NEED HELP ASAP!!!!!! What is another way to say "to the third power"?

Answer:

a) see attached

b) see attached

Step-by-step explanation:

There is existence of outlier in the Ni data and there is none is Cr data.

########################################

# You can try this out in R programming

cr = c(34, 1, 511, 2, 574, 496, 322, 424, 269, 140, 244, 252, 76, 108, 24,

38, 18, 34, 30, 191)

Ni = c(23, 22, 55, 39, 283, 34, 159, 37, 61, 34, 163, 140, 32,

23, 54, 837, 64, 354, 376, 471)

par(mfrow=c(1,2))

hist(cr, col='green')

hist(Ni, col='brown')

par(mfrow=c(1,2))

boxplot(cr, main = 'Boxplot of Cr')

boxplot(Ni, main = 'Boxplot of Ni')

boxplot(cr, Ni)

The question involves constructing histograms and boxplots for two sets of data on soil concentrations of certain metals. After construction, these should be analysed for differences in central tendency, spread, variability, skewness, or presence of outliers.

This question seems to be related to Mathematics, specifically Statistics and Data Analysis. Firstly, to construct a histogram, you need to group the data into 'bins' or 'intervals', count how many data points fall into each bin, and then plot these counts as bars in your histogram. It's similar for both Chromium (Cr) and Nickel (Ni) data.

To construct comparative boxplots, it's essential to: calculate the quartiles (Q1, Q2 aka median, and Q3), identify the minimum and maximum points, and define possible outliers. With these statistical measures, you can draw the boxplots for both sets and analyse them. The dataset Cr seems to have several values clustered in the low range while Ni seems to have more spread, which can be verified by the boxplot comparison.

The interpretation of the boxplots should reflect on any differences in central tendency and spread between the two sets of concentrations. The variability, skewness, or outliers could also be significantly different. Any such inferred details from the boxplots may hint towards different contamination levels or heterogeneity in each of the datasets.

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Answer:

Step-by-step explanation:

Given:

[tex]P(S\cap F)=0.44\\P(S\cap F^{c})=0.13\\P(S^{c}\cap F^{c}) = 0.32\\P(S^{c}\cap F) = 0.11[/tex]

The rule of total probability states that:

[tex]P(A) = P(A\cap B) + P(A\cap B^{c})[/tex]

Compute the individual probabilities as follows:

[tex]P(S) = P(S\cap F) + P(S\cap F^{c})\\=0.44+0.13\\0.57[/tex]

[tex]P(S^{c}) = 1 - P(S)\\=1-0.57\\=0.43[/tex]

[tex]P(F) = P(S\cap F) + P(S^{c}\cap F)\\=0.44+0.11\\=0.55[/tex]

[tex]P(F^{c})=1-P(F)\\=1-0.55\\=0.45[/tex]

Conditional probability of an event A given B is:

[tex]P(A|B)=\frac{P(A\cap B)}{P(B)}[/tex]

         [tex]P(S|F)=\frac{P(S\cap F)}{P(F)}\\=\frac{0.44}{0.55}\\=0.80[/tex]

        [tex]P(S|F^{c})=\frac{P(S\cap F^{c})}{P(F^{c})}\\=\frac{0.13}{0.45}\\=0.289[/tex]

        [tex]P(S^{c}|F)=\frac{P(S^{c}\cap F)}{P(F}\\=\frac{0.11}{0.55}\\=0.20[/tex]

       [tex]P(S^{c}|F^{c})=\frac{P(S^{c}\cap F^{c})}{P(F^{c})}\\=\frac{0.32}{0.45}\\=0.711[/tex]

The probability of successful product given favorable test market is 0.80, and the probability of successful product given unfavorable test market is 0.29. The probability of unsuccessful product given favorable test market is 0.20, and the probability of unsuccessful product given unfavorable test market is 0.71.

To solve this, we first need to determine the total probability of each market test outcome. The probability that the market test is favorable (P(favorable test market)) can be found by summing the probabilities of a successful product with a favorable test market and an unsuccessful product with a favorable test market. Therefore, P(favorable test market) = 0.44 + 0.11 = 0.55. Similarly, P(unfavorable test market) = 0.13 + 0.32 = 0.45.

To find the conditional probabilities, we use the formula P(A|B) = P(A and B) / P(B):

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Answer:

7/12

Step-by-step explanation:

Step 1:  Find common denominators

The common denominator is 12

10/12 is already good

1/4 * 3/3 = 3/12

Step 2:  Subtract

10/12 - 3/12 = 7/12

Answer:  7/12

Final answer:

To subtract the fractions 10/12 and 1/4, find a common denominator, convert the fractions, and then subtract the numerators, resulting in the answer 7/12.

Explanation:

To subtract fractions with unlike denominators, you first need to find a common denominator. For the fractions 10/12 and 1/4, the common denominator is 12. You can convert 1/4 to 3/12, because 4 times 3 equals 12. Now you can subtract the fractions as follows:

10/12 - 3/12 = (10 - 3)/12 = 7/12

So, the answer to 10/12 - 1/4 is 7/12.

This process relies on the intuitive understanding that you can only subtract numerators if the denominators are the same. We never add or subtract the denominators themselves. When your fractions share the same denominator, the subtraction is straightforward as we are simply dealing with parts of the whole that are of the same size.

Answer: m= 18g

Step-by-step explanation:

Let m= miles you can drive

g = g for gallons in tank

We know that there is directly proportional relation between the distance traveled by vehicle and the number of gallons of gas.

