The value of a house is increasing by 1900 per year. If it is worth 180,000 today, what will it be worth in five years?

Answer:

[tex]z=0.842<\frac{a-204}{55}[/tex]

And if we solve for a we got

[tex]a=204 +0.842*55=250.31[/tex]

So the value of height that separates the bottom 80% of data from the top 20% is 250.31.  

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the room hotel rate of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(204,55)[/tex]  

Where [tex]\mu=204[/tex] and [tex]\sigma=55[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.20[/tex]   (a)

[tex]P(X<a)=0.80[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.2

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.842[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.842[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=0.842<\frac{a-204}{55}[/tex]

And if we solve for a we got

[tex]a=204 +0.842*55=250.31[/tex]

So the value of height that separates the bottom 80% of data from the top 20% is 250.31.  

Answer:16.5

Step-by-step explanation:

Answer:

11 7/10

Step-by-step explanation:

Lets make the fractions into common denominators, by mulitplying 4 by 5 and 5 by 4 (see below) and follow PEMDAS

3 1/4(5) +(3 1/4(5) + 5 1/5(4))

3 5/20+ (3 5/20+5 4/20)

Simplify like terms

3 5/20+ (8 9/20)

11 14/20

Simplify fraction

11 7/10

Answer:

1. H0 : p = 0.9

   H1 : p ≠ 0.9

2. The test is two tailed.

3. Reject the null hypothesis

Step-by-step explanation:

We are given that we have to test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.

So, Null Hypothesis, [tex]H_0[/tex] : p = 0.90

Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.9

Here, the test is two tailed because we have given that to test  the claim that the proportion of people who own cats is significantly different than 90% which means it can be less than 0.90 or more than 0.90.

Now, test statistics is given by;

            [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)   , where,  n = sample size = 100

                                                       [tex]\hat p[/tex] = 0.94 (given)

So, Test statistics = [tex]\frac{0.94-0.90}{\sqrt{\frac{0.94(1-0.94)}{100} } }[/tex] = 1.68

Now, P-value = P(Z > 1.68) = 1 - P(Z <= 1.68)

                                           = 1 - 0.95352 = 0.0465 ≈ 0.05 or 5%

Now, our decision rule is that;

       If p-value < significance level  ⇒ Reject null hypothesis

       If p-value > significance level  ⇒ Accept null hypothesis

Since, here p-value is less than significance level as 0.05 < 0.1, so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that proportion of people who own cats is significantly different than 90%.

Answer:

At 0.05 level of significance, there is no difference in the mean number of larvae per stem. This supports the scientist's claim.

(d) H0: MeanMalathion - MeanNoMalathion equals 0

H1: MeanMalathion - MeanNoMalathion not equals 0

Step-by-step explanation:

Test statistic (t) = (mean 1 - mean 2) ÷ sqrt[pooled variance (1/n1 + 1/n2)]

Let the difference between the two means be x and the pooled variance be y

n1 = 5, n2 = 12

t = x ÷ sqrt[y(1/5 + 1/12)] = x ÷ sqrt(0.283y) = x ÷ 0.532√y = 1.88x/√y

Assuming the ratio of x to √y is 0.5

t = 1.88×0.5 = 0.94

n1 + n2 = 5 + 12 = 17

degree of freedom = n1 + n2 - 2 = 17 - 2 = 15

significance level = 0.05 = 5%

critical value corresponding to 15 degrees of freedom and 5% confidence interval is 2.131

The test is a two-tailed test because the alternate hypothesis is expressed using not equal to.

The region of no rejection of the null hypothesis lies between -2.131 and 2.131

Conclusion

Fail to reject the null hypothesis because the test statistic 0.94 falls within the region bounded by the critical values.

The scientist's claim is right.

A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

An alternate hypothesis is also a statement from a population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.

Answer:

750,0000

Step-by-step explanation:

Answer:

Answer is attached

Step-by-step explanation:

6.11

Marginal Distribution of Marital Status

Single 4.1%

Married 93.9%

Divorced 1.5%

Widowed .5%

Marginal Distribution of Job Grade

Job Grade 1 11.6%

Job Grade 2 51.5%

Job Grade 3 30.2%

Job Grade 4 6.7%

Each of the marginal distributions sum up to exactly 100%. Depending on how one rounds the percentages, it is possible that the results will not be exactly 100%.

