done it), 1 - 0.630 = 0.370.

C. The probability of no more than two people having done a one-time fling is found by adding the probabilities of exactly zero, one, and two persons having done it. Using the binomial probability formula, this would amount to 0.630 (for 0 person) + 9(0.05)*(0.95)^8 (for one person) + 36*(0.05)^2*(0.95)^7 (for two people) = 0.853, approximately.

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Answer: 5/36

Step-by-step explanation:

We assume it's a fair die and the probability of any number coming up is 1/6.

Let's denote first die with it's number as A2, that is first die roled number 2

Let's denote second die with it's number as B4, that is Second die rolled 4.

For us to have the sum of those numbers to be 8, then we have possibilities of rolling the numbers

A2 and B6, A3 and B5, A4 and B4, A5 and B3, A6 and B2

This becomes :

[pr(A2) * pr(B6)] + [pr(A3) * pr(AB5)] + [pr(A4) * pr(B4)] + [pr(A5) * pr(3)] + [pr(A6) * pr(B2)]

Which becomes:

[1/6 * 1/6] + [1/6 * 1/6] + [1/6 * 1/6] + [1/6 * 1/6 + [1/6 * 1/6]

Which becomes :

1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36

Hence, the probability of the sum of the two numbers on the dice being 8 is 5/36

Answer: Normal approximation can be used for discrete sampling distributions, such as Binomial distribution and Poisson distribution if certain conditions are met.

Step-by-step explanation: We will give conditions under which the Binomial and Poisson distribitions, which are discrete, can be approximated by the Normal distribution. This procedure is called normal approximation.

1. Binomial distribution: Let the sampling distribution be the binomial distribution [tex]B(n,p)[/tex], where [tex]n[/tex] is the number of trials and [tex]p[/tex] is the probability of success. It can be approximated by the Normal distribution with the mean of [tex]np[/tex] and the variance of [tex]np(1-p)[/tex], denoted by [tex]N(np,np(1-p))[/tex] if the following condition is met:

[tex]n>9\left(\frac{1-p}{p}\right)\text{ and } n>9\left(\frac{p}{1-p}\right)[/tex]

2. Poisson distribution: Let the sampling distribution be the Poisson distribution [tex]P(\lambda)[/tex] where [tex]\lambda[/tex] is its mean. It can be approximated by the Normal distribution with the mean [tex]\lambda[/tex] and the variance [tex]\lambda[/tex], denoted by [tex]N(\lambda,\lambda)[/tex] when [tex]\lambda[/tex] is large enough, say [tex]\lambda>1000[/tex] (however, different sources may give different lower value for [tex]\lambda[/tex] but the greater it is, the better the approximation).

Answer:

Step-by-step explanation:

The relative frequencies are 0.15, 0.08, 0.39, and 0.38 for the respective classes while 38% of the observations are at least 300 but less than 350.

The cumulative relative frequency distribution given can be used to construct a relative frequency distribution. To find the relative frequency for each class, subtract the cumulative frequency of the previous class from the cumulative frequency of the current class. Therefore:

The Total relative frequency is the sum of all the relative frequencies and should equal 1 (or almost equal to 1 due to rounding differences)

For the second part of your question: 'what percent of the observations are at least 300 but less than 350?', since this is the relative frequency distribution, it means that class 300 to 350 has already been expressed as a percentage (since relative frequency is a proportion and can be expressed as a percentage). So, 38% of the observations are at least 300 but less than 350.

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Answer:

[tex]0.23 - 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.172[/tex]

[tex]0.23 + 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.288[/tex]

The 95% confidence interval would be given by (0.172;0.288)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

[tex]X=46[/tex] number of vehicles that were not covered by insurance

[tex]n=200[/tex] random sample taken

[tex]\hat p=\frac{46}{200}=0.23[/tex] estimated proportion of vehicles that were not covered by insurance

[tex]p[/tex] true population proportion of vehicles that were not covered by insurance

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

The confidence interval for the mean is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.23 - 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.172[/tex]

[tex]0.23 + 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.288[/tex]

The 95% confidence interval would be given by (0.172;0.288)

Answer:

Substitution of x+1

Step-by-step explanation:

We are given that

[tex]\int \frac{1}{x^2+2x+2}dx[/tex]

[tex]\int\frac{1}{(x^2+2x+1)+1}dx[/tex]

[tex]\int\frac{1}{(x+1)^2+1^2}dx[/tex]

By using identity

[tex](a+b)^2=a^2+b^2+2ab[/tex]

Substitute x+1=t

Differentiate w.r.t x

dx=dt

Substitute the values

[tex]\int\frac{1}{t^2+1^2}dx[/tex]

[tex]\frac{1}{1}tan^{-1}\frac{t}{1}+C[/tex]

By using formula :[tex]\int\frac{1}{x^2+a^2}dx=\frac{1}{a}tan^{-1}\frac{x}{a}+C[/tex]

[tex]tan^{-1}(x+1)+C[/tex]

To complete the square for the quadratic polynomial x^2 + 2x + 2, we obtain (x + 1)^2 + 1. Substituting u = x + 1 simplifies the integral, which is then recognized as the derivative of arctan(u), resulting in the solution arctan(x + 1) + C.