Then, rate of miles driven by you per gallon = [tex]\dfrac{m}{g}[/tex]

Since , rate of miles driven by you per gallon  = 18 miles per gallon (given)

Then,

[tex]\dfrac{m}{g}=18\\\\ m= 18g[/tex]

Hence, the equation to model this situation : m= 18g

Answer:

[tex]45^{\circ}[/tex]

Step-by-step explanation:

We are given that two lines equation

[tex]x+\sqrt 3y=1[/tex]...(1)

[tex](1-\sqrt 3)y+(1+\sqrt 3)y=8[/tex]

Compare with the equation of line

ax+by+c=0

[tex]a_1=1,b_1=\sqrt 3[/tex]

[tex]a_2=(1-\sqrt 3),b_2=(1+\sqrt 3)[/tex]

The angle between two lines =Angle between two vectors

The angle between two vector

[tex]a_1i+b_1j[/tex] and

[tex]a_2i+b_2j[/tex]

is given by

[tex]cos\theta=\frac{a_1a_2+b_1b_2}{\sqrt{a^2_1+b^2_1}\sqrt{a^2_2+b^2_2}}[/tex]

Using the formula

Therefore, the angle between two lines

[tex]cos\theta=\frac{1(1-\sqrt 3)+\sqrt 3(1+\sqrt 3)}{\sqrt{(1)+(\sqrt 3)^2}\times \sqrt{(1-\sqrt 3)^2+(1+\sqrt 3)^2}}[/tex]

[tex]cos\theta=\frac{1-\sqrt 3+\sqrt 3+3}{\sqrt{1+3}\times\sqrt{1+3-2\sqrt 3+1+3+2\sqrt 3}}[/tex]

[tex]cos\theta=\frac{4}{2\times\sqrt 8}=\frac{2}{2\sqrt 2}[/tex]

[tex]cos\theta=\frac{1}{\sqrt 2}[/tex]

[tex]cos\theta=cos45^{\circ}[/tex]

By using [tex]cos45^{\circ}=\frac{1}{\sqrt 2}[/tex]

[tex]\theta=45^{\circ}[/tex]

Hence, the angle between two lines =45 degree

Answer:

[tex] A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196 [/tex]

Step-by-step explanation:

For this case the solution flows at a rate of 2L/min and leaves at 1L/min. So then we can conclude the volume is given by [tex] V= 10 +t[/tex]

Since the initial volume is 10 L and the volume increase at a rate of 1L/min.

For this case we can define A as the concentration for the salt in the container. And for this case we can set up the following differential equation:

[tex] \frac{dA}{dt}= 4 \frac{gr}{L} *2 \frac{L}{min} - \frac{A}{10+t}[/tex]

Because at the begin we have a concentration of 8 gr/L and would be decreasing at a rate of [tex] \frac{A}{10+t}[/tex]

So then we can reorder the differential equation like this:

[tex] \frac{dA}{dt} +\frac{A}{10+t} =8[/tex]

We find the solution using the integration factor:

[tex] \mu = -\int \frac{1}{10+t} dt = -ln(10+t)[/tex]

And then the solution would be given by:

[tex] A = e^{-ln (10+t)} (\int e^{\int \frac{1}{10+t} dt})[/tex]

And if we simplify this we got:

[tex] A= \frac{1}{10+t} (c + \int (10 +t) 8 dt)[/tex]

And after do the integral we got:

[tex] A= \frac{c}{10+t} +4 (10 +t) [/tex]

And using the initial condition t=0 A= 20 we have this:

[tex] 20 = \frac{c}{10} +40[/tex]

[tex] c= -200[/tex]

So then we have this function for the solution of A:

[tex] A= \frac{-200}{10+t} +4 (10 +t) [/tex]

And now replacinf t= 40 we got:

[tex] A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196 [/tex]

The amount of salt in the tank after 40 minutes is determined by first calculating the total amount of salt added to the tank, then considering the amount of salt-water mixture that exited the tank during this period due to the outgoing flow.

This question relates to a typical problem within the field of differential equations. You are asked to find the amount of salt in the tank after 40 minutes. The initial amount of salt in the tank is 20g, and additional salt-water solution is being pumped in at a rate of 2L/min with a salt concentration of 4g/L, hence adding 8g of salt per minute. At the same time, the well-stirred mixture is flowing out at a rate of 1L/min.

Therefore, after 40 minutes:
Additional salt added = 8g/min * 40 min = 320g.
Total salt content now = initial salt content + added salt = 20g + 320g = 340g.

However, within these 40 minutes,40 L of the mixture has flown out. Since the mixture is well-stirred, it retains the same concentration of salt as the mixture in the tank. Hence, if x is the amount of salt left in the tank, it will be on average the same concentration as that which flowed out and can be represented as follows: x/50 = (340-x)/40.

Solving for x we get the amount of salt left in the tank after 40 minutes.

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Answer:

Step-by-step explanation:

a) probability a person admitted to the hospital will suffer a treatment-caused injury due to negligence

P(injury) = 4%

P(negligence) = 1/4 = 0.25

We need to find probability (injury)(negligence)

P(injury) * P(negligence) = 0.04*0.25 = 0.01

b) probability a person admitted to the hospital will die from a treatment-caused injury

P(injury) = 4%

P(death) = 1/7

P(Injury) *P(death) = 0.04/7 = 0.00571

c) In the case of a negligent treatment-caused injury, what is the probability a malpractice claim will be paid

P(claim) = 1/7.5

P(payment) = 1/2

P(claim)*P(payment) = 1/7.5 * 1/2 = 0.06

Answer:

a.) 0.01

b.) 0.006

c.)0.00067

Step-by-step explanation:

Answer:

[tex]1332.1 \:cm^3[/tex] is neither significantly low nor significantly high.