Question

A salesperson obtained a systematic sample of size 25 from a list of 500 clients. To do​ so, he randomly selected a number from 1 to 20, obtaining the number 13. He included in the sample the 13th client on the list and every 20th client thereafter. List the numbers that correspond to the 25 clients selected.

Answer:

13, 33, 53, ...... 393

Step-by-step explanation:

The Question is making reference to arithmetic progression

Where the first term is 13

And the difference between each interval is 20 (i.e number of clients)

Using the following formula, well obtained the client in any position

Tn = a + (n - 1) d

Where Tn = nth term

a = T1 = 1st term = 13

n = number of terms = 1 to 25

d = common difference = 20

Calculating T2

T2 = 13 + (2 - 1) * 20

T2 = 13 + 20

T2 = 33

Calculating T3

T3 = 13 + (3 - 1) * 20

T3 = 53

.......

Calculating the 20th term

T3 = 13 + (20 - 1) * 20

T20 = 393

Answer:

79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

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 = 50, p = 0.13[/tex]

So

[tex]\mu = E(X) = np = 50*0.13 = 6.5[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.13*0.87} = 2.38[/tex]

Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

This is 1 subtracted by the pvalue of Z when X = 4. So

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

[tex]Z = \frac{4 - 6.5}{2.38}[/tex]

[tex]Z = -0.84[/tex]

[tex]Z = -0.84[/tex] has a pvalue of 0.2005

1 - 0.2005 = 0.7995

79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

Using the normal approximation to the binomial distribution, the probability that at least 5 out of 50 motorists will be uninsured when 13% of motorists are uninsured is approximately 74.22%.

First, we calculate the mean (μ) and the standard deviation (σ) of the number of uninsured motorists in a sample of 50. The mean number of uninsured motorists is μ = np, where n is the sample size and p is the probability of a motorist being uninsured. So, μ = 50 * 0.13 = 6.5.

The standard d eviation is σ = √(np(1-p)). Therefore, σ = √(50 * 0.13 * (1 - 0.13)) ≈ 2.31.

To find the probability of at least 5 uninsured motorists, we want P(X ≥ 5). We need to standardize this to the standard normal distribution to find the z-score. The z-score is z = (X - μ) / σ.

For X = 5, the z-score is z = (5 - 6.5) / 2.31 ≈ -0.65. We use the z-table or a calculator to find the probability corresponding to z = -0.65.

Since we want at least 5 uninsured, we find P(Z ≥ -0.65), which equals 1 - P(Z < -0.65). The value from the z-table for -0.65 is approximately 0.2578. Therefore, the probability we're looking for is 1 - 0.2578 = 0.7422, or 74.22%.

Final answer:

Nickel does not crystallize in a simple cubic structure because the hypothetical density of a simple cubic arrangement is 2.23 g/cm³, significantly less than the actual density of nickel, which is 8.90 g/cm³.

Explanation:

To determine if nickel crystallizes in a simple cubic structure, we can compare the calculated density of a simple cubic arrangement of nickel atoms to the actual density of nickel. Given the edge length of the unit cell for nickel is 0.3524 nm (3.524 x 10⁻⁸ cm) and knowing the properties of a simple cubic lattice, the volume of a single cubic unit cell would be V = a³ = (3.524 x 10⁻⁸ cm)³ = 4.376 x 10⁻²³ cm³. With the mass of a single nickel atom (58.693 g/mol divided by Avogadro's number, 6.022 x 10²³ atoms/mol), if the nickel crystallized in a simple cubic structure, the predicted density would be:

density = mass/volume = (9.746 x 10⁻²³ g) / (4.376 x 10⁻²³ cm³) = 2.23 g/cm³. However, because the actual density of nickel is 8.90 g/cm³, we can conclude that nickel does not form a simple cubic structure.

Answer:

0.62 = 62% probability that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Desired outcomes:

Adults who consider the occupation of firefighter to have very great prestige. So the number of desired outcomes is [tex]D = 627[/tex]

Total outcomes.