The student asked to complete the square for the integrand x2 + 2x + 2 and to propose a substitution to compute the integral ∫ 1 / (x2 + 2x + 2) dx.

First, we complete the square for the quadratic polynomial x2 + 2x + 2. That gives us:

(x + 1)2 + 1 = x2 + 2x + 1 + 1 = x2 + 2x + 2.

We have now written our quadratic in the form of a perfect square plus a constant, which simplifies our integral:

∫ 1 / ((x + 1)2 + 1) dx.

We can substitute u = x + 1, which implies du = dx.

This transforms our integral into:

∫ 1 / (u2 + 1) du,

which can be recognized as the derivative of arctan(u), hence the integral is:

arctan(u) + C, where C is the constant of integration.

Substituting back our original variable, the solution is:

arctan(x + 1) + C.

Answer:

Step-by-step explanation:

i) Confidence level = 90%

ii) therefore significance level = [tex]\alpha[/tex]  =  [tex]\frac{100 - 90}{100}\hspace{0.1cm} = \hspace{0.1cm} 0.1[/tex]

iii) this problem is for a two tailed test.

iv) therefore significance level = [tex]\frac{\alpha }{2}[/tex] = [tex]\frac{0.1}{2}[/tex] = 0.05

iii) sample size = 15

iv) therefore degrees of freedom  = sample size  - 1   =   15 - 1    =  14

v) the upper ( or right) chi squared value, [tex]\chi^2_{R}[/tex]  from Chi squared critical value tables = 23.685

The correct critical value χR is 23.685.

To find the critical value χR corresponding to a sample size of 15 and a confidence level of 90%, we first determine the degrees of freedom.

The degrees of freedom (df) for a chi-square distribution is calculated as n - 1, where n is the sample size. Here, n = 15, so df = 14.

Next, look up the critical value for the chi-square distribution with 14 degrees of freedom and a 90% confidence level.

In statistical tables, the value corresponding to a 90% confidence level (or equivalently, 10% significance level, which leaves 5% in each tail of the distribution) for 14 degrees of freedom is 23.685.

Therefore, the correct critical value χR is 23.685.

Answer: m1 = 4

m2 = 5

m3 = 2

Step-by-step explanation:

given (21/11, 6/11) = m1 (-1/3) + m2 (3, -2) + m3 (5, 2)

= (-m1 + 3m2 + 5m3) / 11 = 21/11

= (3m1 + (-2)m2 + 2m3) / 11 = 6/11

so that m1 + m2 +m3 = 11

-m1 + 3m2 + 5m3 = 21

3m1 - 2m2 + 2m3 = 6

from this, we get the augmented matrix as

\left[\begin{array}{cccc}-1&1&1&11\\-1&3&5&21\\3&-2&2&6\end{array}\right]

= \left[\begin{array{cccc}-1&1&1&11\\0&4&6&32\\0&-5&-1&-27\end{array}\right]  \left \{ {{R2=R2 + R1} \atop {R3=R3 -3R1 }]} \right.

= \left[\begin{array}{cccc}-1&1&1&11\\0&1&3/2&8\\0&-5&-1&-27\end{array}\right]

= \left[\begin{array}{cccc}-1&1&1&11\\0&1&3/2&8\\0&0&13/2&13\end{array}\right]

(R3 = R3 + 5R2)

this gives m1 + m2 + m3 = 11

m2 + 3/2 m3 = 8

13/2 m3 = 8

13/2 m3 = 13

m3 = 2

m2 = 8 -3/2 (2) = 5

= m1 = 11- 5 - 2 = 4

this gives

m1 = 4

m2 = 5

m3 = 2

To achieve the given center of mass, 4 kg of mass should be allocated to the vector u1, 6 kg to the vector u2, and 1 kg to the vector u3.

The provided equation for the center of mass expresses a weighted average of the position vectors, with each mass acting as the weight on its respective vector. We are given three vectors and a total mass, and must find the weights to apply to each vector in order to achieve a particular center of mass. Setting up equations for each component of the center of mass, we get two equations:

21/11 = (-1m1 + 3m2 + 5m3) / 11 and 6/11 = (3m1 - 2m2 + 2m3) / 11.

These equations can then be solved simultaneously to determine the values of m1, m2, and m3. Solving these equations gives m1 = 4 kg, m2 = 6 kg, and m3 = 1 kg. So, four kilograms of mass should be allocated to the vector u1, six kilograms to the vector u2, and one kilogram to the vector u3.

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Answer:

Population: All undergraduates who attend a university and live in the United States  

Step-by-step explanation:

We are given the following situation in the question:

 "A polling organization contacts 1835 undergraduates who attend a university and live in the United States and asks whether or not they had spent more than nbsp $200 on food nbsp in the last month."

For the given situation, we have

Sample size, n = 1835

Sample: 1835 undergraduates who attend a university and live in the United States

Variable: whether or not they had spent more than $200 on food in the last month

Thus, the population for this scenario will be

Population: All undergraduates who attend a university and live in the United States

The population in the study is 1835 undergraduates.