Step-by-step explanation:

From the information given, we know that:

The mean is [tex]\bar{x}=1105.1 \:cm^{3}[/tex] and the standard deviation is [tex]\sigma=123.5 \:cm^{3}[/tex].

The range rule of thumb says that:

[tex]minimum \:usual \:value=\bar{x}-2\sigma\\\\maximum \:usual \:value=\bar{x}+2\sigma[/tex]

Applying these definitions, we get that

[tex]minimum \:usual \:value=1105.1-2(123.5)=858.1 \:cm^3\\\\maximum \:usual \:value=1105.1+2(123.5)=1352.1 \:cm^3[/tex]

We note that [tex]1332.1 \:cm^3[/tex] is between [tex]858.1 \:cm^3[/tex] and [tex]1352.1 \:cm^3[/tex], which indicates  [tex]1332.1 \:cm^3[/tex] is neither significantly low nor significantly high.

Answer:

a) [tex]P(F) = \frac{672}{1254}=\frac{112}{209}=0.536[/tex]

b) [tex]P(M) = \frac{582}{1254}= \frac{97}{209}=0.464[/tex]

c) [tex] P(A') = 1-P(A) = 1- \frac{124}{672}= \frac{137}{168}=0.815[/tex]

Step-by-step explanation:

For this case we have a total of 1254 people. 672 are women and 582 are female.

We know that 124 women wnat on to graduate school.

And 198 male want on to graduate school

We can define the following events:

F = The alumnus selected is female

M= The alumnus selected is male

A= Female and attend graduate school

And we can find the probabilities required using the empirical definition of probability like this:

Part a

[tex]P(F) = \frac{672}{1254}=\frac{112}{209}=0.536[/tex]

Part b

[tex]P(M) = \frac{582}{1254}= \frac{97}{209}=0.464[/tex]

Part c

For this case we find the probability for the event A: The student selected is female and did attend graduate school

[tex] P(A) =\frac{124}{672}=\frac{31}{168}=0.185[/tex]

And using the complement rule we find P(A') representing the probability that the female selected did not attend graduate school like this:

[tex] P(A') = 1-P(A) = 1- \frac{124}{672}= \frac{137}{168}=0.815[/tex]

Answer:

2,929,687 pictures

Step-by-step explanation:

1 pixel requires 2 bytes of storage

A color digital photo uses 512 pixels (512×2 bytes = 1024 bytes) per row and 512 pixels (1024) bytes per column

Total storage used by 1 picture = 1024 + 1024 = 2048 bytes

Storage capacity my camera = 6GB = 6×10^9 bytes

Number of pictures I can have in my camera = 6×10^9/2048 = 2,929,687 pictures

By calculating the storage required for one 512 x 512 pixel photo and dividing the total storage capacity by this amount, we find that a camera with 6 GB of storage can hold approximately 12,288 such photos.

To calculate how many 512 x 512 pixel color digital photos can be stored in a camera with 6 GB of storage, where each pixel requires two bytes of storage, we follow these steps:

First, find the total number of pixels in one photo:

512 pixels/row × 512 pixels/column = 262,144 pixels/photo.

Next, calculate the storage required for one photo:

262,144 pixels/photo × 2 bytes/pixel = 524,288 bytes/photo.

Since there are 1,024 bytes in one kilobyte (KB) and 1,024 KB in one megabyte (MB), we convert the photo size to megabytes:

524,288 bytes/photo / 1,024 bytes/KB / 1,024 KB/MB
≈ 0.5 MB/photo.

Now, convert 6 GB of storage to MB:

6 GB × 1,024 MB/GB = 6,144 MB.

Finally, divide the total storage by the size of one photo to determine the number of photos:

6,144 MB / 0.5 MB/photo = 12,288 photos.

Therefore, you could store approximately 12,288 such pictures on a camera with 6 GB of storage.

Answer:

0.0202

Step-by-step explanation:

I got this because there is 1000 millimeters in a meter. if you divide 20.2 by 1000, you get 0.0202.

Have a good day :3

Answer:

The number of rookies who scored 4 or better is 24.

Step-by-step explanation:

the frequency distribution of for the ranking of pro-football rookies is:

Ranking   Frequency

    1                  3

    2                 7

    3                16

    4                21

    5                  3

Compute the number of rookies who scored 4 or better as follows:

No. of rookies with rankings 4 or more = No. of rookies with rank 4 +

                                                                           No. of rookies with rank 5

                                                                 = 21 + 3

                                                                 = 24

Thus, the number of rookies who scored 4 or better is 24.

Answer:

a) P=0.0225

b) P=0.057375

Step-by-step explanation:

From exercise we have that  15​% of items produced are defective, we conclude that  probabiity:

P=15/100

P=0.15

a)  We calculate the probabiity  that two items are​ defective:

P= 0.15 · 0.15

P=0.0225

b)  We calculate the probabiity  that  two  of  three items are​ defective:

P={3}_C_{2} · 0.85 · 0.15 · 0.15

P=\frac{3!}{2!(3-2)!} · 0.019125

P=3 · 0.019125

P=0.057375

Using the binomial distribution, it is found that there is a:

a) 0.0225 = 2.25% probability that both are​ defective.

b) 0.0574 = 5.74% probability that two are​ defective.