All adults sampled. So [tex]T = 1010[/tex]

Probability:

[tex]P = \frac{D}{T} = \frac{627}{1010} = 0.62[/tex]

0.62 = 62% probability that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.

Final answer:

The question involves using Green's Theorem to calculate work done by a specific force field along a triangular path. Green's Theorem links the line integral around a closed curve to a double integral over the region it encloses. However, direct calculation of line integrals may be more applicable for this problem.

Explanation:

The question asks to use Green's Theorem to calculate the work done by a force F(x, y) = x(x + y) i + xy^2 j moving a particle along a specified path. Green's Theorem relates a line integral around a simple closed curve C, to a double integral over the plane region D bounded by C. It's expressed in the form ∑_C (M dx + N dy) = ∫∫_D (∂N/∂x - ∂M/∂y) dA where M and N are components of a vector field.

To use Green's Theorem, we identify M = x(x + y) and N = xy^2. Thus, ∂N/∂x = y^2 and ∂M/∂y = x. The path described (origin to (6,0), to (0,6), and back to origin) encloses a triangle, the area of which is easily integrated over. However, it is crucial to note that direct integration methods or alternative strategies may sometimes be more straightforward for such paths, especially when they do not form a typical closed curve as in standard Green's Theorem applications.

In the context of this problem, the key would be computing the line integrals directly due to the specific nature of the force field and the path involved. The use of Green's Theorem hints at setting up an integration over the area, but with the given vector field and triangular path, the execution would involve careful calculation of respective line integrals or assessing the area directly since the theorem simplifies to a calculation over the region enclosed by the path.

Answer:

The percent of the people who tested positive actually have the disease is 38.64%.

Step-by-step explanation:

Denote the events as follows:

X = a person has the disease

P = the test result is positive

N = the test result is negative

Given:

[tex]P(X)=0.01\\P(P|X^{c})=0.15\\P(N|X)=0.10[/tex]

Compute the value of P (P|X) as follows:

[tex]P(P|X)=1-P(P|X^{c})=1-0.15=0.85[/tex]

Compute the probability of a positive test result as follows:

[tex]P(P)=P(P|X)P(X)+P(P|X^{c})P(X^{c})\\=(0.85\times0.10)+(0.15\times0.90)\\=0.22[/tex]

Compute the probability of a person having the disease given that he/she was tested positive as follows:

[tex]P(X|P)=\frac{P(P|X)P(X)}{P(P)}=\frac{0.85\times0.10}{0.22} =0.3864[/tex]

The percentage of people having the disease given that he/she was tested positive is, 0.3864 × 100 = 38.64%.

Answer:

75%

Step-by-step explanation:

Find the percentage corresponding to 42/56:  (0.75)(100%) = 75%

Answer:

75%

Step-by-step explanation:

Becauuuuuuseeee if you divide 42  by 56 it would be 0.75 and yeah. so take out the 0 and its 75%

Answer:

a. P (X=0) = 0.223

b. P(X≤2) = 3.93 * 10^-5

C. No

D. t=4.6 miutes

Step-by-step explanation:

a. P (X=0) = [tex]e^{-1.5}[/tex] = 0.223

b. P(X≤2) =[tex]e^{1.5 * 10} (1 + 1.5*10 + \frac{1.5^{2} * 10^{2} }{2} )[/tex]

   = 3.93 * 10^-5

C. No, the answer to the previous part does not depend on whether the 10-minute period is an uninterrupted interval, it only depend on the uniformity of the density of views per minute ad independency of the disjoint time intervals.

D. P (X=0) =0.001

  [tex]e^{-1.5t} = 0.001\\\\-1.5t=ln(0.001)\\\\t =\frac{-6.9}{-1.5}[/tex]

t=4.6 miutes

The Poisson distribution is used to calculate the probabilities of certain counts of discrete events occurring in a given time interval. By using the average rate of occurrence, we can determine probabilities relevant to web traffic, call intervals, or urgent care visits.