The population in the study is the total number of undergraduates who attend a university and live in the United States.

According to the information given, the polling organization contacted 1835 undergraduates who meet these criteria and asked them about their food expenses.

Therefore, the population in the study is 1835 undergraduates.

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Answer:

A researcher studies length of time in college, first through fourth year, and its relation to academic motivation. To get the most detail out of her measures, she assesses each student in both the fall and spring semesters of each of their four years in school. She finds that students have increasingly higher motivation from their first semester to their seventh semester (the start of their fourth year), with a trailing off in the last semester. The independent variable are:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Triangle VWX is a right angle triangle.

From the given right angle triangle

VX represents the hypotenuse of the right angle triangle.

With 39 degrees as the reference angle,

WX represents the adjacent side of the right angle triangle.

VW represents the opposite side of the right angle triangle.

1) To determine VX, we would apply trigonometric ratio

Sin θ = opposite side/hypotenuse Therefore,

Sin 39 = 4/VX

VX Sin39 = 4

VX = 4/Sin39 = 4/0.6293

VX = 6.4

2) To determine WX, we would apply trigonometric ratio

Tan θ = opposite side/adjacent side. Therefore,

Tan 39 = 4/WX

WX Tan39 = 4

WX = 4/Tan 39 = 4/0.8089

WX = 4.9

3) the sum of the angles in a triangle is 180 degrees. Therefore

∠V + 90 + 39 = 180

∠V = 180 - (90 + 39)

∠V = 51 °

Step-by-step explanation:

The largest whole number can be 9,999,999,999 if it is  decimal representation is simple. The device with me  supports up to 10 ^ 100 exponents, so that 9.9999999E99 could be a candidate too. If your exponents are not limited, then 9E9999999 is the largest (above what the calculator can demonstrate).

Answer:

[tex]f(t) = -0.025t + 0.61[/tex]

In which t is measured in decades.

Step-by-step explanation:

A linear function f(t) has the following format:

[tex]f(t) = at + b[/tex]

We are given two points of the function, so we can solve a system of equations to find the values for a and b.

61% of all adults in 1971 (t = 0)

This means that [tex]f(0) = 0.61[/tex]

So

[tex]f(t) = at + b[/tex]

[tex]0.61 = a*(0) + b[/tex]

[tex]b = 0.61[/tex]

51% in 2011 (t = 4)

This means that [tex]f(4) = 0.51[/tex]

So

[tex]f(t) = at + 0.61[/tex]

[tex]0.51 = 4a + 0.61[/tex]

[tex]4a = -0.10[/tex]

[tex]a = -0.025[/tex]

So the linear function is:

[tex]f(t) = -0.025t + 0.61[/tex]

In which t is measured in decades.

Answer:the total discounted cost for your stay is $1948.2

Step-by-step explanation:

If you book your room six months in advance, you will save 15% off the price of the room. You decide to take advantage of the offer and book a room for 6 nights. If the original price per night for the room is $382, this means that the value of the discount would be

15/100 × 382 = 0.15 × 382 = $57.3

The cost of the room for a night would be

382 - 57.3 = $324.7

Since you are staying for six nights,

the total discounted cost for your stay would be

324.7 × 6 = $1948.2

Answer:

[tex]r=\sqrt{6/sinucosu}[/tex]

Step-by-step explanation:

To convert to polar form:

[tex]x=rcosu[/tex]

[tex]y=rsinu[/tex]

We substitute into our function:

[tex]y=6/x[/tex]

[tex]rsinu=6/rcosu[/tex]

multiply both sides by r:

[tex]r^2sinu=6/cosu[/tex]

solve for r by dividing by sinu:

[tex]r=\sqrt{6/sinucosu}[/tex]

Answer: the integers that is a Pythagorean triple and are the side lengths of a right triangle is

C. 9, 40, 41

Step-by-step explanation:

A Pythagorean triple is a set of three numbers which satisfy the Pythagoras theorem. The Pythagoras theorem is expressed as

Hypotenuse^2 = opposite side^2 + adjacent side^2

Let us try each set of numbers.

A. 20,23,28

28^2 = 20^ + 23^2

784 = 400 + 529 = 929

Since both sides of the equation are not equal, the set of numbers is not a Pythagoras triple.

B. 18, 26, 44

44^2 = 18^ + 26^2

1936 = 324 + 676 = 1000

Since both sides of the equation are not equal, the set of numbers is not a Pythagoras triple.

C. 9, 40, 41

41^2 = 9^ + 40^2

1681 = 81 + 1600 = 1681

Since both sides of the equation are equal, the set of numbers is a Pythagoras triple.

D. 8, 20, 32

32^2 = 20^ + 8^2

1024 = 400 + 64 = 464

Since both sides of the equation are not equal, the set of numbers is not a Pythagoras triple.

Answer:

17.32

Step-by-step explanation:

to get TU

sinФ = opposite/hypothenus

sin 60° = TU/20

√3/2=TU/20

( √3/2 )  x 20 = 10√3 = TU = 17.3205080757 = 17.32