For each item, there are only two possible outcomes, either it is defective, or it is not. The probability of an item being defective is independent of any other item, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

In this problem, 15​% of items produced are defective, hence [tex]p = 0.15[/tex].

Item a:

Two items, hence [tex]n = 2[/tex].

The probability is P(X = 2), then:

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

[tex]P(X = 2) = C_{2,2}.(0.15)^{2}.(0.85)^{0} = 0.0225[/tex]

0.0225 = 2.25% probability that both are​ defective.

Item b:

Three items, hence [tex]n = 3[/tex].

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

[tex]P(X = 2) = C_{3,2}.(0.15)^{2}.(0.85)^{1} = 0.0574[/tex]

0.0574 = 5.74% probability that two are​ defective.

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Answer:

a) Selling price y= a + b (age x)

b)

a= 728.025

b= -38.217

c)

Selling price y = 728.025 - 38.217 age x

d)

SSE=8280.25

Step-by-step explanation:

a)

The regression model can be written as

y=a+bx

Here y=selling price and x is age.

So, the regression model will be

Selling price y= a + b (age x)

b)

We have to find the values of "a" and "b"

[tex]b=\frac{nsumxy-(sumx)(sumy)}{nsumx^{2} -(sumx)^2}[/tex]

sumx=5+10+12+14+15=56

sumy=500+400+300+200+100=1500

sumxy=5*500+10*400+12*300+14*200+15*100=14400

sumx²=5²+10²+12²+14²+15²=690

n=5

[tex]b=\frac{5(14400)-(56)(1500)}{5(690) -(56)^2}[/tex]

b=-12000/314

b=-38.217

ybar=a+bxbar

a=ybar-bxbar

ybar=sumy/n=1500/5=300

xbar=sumx/n=56/5=11.2

a=300-(-38.217)(11.2)

a=300+428.025

a=728.025

c)

Selling price y = a - b(age x)

Selling price y = 728.025 - 38.217 age x

d)

SSE known as sum of square of error can be calculated as

SSE=sum(y-yhat)²

y  500 400 300 200 100

x   5      10    12     14    15

yhat= 728.025 - 38.217 age x 536.940  345.855 269.421  192.987  154.770

y-yhat  -36.940  54.145  30.579 7.013  -54.770

(y-yhat)² 1364.56  2931.68  935.08  49.18  2999.75

SSE=sum(y-yhat)²

SSE=1364.56 +2931.68 +935.08 +49.18 +2999.75

SSE =8280.25

Answer:

The measure of dispersion which is likely to vary most between your first and second samples is the range.

Step-by-step explanation:

The range and standard deviation of a data are measures of dispersion, i.e. they measure the degree to which the data is dispersed.

The formula to compute the range is:

[tex]Range=X_{max}-X_{min}[/tex]

The formula to compute the sample standard deviation is:

[tex]s=\sqrt{\frac{1}{n-1}\sum (X-\bar X)^{2} }[/tex]

The sample size is: n = 50.

Thus, the measure of dispersion which is likely to vary most between your first and second samples is the range.

Final answer:

The range is more likely to vary between two samples from a population due to its sensitivity to extreme values, whereas the standard deviation is a more stable measure of dispersion that accounts for all values in the dataset.

Explanation:

The measure of dispersion most likely to vary between the first and second samples from the population is the range. The range is sensitive to the particular values in the dataset because it only takes into account the smallest and largest values. Since each sample may have a different set of extreme values, the range can vary greatly from sample to sample. In contrast, the standard deviation is less likely to vary significantly between samples, as it measures the amount of variation or dispersion from the mean of a dataset and takes into account all the values in the set.

When drawing multiple samples from a population, the distribution of the sample means will tend to form a normal distribution around the population mean according to the Central Limit Theorem. In the case of a normally distributed population with a known mean and standard deviation, the sampling distribution of the sample mean will have a standard deviation equal to the population standard deviation divided by the square root of the sample size (known as the standard error).

Answer:

No solution.

Step-by-step explanation:

A system has only one solution if it ends at ax = b, in which a is different than 0.

If it ends at 0x = 0, the system has infinitely many solutions.

If it ends at a division by 0, or 0 = constant(different than 0), the system is inconsistent.

Our system is:

−20x + 30y = −3

8x − 12y = −3

I am going to multiply the top equations by 2 and the bottom equation by 5, and add them. So

-40x + 60y = -6

40x - 60y = -15

So

-40x + 40x + 60y - 60y = -6 - 15

0x + 0y = -21

We cannot divide 21 by 0, which means that this system of equations has no solution.

Answer:

There is not enough evidence to support the claim that interested consumers would be willing, on average, to pay $20.00 for the product.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $20.00

Sample mean, [tex]\bar{x}[/tex] = $18.14

Sample size, n = 315

Population standard deviation, σ = $2.98

a) Reference value

[tex]\mu = 20[/tex]

b) First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 20.00\text{ dollars}\\H_A: \mu \neq 20.00\text{ dollars}[/tex]

We use Two-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{18.14 - 20.00}{\frac{2.98}{\sqrt{315}} } = -11.078[/tex]

c) Calculating the p-value at 5% significance level

P-Value < 0.00001

Thus,

p<0.001

d) Calculating the p-value at 1% level of significance

P-Value < 0.00001

Thus,

p<0.001

Thus, at both 5% and 1% level of significance, the p value is lower than the significance level, we fail to accept the null hypothesis and reject it.