The Poisson distribution is used to model the number of events that occur in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. Various scenarios using Poisson distributions involve understanding the probability of a certain number of events occurring over a specified period.

Answer:

[tex]t=1.60\ sec[/tex]

Step-by-step explanation:

Functions

We are given the function that relates the height of the ball and the time elapsed t. The function is:

[tex]h = 5 + 25t - 16t^2[/tex]

We need to know the time needed for the height to reach 4 feet, that is h=4

[tex]5 + 25t - 16t^2=4[/tex]

Rearranging

[tex]16t^2-25t -1=0[/tex]

Solving for t

[tex]\boxed{t=1.60\ sec}[/tex]

The other solution is negative and is discarded because t cannot be negative.

Answer:

The base of the triangle is decreasing at a rate 8 centimeter per minute.      

Step-by-step explanation:

We are given the following in the question:

[tex]\dfrac{dh}{dt} = 2.5\text{ cm per minute}\\\\\dfrac{dA}{dt} = 2\text{ square cm per minute}[/tex]

Instant height = 7.5 cm

Instant area = 96 square centimeters

Area of triangle =

[tex]A=\dfrac{1}{2}\times b\times h[/tex]

where b is the base and h is the height of the triangle.

[tex]96 = \dfrac{1}{2}\times b \times 7.5\\\\b = \dfrac{96\times 2}{7.5} = 25.6[/tex]

Rate of change of area of triangle =

[tex]\dfrac{dA}{dt} = \dfrac{d}{dt}(\dfrac{bh}{2})\\\\\dfrac{dA}{dt} =\dfrac{1}{2}(b\dfrac{dh}{dt} + h\dfrac{db}{dt})[/tex]

Putting values, we get,

[tex]2 = \dfrac{1}{2}(7.5\dfrac{db}{dt}+25.6(2.5))\\\\4 = 7.5\dfrac{db}{dt} + 64\\\\-60 = 7.5\dfrac{db}{dt} \\\\\Rightarrow \dfrac{db}{dt} = \dfrac{-60}{7.5} = -8[/tex]

Thus, the base of the triangle is decreasing at a rate 8 centimeter per minute.

Answer:

a) Mean 6, standard deviation 2.42

b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c) 14.85% probability that we find less than 4 of the seniors are married.

d) 99.77% probability that we find at least 1 of the seniors are married

Step-by-step explanation:

For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. So we use the binomial probability distribution to solve this question.

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 combinations 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.

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]

In this problem, we have that:

[tex]n = 300, p = 0.02[/tex]

a.) Find the mean and standard deviation of the sample count X who are married.

Mean

[tex]E(X) = np = 300*0.02 = 6[/tex]

Standard deviation

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42[/tex]

b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?

This is P(X = 8).

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

[tex]P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040[/tex]

10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c.) What is the probability that we find less than 4 of the seniors are married?

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

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

[tex]P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023[/tex]

[tex]P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143[/tex]

[tex]P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436[/tex]

[tex]P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883[/tex]

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485[/tex]

14.85% probability that we find less than 4 of the seniors are married.

d.) What is the probability that we find at least 1 of the seniors are married?

Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So

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

From c), we have that [tex]P(X = 0) = 0.0023[/tex]. So

[tex]0.0023 + P(X \geq 1) = 1[/tex]

[tex]P(X \geq 1) = 0.9977[/tex]

99.77% probability that we find at least 1 of the seniors are married

In this problem, the mean and standard deviation of a binomial distribution, with probability of success 0.02 and sample size 300, are found. Subsequently, the probabilities that 8, less than 4, and at least 1 of the seniors are married are computed using the binomial formula.