There is not enough evidence to support the claim that interested consumers would be willing, on average, to pay $20.00 for the product.

Answer:

The value of sensitivity obtained is equal to the one given in statement.

Step-by-step explanation:

Sensitivity is the ratio of number of positive test results to the number of individuals having disease. It is the ability of the test to detect all those with disease in the screened population.

Its formula is written as:

Sensitivity = [tex]\frac{number of true positives}{total with disease}[/tex]

So, in this case we have values

sensitivity = 14/16 = 0.875 So, the given statement is true.

Answer:

$7,000,000

Step-by-step explanation:

The taxable income is the amount of money (income earned or unearned) by an individual or an organization that creates a potential tax liability.

The formula is shown below:

Taxable Income Formula = Gross Total Income – Total Exemptions – Total Deductions

Note that $5,000,000 for 2015 is an operating loss which is a deduction.

Gross Total Income = $4000000+$4000000+$4000000 = $12,000,000

Total Exemption = 0

Total Deductions = $5,000,000

Taxable Income Formula = $12,000,000 - $5,000,000 = $7,000,000

The effective taxable income for 2017 after carrying forward the operating loss is -$1 million.

The question pertains to a business scenario and involves calculating the effective taxable income after carrying forward an operating loss. To calculate the effective taxable income for 2017, we need to consider the operating losses from previous years. In this case, your firm had a $5 million operating loss in 2015. This loss can be carried forward to offset taxable income in future years.

Therefore, the effective taxable income for 2017 after carrying forward the operating loss is -$1 million.

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Answer:

parameter tested=σ(population standard deviation)

two tailed

Step-by-step explanation:

We are given that null hypothesis that σ is equal to 88 and alternative hypothesis that σ is not equal to 88. The parameter that is tested here is σ.

σ denotes the population standard deviation. Thus, the population parameter σ is tested here.

The type of test depends on the alternative hypothesis we have taken. The alternative hypothesis states that σ is not equal to 88. It means that σ can either be greater than 88 or less than 88. Thus, the test can either be right tailed or left tailed. This type of test is called as two tailed test.

Answer:

option b

-x^2 + 3x - 11

Step-by-step explanation:

we have to add the X of the same order

x^2

-3x^2 + 2x^2 = -x^2

x^1

4x + ( -x ) = 3x

x^0

0 + (-11) = -11

finally we accommodate and get the final result

-x^2 + 3x - 11

Answer:-x^2 + 3x - 11

Step-by-step explanation:

I got it right on the test

Answer:

See explanation

Step-by-step explanation:

In Saras class, [tex]\frac{2}{5}[/tex] of the students ride a bus, and [tex]\frac{1}{3}[/tex] ride a car to school.

The rest

[tex]1-\dfrac{2}{5}-\dfrac{1}{3}=\dfrac{15-6-5}{15}=\dfrac{4}{15}[/tex]

walk to school (simply subtract from one whole the fractions of students).

The fraction of students who ride a bus or car to school is

[tex]\dfrac{2}{5}+\dfrac{1}{3}=\dfrac{6+5}{15}=\dfrac{11}{15}.[/tex]

See attached diagram for pictorial representations (red diagram shows the fraction of students who ride a bus, blue - ride a car, green - walk to school)

The fraction of students who walk to school is [tex]$\boxed{\frac{4}{15}}$[/tex].

To find the fraction of students who walk to school, we first consider the fractions of students who ride a bus or car to school. According to the information given:

- The fraction of students who ride a bus is [tex]$\frac{2}{5}$[/tex].

- The fraction of students who ride a car is [tex]$\frac{1}{3}$[/tex].

To find the total fraction of students who use transportation (bus or car), we add these two fractions together. However, we must first find a common denominator to add them properly. The least common multiple (LCM) of 5 and 3 is 15, so we convert both fractions to have a denominator of 15:

[tex]- $\frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15}$ (for the bus riders)[/tex]

[tex]- $\frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15}$ (for the car riders)[/tex]

Now we can add these fractions:

[tex]- $\frac{6}{15} + \frac{5}{15} = \frac{11}{15}$[/tex]

This represents the fraction of students who either ride a bus or a car to school. To find the fraction of students who walk to school, we subtract this fraction from the whole, which is 1 (since the whole class is represented by 1, or [tex]$\frac{15}{15}$):[/tex]

[tex]- $1 - \frac{11}{15} = \frac{15}{15} - \frac{11}{15} = \frac{4}{15}$[/tex]

Therefore, the fraction of students who walk to school is [tex]$\frac{4}{15}$[/tex].

To visualize this with a diagram, we can imagine a whole circle representing the entire class. We divide this circle into 15 equal parts (since our denominators are 15). We shade 6 parts to represent the bus riders and another 5 parts to represent the car riders. The remaining unshaded parts, which are 4 in number, represent the students who walk to school.

The equation representing the total fraction of students who walk or use transportation is:

[tex]\[ \text{Bus riders} + \text{Car riders} + \text{Walkers} = \text{Whole class} \] \[ \frac{6}{15} + \frac{5}{15} + \text{Walkers} = 1 \][/tex]

Solving for Walkers:

[tex]\[ \text{Walkers} = 1 - \left( \frac{6}{15} + \frac{5}{15} \right) \] \[ \text{Walkers} = 1 - \frac{11}{15} \] \[ \text{Walkers} = \frac{15}{15} - \frac{11}{15} \] \[ \text{Walkers} = \frac{4}{15} \][/tex]

Thus, the fraction of students who walk to school is[tex]$\boxed{\frac{4}{15}}$.[/tex]

Answer:

1. 1/12 or 0.083

2. 5/36 or 0.1389

3. 7/12 or 0.5833

4. 5/12 or 0.4167

5. 11/12 or 0.9167

6. 1/9 or 0.1111

7. 1/3 or 0.3333

8. 5/12 or 0.4167

Step-by-step explanation:

This problem can easily be understood by making sample space first.

Outcomes in sample space=S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Total outcomes=36

The probability in the following scenario is calculated as

Probability=number of outcomes/Total outcomes

1.

The sum is 4

The sum is 4 ={(1,3),(2,2),(3,1)}

number of outcomes=3

P(The sum is 4)=3/36=1/12 or 0.083

2.

The sum is 6

The sum is 6 ={(1,5),(2,4),(3,3),(4,2),(5,1)}

number of outcomes=5

P(The sum is 6)=5/36 or 0.1389

3.

The sum is at least 7

The sum is at least 7=The sum is greater than or equal to 7

The sum is at least 7={(1,6),(2,5),(2,6),(3,4),(3,5),(3,6),(4,3),(4,4),(4,5),(4,6),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

number of outcomes=21

P(The sum is at least 7)=21/36

P(The sum is at least 7)=7/12 or 0.5833

4.

The sum is at least 8

The sum is at least 8=The sum is greater than or equal to 8

The sum is at least 8={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

number of outcomes=15

P(The sum is at least 8)=15/36

P(The sum is at least 8)=5/12 or 0.4167

5.

The sum is less than 11

The sum is less than 11={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(6,1),(6,2),(6,3),(6,4)}

number of outcomes=33

P(The sum is less than 11)=33/36

P(The sum is less than 11)=11/12 or 0.9167

6.

The sum is 2, 3, or 12

The sum is 2, 3, or 12={(1,1),(1,2),(2,1),(6,6)}

number of outcomes=4

P(The sum is 2, 3, or 12)=4/36

P(The sum is 2, 3, or 12)=1/9 or 0.1111

7.

The sum is odd and no more than 7

The sum is odd and no more than 7={(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(6,1)}

number of outcomes=12

P(The sum is odd and no more than 7)=12/36

P(The sum is odd and no more than 7)=1/3 or 0.3333

8.

The sum is odd or prime

The sum is odd or prime={(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)}

number of outcomes=15

P(The sum is odd or prime)=15/36

P(The sum is odd or prime)=5/12 or 0.4167

Probabilities are,

1. The sum is 4 = 1/12

2. The sum is 6 = 5/36

3. The sum is at least 7 = 7/12

4. The sum is at least 8 = 5/12

5. The sum is less than 11 = 11/12

6. The sum is 2, 3, or 12 = 1/9

7. The sum is odd and no more than 7 = 1/3

8. The sum is odd or prime = 19/36

When six sided die toss two times, Then total possible Outcomes shown below

[tex](1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}[/tex]

Total number of  outcomes=36

Probability is calculated by dividing number of favourable outcomes by Total number of  outcomes

1. Favourable outcomes for the sum is 4 [tex]={(1,3),(2,2),(3,1)}[/tex]

number of outcomes=3

The probability of The sum is 4. = [tex]3/36=1/12 = 0.083[/tex]

2.Favourable outcomes for the sum is 6[tex]={(1,5),(2,4),(3,3),(4,2),(5,1)}[/tex]

number of outcomes=5

The probability of The sum is 6, =5/36  

3.The sum is at least 7,

={(1,6),(2,5),(2,6),(3,4),(3,5),(3,6),(4,3),(4,4),(4,5),(4,6),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

number of outcomes=21

Probability of the sum is at least 7 =21/36=7/12 = 0.5833

4.The sum is at least 8

The sum is at least 8={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

number of outcomes=15

The sum is at least 8=15/36 = 5/12 = 0.4167

5.The sum is less than 11

The sum is less than 11={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(6,1),(6,2),(6,3),(6,4)}

number of outcomes=33

The sum is less than 11=33/36 = 11/12

6.The sum is 2, 3, or 12={(1,1),(1,2),(2,1),(6,6)}

number of outcomes=4

The sum is 2, 3, or 12=4/36 = 1/9

7.The sum is odd and no more than 7={(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(6,1)}

number of outcomes=12

The sum is odd and no more than 7=12/36 = 1/3

8.The sum is odd or prime={(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5), (3, 6), (6, 3), (4,5), (5, 4)

number of outcomes=19

The sum is odd or prime=19/36

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Answer:

a. The angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(\sqrt{3}, 1\right)[/tex] and [tex]\mathbf{v}=\left(0, 5\right)[/tex] is [tex]\phi=\frac{\pi}{3}[/tex] or [tex]60\º[/tex].

b. The angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(0, 4, 4\right)[/tex] and [tex]\mathbf{v}=\left(3, -3, 0\right)[/tex] is [tex]\phi=\frac{2 \pi}{3}[/tex] or [tex]120\º[/tex].

Step-by-step explanation:

a. To calculate the angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(\sqrt{3}, 1\right)[/tex] and [tex]\mathbf{v}=\left(0, 5\right)[/tex] you must:

Step 1: Calculate the dot product.