This problem deals with the Binomial distributions in statistics. Since we know the proportion of high school seniors who are married is 0.02, and the sample size is 300, we can use these values to calculate the mean and the standard deviation.

a.) The mean (mean = np) of the sample count X who are married is 0.02*300=6, and the standard deviation would be sqrt(n*p*(1-p)) = sqrt(300*0.02*0.98) = √5.88≈2.43.

b.) The probability that, in our sample of 300, we find that 8 of the seniors are married is given by the binomial formula P(X=k) = C(n,k)(p^k)(1-p)^(n-k). Plugging n=300, k=8, p=0.02 into the formula, we get the desired probability.

c.) The probability that we find less than 4 of the seniors are married is sum of P(X=k) from k=0 to 3. This could be computed using the aforementioned binomial formula. Remember, you're summing the probabilities for each k.

d.) The probability that we find at least 1 of the seniors are married can be found by subtracting the probability that none of the seniors are married from 1 (i.e., P(X >=1) = 1 - P(X=0)).

https://brainly.com/question/39749902

#SPJ11

Answer:

64.69e221

Step-by-step explanation:

When choosing, the combination formula for selection is used. That is when selecting "r" number of items from a possible "n" items, then the number of ways is denoted as:

nCr = n! / (n-r)! × r!

Since 66 string quartet have to be chosen and the 3 genres must be equally represented in the string quartet, then we must have 22 number of each genre in it.

Number of ways to select 22 mendelssohn from possible 66 = 66C22 = 1.82 × 10^17

Number of ways to select 22 Beethoven from possible 1616 = 1616C22 = 2.97 × 10^49

Number of ways to select 22 Haydn from possible 6868 = 6868C22 = 2.2 × 10^63

Total number of ways to arrange these 66 schedules = 66! = 5.44 × 10^92

Number of possible schedule = [1.82 * 10^17] * [2.97*10^49] * [2.2*10^63] * [5.44*10^92]

=64.69 ×10^221. ≈64.69e221

Answer:

An insurance company reports that 75% of its claims are settled within two months of being filed. In order to test that the percent is less than seventy-five, a state insurance commission randomly selects 35 claims and determines that 23 of the 35 were settled within two months.

a) Write out the hypotheses.

Fewer than 75% of claims settled within two months of filing.

b) Calculate the test statistic.

Test statistic = percent of claims settled in two months = 23/35 = 65.7%

c) Find the p-value.

we need to use the z-score with a table

=  

standard deviation = s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2}, =

√(1/34) [(0.657)(12) + (0.343)(23)] = √0.463911765 = 0.681

let 1 = "solved" and 0 = "unresolved"

thus our mean is 0.657

z = (0.75 - 0.657)/(0.681) - 0.137

p = 0.55172

d) Do we reject the null hypothesis? Explain.

Yes, because our p value is not below 0.05 and is not substantial to prove our null hypothesis

e) What can you conclude based on this evidence?

That further testing is needed.

Answer:

x + 2

3x − 1

x − 2

Step-by-step explanation:

The order of the polynomial is 4, so there are 4 roots.

Use rational root theorem to find possible rational roots.

±1, ±2, ±4, ±8, ±1/3, ±2/3, ±4/3, ±8/3

By trial and error, -2, ⅓, and 2 are all roots.  The fourth root can't be imaginary (complex roots come in pairs), so one of the real roots must be repeated.

Answer: the factors are (x - 2)(x + 2)(3x - 1)

Step-by-step explanation:

The given polynomial function is expressed as

f(x) = 3x⁴ - 7x³ - 10x² + 28x - 8

The first step is to test for any value of x that satisfies the polynomial when

3x⁴ - 7x³ - 10x² + 28x - 8 = 0

Assuming x = 2, then

3(2)⁴ - 7(2)³ - 10(2)² + 28 × 2 - 8 = 0

48 - 56 - 40 + 56 - 8 = 0

0 = 0

It means that x - 2 is a factor.

To determine the other factors, we would apply the long division method. The steps are shown in the attached photo. Looking at the photo, we would factorize the quadratic equation which is expressed as

3x² + 5x - 2 = 0

3x² + 6x - x - 2 = 0

3x(x + 2) - 1(x + 2)

(3x - 1)(x + 2) = 0

Given that the equation of the line is [tex]y=4x-3[/tex]

We need to determine the point (2,7) lies on the line.

For a point to lie on the line, substituting the point in the equation of the line makes it valid.

Thus, substituting the point (2,7) in the equation of the line [tex]y=4x-3[/tex], we get;

[tex]7=4(2)-3[/tex]

[tex]7=8-3[/tex]

[tex]7\neq 5[/tex]

Thus, the both sides of the equation are not equal.

Substituting the point (2,7) in the equation makes the equation invalid.

Thus, the point (2,7) does not lie on the line  [tex]y=4x-3[/tex]

Let the width = X

Then the length would be 2X ( twice the width).

Area is Length x width, so you have X * 2X = 18

x *2x = 2x^2

Now you have 2x^2 = 18

Divide both sides by 2:

x^2 = 9

Take the square root of both sides:

x = sqrt(9)

x = 3

The width is 3 inches.

The width of the rectangle is determined by establishing an equation from the given area (18 square inches) and the fact that the length is twice the width, which leads to the solution that the width is 3 inches.

To find the width of a rectangle when the area is 18 square inches, and the length is twice the width, we need to set up an equation. Let the width be w inches. Then the length would be 2w inches, since it's twice the width. We know that the area (A) of a rectangle is the multiplication of its length and width, so we have:

A = length  imes width

18 = 2w  imes w

18 = 2w2

Now we solve for w:

w2 = 18 / 2

w2 = 9

w = 3

The width of the rectangle is 3 inches.

Answer:

(a) The probability that the individual likes both vehicle #1  and vehicle #2 is 0.40.

(b) The value of P (A₂ | A₃) is 0.7143.

(c) The events A₂ and A₃ are not independent.

(d) The probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.

Step-by-step explanation:

The events are defined as follows:

A₁ = an individual like vehicle #1

A₂ = an individual like vehicle #2

A₃ = an individual like vehicle #3

The information provided is:

[tex]P(A_{1})=0.55\\P(A_{2})=0.65\\P(A_{3})=0.70\\P(A_{1}\cup A_{2})=0.80\\P(A_{2}\cap A_{3})=0.50\\P(A_{1}\cup A_{2}\cip A_{3})=0.88\\[/tex]

(a)

Compute the probability that the individual likes both vehicle #1  and vehicle #2 as follows:

[tex]P(A_{1}\cap A_{2})=P(A_{1})+P(A_{2})-P(A_{1}\cup A_{2})\\=0.55+0.65-0.80\\=0.40[/tex]

Thus, the probability that the individual likes both vehicle #1  and vehicle #2 is 0.40.

(b)

Compute the value of P (A₂ | A₃) as follows:

[tex]P(A_{2}|A_{3})=\frac{P(A_{2}\cap A_{3})}{P(A_{3}}\\=\frac{0.50}{0.70}\\=0.7143[/tex]

Thus, the value of P (A₂ | A₃) is 0.7143.

(c)

If two events X and Y are independent then,

[tex]P(X\cap Y)=P(X)\times P(Y)\\P(X|Y)=P(X)[/tex]

The value of P (A₂ ∩  A₃) is 0.50.

The product of the probabilities, P (A₂) and P (A₃) is:

[tex]P(A_{2})\times P(A_{3})=0.65\times0.70=0.455[/tex]

Thus, P (A₂ ∩ A₃) ≠ P (A₂) × P (A₃)

The value of P (A₂ | A₃) is 0.7143.

The value of P (A₂) is 0.65.

Thus, P (A₂ | A₃) ≠ P (A₂).

The events A₂ and A₃ are not independent.

(d)

Compute that probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ as follows:

[tex]P(A_{2}\cup A_{3}|A_{1}^{c})=\frac{P((A_{2}\cup A_{3})\cap A_{1}^{c})}{P(A_{1}^{c})}\\=\frac{P((A_{2}\cup A_{3}\cup A_{1})-P(A_{1})}{1-P(A_{1})} \\=\frac{0.88-0.55}{1-0.55}\\=0.7333[/tex]

Thus, the probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.

Final answer:

The probability that the individual likes both vehicle #1 and vehicle #2 is 0.40. Events A2 and A3 are not independent as P(A2 and A3) does not equal P(A2)P(A3). If the individual did not like vehicle #1, the probability of liking at least one of the other two vehicles is 0.73.