The dot product is given as [tex]\displaystyle{\large{{\left({u}_{{x}},{u}_{{y}}\right)}\cdot{\left({v}_{{x}},{v}_{{y}}\right)}={u}_{{x}}\cdot{v}_{{x}}+{u}_{{y}}\cdot{v}_{{y}}}}[/tex].

So,

[tex]\left(\sqrt{3}, 1\right)\cdot\left(0, 5\right)=\left(\sqrt{3}\right)\cdot\left(0\right)+\left(1\right)\cdot\left(5\right)=5[/tex]

Step 2: Find the lengths of the vectors.

[tex]\left|\mathbf{u}\right|=\sqrt{\left(u_x\right)^2+\left(u_y\right)^2}=\sqrt{\left(\sqrt{3}\right)^2+\left(1\right)^2}=2[/tex]

[tex]\left|\mathbf{v}\right|=\sqrt{\left(v_x\right)^2+\left(v_y\right)^2}=\sqrt{\left(0\right)^2+\left(5\right)^2}=5[/tex]

Step 3: The angle is given by [tex]\cos\left(\phi\right)=\frac{\mathbf{u} \cdot \mathbf{v}}{\left|\mathbf{u}\right| \cdot \left|\mathbf{v}\right|}[/tex]

[tex]\frac{5}{2 \cdot 5}=\frac{1}{2}[/tex]

[tex]\phi=cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}=60^0[/tex]

b. To calculate the angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(0, 4, 4\right)[/tex] and [tex]\mathbf{v}=\left(3, -3, 0\right)[/tex] you must:

Step 1: Calculate the dot product.

[tex]\left(0, 4, 4\right)\cdot\left(3, -3, 0\right)=\left(0\right)\cdot\left(3\right)+\left(4\right)\cdot\left(-3\right)+\left(4\right)\cdot\left(0\right)=-12[/tex]

Step 2: Find the lengths of the vectors.

[tex]\left|\mathbf{u}\right|=\sqrt{\left(u_x\right)^2+\left(u_y\right)^2+\left(u_z\right)^2}=\sqrt{\left(0\right)^2+\left(4\right)^2+\left(4\right)^2}=4 \sqrt{2}[/tex]

[tex]\left|\mathbf{v}\right|=\sqrt{\left(v_x\right)^2+\left(v_y\right)^2+\left(v_z\right)^2}=\sqrt{\left(3\right)^2+\left(-3\right)^2+\left(0\right)^2}=3 \sqrt{2}[/tex]

Step 3: The angle is given by [tex]\cos\left(\phi\right)=\frac{\mathbf{u} \cdot \mathbf{v}}{\left|\mathbf{u}\right| \cdot \left|\mathbf{v}\right|}[/tex]

[tex]\cos\left(\phi\right)=\frac{\mathbf{u} \cdot \mathbf{v}}{\left|\mathbf{u}\right| \cdot \left|\mathbf{v}\right|}=\frac{-12}{4 \sqrt{2} \cdot 3 \sqrt{2}}=- \frac{1}{2}[/tex]

[tex]\phi=cos^{-1}\left(- \frac{1}{2}\right)=\frac{2 \pi}{3}=120^0[/tex]

To find the angle between two vectors, use the dot product formula. For the vectors (√3,1) and <0, 5 >, the angle is π/3 radians. For the vectors <0,4,4> and <3,-3,0>, the angle is 2π/3 radians.

To find the angle between two vectors, we can use the dot product formula. Let's start with the first set of vectors (√3,1) and <0, 5>. We can calculate the dot product by multiplying the corresponding components and adding them up: (√3)(0) + (1)(5) = 5. Next, we need to calculate the magnitudes of each vector. The magnitude of (√3,1) is √(√3)^2 + 1^2 = √4 = 2. The magnitude of <0, 5> is √0^2 + 5^2 = √25 = 5. Now, we can calculate the angle between the vectors using the formula: cosθ = dot product / (magnitude 1 * magnitude 2). Plugging in the values, cosθ = 5 / (2 * 5) = 1/2. Taking the inverse cosine of 1/2 gives us the angle in radians: θ = acos(1/2) = π/3 ≈ 1.047.

For the second set of vectors <0,4,4> and <3,-3,0>, we will follow the same steps. The dot product is (0)(3) + (4)(-3) + (4)(0) = -12. The magnitude of <0,4,4> is √0^2 + 4^2 + 4^2 = √32 = 4√2. The magnitude of <3,-3,0> is √3^2 + (-3)^2 + 0^2 = √18 = 3√2. Plugging in the values, cosθ = -12 / (4√2 * 3√2) = -1/2. Taking the inverse cosine of -1/2 gives us the angle in radians: θ = acos(-1/2) = 2π/3 ≈ 2.094.

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Answer:

60th percentile.

Step-by-step explanation:

When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.

In this problem, we have that:

300 observations, there are 180 points less than 81.2.

180/300 = 0.6

So 60% of the observation are lower than 81.2, which means that 81.2 is the 60th percentile.

Answer:

a) So 25% probability that exactly two heads occur, given that the first outcome was a tail.

b) 50% probability that exactly two heads occur, given that the first two outcomes were heads.

c) 0% probability that exactly two heads occur, given that the first two outcomes were tails.

Step-by-step explanation:

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

We have the following sample space, that is, the possible outcomes:

In which h is heads and t is tails

h - h - h

h - h - t

h - t - h

h - t - t

t - h - h

t - h - t

t - t - h

t - t - t

What is the probability that exactly two heads occur, given that

a. the first outcome was a tail

Four outcomes in which the first outcome was a tail. They are:

t - h - h

t - h - t

t - t - h

t - t - t

In only 1 one them, exactly two heads occur.