Explanation:

Probability of Likes and Independency of Events

To find the probability that the individual likes both vehicle #1 and vehicle #2, which is P(A1 ∩ A2), we use the formula: P(A1 ∩ A2) = P(A1) + P(A2) - P(A1 ∪ A2). Given that P(A1) = 0.55, P(A2) = 0.65, and P(A1 ∪ A2) = 0.80, we can calculate P(A1 ∩ A2):

P(A1 ∩ A2) = 0.55 + 0.65 - 0.80 = 0.40.

The conditional probability P(A2|A3) represents the probability of liking vehicle #2 given that the individual already likes vehicle #3. Using the definition of conditional probability:

P(A2|A3) = P(A2 ∩ A3) / P(A3) = 0.50 / 0.70 = approximately 0.714.

To determine if A2 and A3 are independent events, we can check if P(A2 ∩ A3) equals P(A2)P(A3) which would be 0.65 * 0.70 = 0.455. Since P(A2 ∩ A3) = 0.50, which is not equal to 0.455, A2 and A3 are not independent.

Moreover, if A2 and A3 were independent, then knowing that A3 occurred would not change the probability of A2 happening. However, P(A2|A3) is not equal to P(A2), indicating their dependence. Lastly, when the individual did not like vehicle #1, the probability that they liked at least one of the other two vehicles can be found using the probability of the union of events, subtracting the probability of liking vehicle #1:

P(A2 ∪ A3 | not A1) = P(A2 ∪ A3) - P(A1) + P(A1 ∩ A2 ∩ A3) = 0.88 - 0.55 + 0.40 = 0.73.

8 pounds of chamomile tea must be mixed to make a mixture that costs 14.63 per pound.              

Step-by-step explanation:

We are given the following in the question:

Chamomile tea:

Unit cost = 18.20 per pound

Amount = x pounds

Total cost =

[tex]18.20\times x = 18.20x[/tex]

Orange tea:

Unit cost = 12.25 per pound

Amount = 12 pounds

Total cost =

[tex]12.25\times 12 = 147[/tex]

Total mixture:

Unit cost = 14.63 per pound

Amount = (12+x) pounds

Total cost =

[tex]14.63\times (12+x) = 175.56 + 14.63x[/tex]

We can write the equation that cost of mixture is equal to cost of chamomile tea and orange tea.

[tex]18.20x + 147 = 175.56 + 14.63x\\\Rightarrow 18.20x- 14.63x = 175.56-147\\\Rightarrow 3.57x = 28.56\\\Rightarrow x = 8[/tex]

Thus, 8 pounds of chamomile tea must be mixed to make a mixture that costs 14.63 per pound.

Answer:

Please find attached answer.

Step-by-step explanation:

The Naive Bayes model is likely to classify the document 'machine learning is fun' and 'christmas family fun' as 'entertainment', even when using Laplace smoothing.

The Naive Bayes model classifies texts based upon the likelihood of a particular term belonging to a given class. To find the target level for the document, you multiply the prior probability of the document being in a given class (either entertainment or education) with the likelihoods of the terms in the query document being in that class and you choose the class with the highest probability.

https://brainly.com/question/21507963

#SPJ3

Answer: the American car gets better gas mileage.

Step-by-step explanation:

American car travels 32 miles on 1 gallon of gas. This means that its gas mileage is 32 miles per gallon.

European car travels 12.7 km on 1 L of gas. We would convert 12.7 km to miles.

1 kilometer = 0.621 mile

12.7 kilometer = 12.7 × 0.621 = 7.8867 miles

We would also convert 1 L to gallons.

1 liter = 0.264 us gallons

Therefore, the gas mileage of the European car in miles per gallon would be

7.8867/0.264 = 29.87 miles per gallon.

Therefore, since 32 miles per gallon is greater than 29.87 miles per gallon, it means that the American car gets better gas mileage.

Answer:

Total Cost = Fixed Cost as x --> ∞

Step-by-step explanation:

C'(x) = 4500 x⁻¹•⁹ where x ≥ 1

Marginal Cost = C'(x) = (dC/dx)

C(x) = ∫ (marginal cost) dx

C(x) = ∫ (4500 x⁻¹•⁹)

C(x) = (-5000 x⁻⁰•⁹) + k

where k = constant of integration or in economics term, K = Fixed Cost.