1/4 = 0.25

So 25% probability that exactly two heads occur, given that the first outcome was a tail.

b. the first two outcomes were heads

Two possibilities in which the first two outcomes were heads.

h - h - h

h - h - t

In 1 of them, we have exactly two heads.

1/2 = 0.5

So 50% probability that exactly two heads occur, given that the first two outcomes were heads.

c. the first two outcomes were tails

Two possibilities in which the first two outcomes were tails.

t - t - h

t - t - t

In none of them we have exactly 2 heads.

0% probability that exactly two heads occur, given that the first two outcomes were tails.

The probability of getting exactly two heads in three coin tosses is 5/8.

To find the probability of exactly two heads occurring in three coin tosses, we need to consider three different scenarios:

If the first outcome was a tail: In this case, we have two remaining tosses and we need both of them to be heads. The probability of getting a head on any single toss is 1/2, so the probability of getting two heads after a tail is (1/2) * (1/2) = 1/4.

If the first two outcomes were heads: In this case, we have one remaining toss and we need it to be a tail. The probability of getting a tail on the remaining toss is 1/2, so the probability of getting exactly two heads after two heads is (1/2) = 1/2.

If the first two outcomes were tails: In this case, we again have one remaining toss and we need it to be a head. The probability of getting a head on the remaining toss is 1/2, so the probability of getting exactly two heads after two tails is (1/2) = 1/2.

Adding up the probabilities of these three scenarios, the overall probability of getting exactly two heads in three coin tosses is 1/4 + 1/2 + 1/2 = 5/8.

The area of a triangle is 1/2 x base x height

Area = 1/2 x 24 x 28 = 336 square inches.

Because the corners are rounded the are would actually be just under 336 but no radius was given to get the exact area.

Answer:

yes

Step-by-step explanation:

Answer:

a) 52.15

b) 58.53

c) 62.98        

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 56

Standard Deviation, σ = 3

We are given that the distribution of x is a bell shaped distribution that is a normal distribution.

Formula:

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

a) We have to find [tex]x_0[/tex] such that

P(X < x)  = 0.1

[tex]P( X < x) = P( z < \displaystyle\frac{x_0 - 56}{3})=0.1[/tex]  

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

[tex]P(z < -1.282) = 0.1[/tex]

[tex]\displaystyle\frac{x_0 - 56}{3} = -1.282\\\\x_0 = 52.15[/tex]

b) We have to find [tex]x_0[/tex] such that

P(X < x)  = 0.8

[tex]P( X < x) = P( z < \displaystyle\frac{x_0 - 56}{3})=0.8[/tex]  

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

[tex]P(z < 0.842) = 0.8[/tex]

[tex]\displaystyle\frac{x_0 - 56}{3} = 0.842\\\\x_0 = 58.53[/tex]

c) We have to find [tex]x_0[/tex] such that

P(X > x)  = 0.01

[tex]P( X > x) = P( z > \displaystyle\frac{x - 56}{3})=0.01[/tex]  

[tex]= 1 -P( z \leq \displaystyle\frac{x - 56}{3})=0.01 [/tex]  

[tex]=P( z \leq \displaystyle\frac{x - 56}{3})=0.99 [/tex]  

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

[tex]P(z < 2.326) = 0.99[/tex]

[tex]\displaystyle\frac{x_0 - 56}{3} = 2.326\\\\x_0 = 62.98[/tex]

To find specific percentiles for a normally distributed variable x with mean 56 and standard deviation 3, you must use a z-score table or calculator to find the corresponding z-scores for the 10%, 80%, and top 1% percentages, then use the formula x0 = µ + z*σ to calculate x0.

To find the value x0 of the normally distributed random variable x with mean µ = 56 and standard deviation σ = 3, that corresponds to specific percentiles, one would typically use a z-score table (or a calculator with a normal distribution function). However, in this scenario, the values provided for the x variable and the associated z-scores in the question are incorrect and would not provide a reasonable answer. Instead, I will explain the steps needed to find x0 without using the incorrect values provided.

After finding each z-score, use the formula x0 = µ + z*σ to calculate the corresponding x0 for each case.

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Answer:

skew lines

Step-by-step explanation:

Given are two lines in 3 dimension as

[tex]L1: x=23+6t, y=12+3t, z=19+5t\\L2: x=-9+7t, y=-7+5t, z=-12+8t[/tex]

To find out whether parallel or skew or intersect

If parallel direction ratios should be proportional

Direction ratios of I line are 6,3,5 and not proportional to that of II line (7,5,8)

So not parallel

If intersect we must have same point for the two lines

Let us change parameter for II line to s to avoid confusion.  If intersecting, then

[tex]23+6t = -9+7s\\12+3t = -7+5s\\19+5t =12+8s\\[/tex]

6t-7s=-32 and 3t-5s = -19

Solving

t=-3 and s =2

Check this with III equation

Left side = 19-15 =4 and right side = 12+16 =28

not equal

So there cannot be any point of intersection.

These two are skew lines

Answer:

d. field notes recorded during participant observation

Step-by-step explanation:

Qualitative data are data that can be gathered and described by observation, interviews and evaluation. Such data are not numerically based but are based on features of the phenomenon under study. It is also known as categorical data.

Methods of Qualitative data includes :

Focus groups, keeping records, direct interviews, case studies etc.