C(x) = [-5000/(x⁰•⁹)] + Fixed Cost

The company wants to make infinitely many units, that is, x --> ∞

C(x --> ∞) = [-5000/(∞⁰•⁹)] + Fixed Cost

(∞⁰•⁹) = ∞

C(x --> ∞) = [-5000/(∞)] + Fixed Cost

But mathematically, any number divide by infinity = 0;

(-5000/∞) = 0

C(x --> ∞) = 0 + Fixed Cost = Fixed Cost.

Total Cost of producing infinite number of units for this cost function is totally the Fixed Cost.

Answer:

a. First four terms of the sequential partial sums

[tex]S_1=\frac{1}{7}, S_2=\frac{7^2-1}{6*7^2}, S_3 =\frac{7^3-1}{6*7^3}, S_4 =\frac{7^4-1}{6*7^4}[/tex]

b. The formula for Sn is option A = [tex]\frac{7^n-1}{6*7^n}[/tex]

c. Value of the series = [tex]\frac{1}{6}[/tex]

Step-by-step explanation:

a. First four terms of the sequential partial sums

[tex]\sum^{\infty}_{k=1}\\S_1=\frac{1}{7}\\ S_2=\frac{1}{7}+ \frac{1}{7^2}= \frac{8}{49} =\frac{7^2-1}{6*7^2}\\S_3=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3} = \frac{57}{343} =\frac{7^3-1}{6*7^3}\\S_4=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4} = \frac{400}{2401} =\frac{7^4-1}{6*7^4}[/tex]

b. Formula for Sn

The sum of n terms

[tex]S_n=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3}+ \frac{1}{7^4}+.....+ \frac{1}{7^n}= \frac{7^n-1}{6*7^n}[/tex]

c. Value for the series

This can be given as the Sum of infinite terms;

[tex]S_{\infty}=\frac{1}{7}+ \frac{1}{7^2}+ \frac{1}{7^3}+ \frac{1}{7^4}+....+ \frac{1}{7^n}+ ....= \lim_{n \to \infty} \frac{7^n-1}{6*7^n}= \frac{1}{6}[/tex]

Answer:

The baby must weigh approximately 8.31 pounds.      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.5 pounds

Standard Deviation, σ = 1.2 pounds

We are given that the distribution of weights of newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find the third quartile or [tex]Q_3[/tex]

We have to find the value of x such that the probability is 0.75

[tex]P( X < x) = P( z < \displaystyle\frac{x - 7.5}{1.2})=0.75[/tex]  

Calculation the value from standard normal z table, we have,  

[tex]\displaystyle\frac{x - 7.5}{1.2} = 0.674\\\\x = 8.3088 \approx 8.31[/tex]

Thus, the baby must weigh approximately 8.31 pounds.

Answer:

95% confidence interval for the true percent of NAU students in Flagstaff who love their MAT114 class is (60.77% , 65.23%)

Step-by-step explanation:

Among 1800 NAU students, 1134 students love their class. We have to find the 95% confidence interval of students who love their class.

We will use the concept of confidence interval of population proportion for this problem.

The proportion of students who love the class = p = [tex]\frac{1134}{1800}=0.63[/tex]

Proportion of students who do not love the class = q = 1 - p = 1 - 0.63 = 0.37

Total number of students in the sample = n = 1800

Confidence Level = 95%

The z values associated with this confidence level(as seen from z table) = 1.96

The formula to calculate the confidence interval for population proportion is:

[tex](p-z\times\sqrt{\frac{p \times q}{n}},p+z\times\sqrt{\frac{p \times q}{n}})[/tex]

Using the values in this expression gives:

[tex](0.63-1.96 \times \sqrt{\frac{0.63 \times 0.37}{1800}}, 0.63+1.96 \times \sqrt{\frac{0.63 \times 0.37}{1800}})\\\\ =(0.6077,0.6523)[/tex]

Thus, 95% confidence interval for the true percent of NAU students in Flagstaff who love their MAT114 class is (0.6077 ,0.6523) or (60.77% , 65.23%