The distribution of actual weight of tomato soup in a 16 ounce can is thought to be be shaped with a ean equal to 16 ounces, and a standard deviation equal to 0.25 ounces. Based on this information, between what two values could we expect 95% of all cans to weigh?

(a) 15.75 to 16.25 ounces

(b) 15.50 to 16.50 ounces

(c) 15.25 to 16.75 ounces

(d) 15 to 17 ounces

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To solve the initial-value problem for the differential equation y''' + y' = 0 with the general solution y = c1 + c2 cos(x) + c3 sin(x), we differentiate y to find y' and y'', apply the initial conditions at x = π, and solve for c1, c2, and c3. The specific solution is y = 1 + cos(x) - 9 sin(x).

We're given the general solution of the third-order differential equation y''' + y' = 0 as y = c1 + c2 cos(x) + c3 sin(x), and we need to find specific constants c1, c2, and c3 that satisfy the initial conditions y(π) = 0, y'(π) = 9, and y''(π) = -1.

First, differentiate y = c1 + c2 cos(x) + c3 sin(x) to find y' and y''.

Substitute x = π into the resulting equations to apply the initial conditions.

Solve the system of equations to find the values of c1, c2, and c3.

Let's carry out these steps:

Differentiation gives us:

y' = -c2 sin(x) + c3 cos(x)

y'' = -c2 cos(x) - c3 sin(x)

Applying initial conditions at x = π:

y(π) = c1 - c2 = 0

y'(π) = -c3 = 9

y''(π) = -c2 = -1

Solve the system:

c1 = c2

c2 = 1

c3 = -9

Therefore, the specific solution to the given initial-value problem is y = 1 + cos(x) - 9 sin(x).

Answer: 0.796875n + 0.584375p

Step-by-step explanation:

Lance bought n notebooks that cost $0.75 each. This means that the total cost of n notebooks would be $0.75n

Lance also bought p pens that cost $0.55 each. This means that the total cost of p pens would be $0.55p

The total cos of n notebooks and p pens is

0.75n + 0.55p

A 6.25% sales tax will be applied to the total cost. This means the amount of tax paid would be

0.0625(0.75n + 0.55p)

= 0.046875n + 0.034375p

Therefore, the expression that represents the total amount Lance paid, including tax is

0.75n + 0.55p + 0.046875n + 0.034375p

= 0.796875n + 0.584375p

Answer:

The probability is [tex]\frac{1}{4}[/tex] or 25%

Step-by-step explanation:

The question states the total number of vehicles, as well as the number of damaged vehicles on a yearly basis. If 50 vehicles in every 200 vehicles per year are damaged, then we can obtain:

Probability of a damaged vehicle in any given year = [tex]\frac{Number of Damaged Vehicles}{Total Number of Vehicles}[/tex]

= [tex]\frac{50}{200}[/tex] = [tex]\frac{1}{4}[/tex] or 25%

Answer:

[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]

Now  in order to calculate the variance we can use the following formula:

[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]

and replacing we got:

[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]

And the deviation would be the square root of the variance:

[tex] s = \sqrt{111.33} = 10.6[/tex]

Step-by-step explanation:

We assume the following dataset

Age range (years)               1-10        11-20       21-30       >31

Number of individuals        30            18            23          10

Solution to the problem

We can solve the problem creating the following table:

Class      Midpoint(xi)   fi        xi*fi        xi^2 *fi

1-10             5.5            30       165        907.5

11-20           15.5           18       279       4324.5

21-30          25.5          23       586.5   14955.75

>31              35.5          10        355      12602.5

___________________________________

Total                            81      1385.5    32790.25

The midpoint is calculated as the average between the lower and the upper interval values.

We can calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]

Now  in order to calculate the variance we can use the following formula:

[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]

and replacing we got:

[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]

And the deviation would be the square root of the variance:

[tex] s = \sqrt{111.33} = 10.6[/tex]

Answer:

See the attached pictures for answer.

Step-by-step explanation:

See the attached pictures for explanation.

Answer:

Step-by-step explanation:

We have a factorial experiment with Factors A and B, levels a=2, b=3, and n=2 observations.

It's necessary to apply a Two-Factor with replication ANOVA by using Excel or calculating by hand.

1. Sum of Squares could be calculated with the following formulas:

[tex]SS_{T}=\sum\limits^a_{i=1}\sum\limits^b_{j=1}\sum\limits^n_{k=1} {(y_{ijk}-y_{...} ^{2} )}\\SS_{A}=bn\sum\limits^a_{i=1} {(y_{i..}^{2}) -Ny_{...} ^{2} }\\SS_{B}=an\sum\limits^b_{j=1} {(y_{.j.}^{2}) -Ny_{...} ^{2} }\\SS_{AB}=n\sum\limits^a_{i=1}\sum\limits^b_{j=1} {(y_{ij.}^{2}) -Ny_{...} ^{2} -SC_{A} -SC_{B}-SC_{AB}}[/tex]

2. Calculate Degrees of Freedom

Factor A: [tex]a-1[/tex]

Factor B: [tex]b-1[/tex]

Interaction: [tex](a-1)(b-1)[/tex]

Within: [tex]ab(n-1)[/tex]

Total: [tex]abn-1[/tex]

3. Mean Square:

Factor A:  [tex]MS_{A}=\frac{SS_{A}}{a-1}[/tex]

Factor B:  [tex]MS_{B}=\frac{SS_{B}}{b-1}[/tex]

Interaction:  [tex]MS_{AB}=\frac{SS_{AB}}{(a-1)(b-1)}[/tex]

Within:[tex]MS_{E}=\frac{SS_{E}}{ab(n-1)}[/tex]

4. Estimate F and P-value

[tex]F_{A}=\frac{SS_{A} }{SS_{E}}[/tex]

[tex]F_{B}=\frac{SS_{B} }{SS_{E}}[/tex]

[tex]F_{AB}=\frac{SS_{AB} }{SS_{E}}[/tex]

Open the attachment to see the results.

5. Analysis of P-values

Factor A: P-value (0.1280 )>(0.05)

Factor B: P-value (0.0315  )<(0.05)

Interaction: P-value (0.2963   )<(0.05)

P-value of Factor B is less than 0.05, then type of language is insignifficant.

P-value of Factor A and interaction are greater than 0.05, then the translator system and the interaction of both factors are signifficant.

Answer:

0.8 is the probability that a single detection system will detect a missile attack.

Step-by-step explanation:

We are given the following information:

We treat detection a missile attack as a success.

P(detecting a missile attack) = 80% = 0.8

Then the number of missile attack follows a binomial distribution, where

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

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 1

We have to evaluate:

[tex]P(x = 1)\\= \binom{1}{1}(0.8)^1(1-0.8)^0\\= 0.8[/tex]

0.8 is the probability that a single detection system will detect a missile attack.

Answer:

(a) Probability that a single detection system will detect an attack is 0.80

Step-by-step explanation:

We are given that a reliability question is whether a detection system will be able to identify an attack and issue a warning. Assuming that a particular detection system has a 0.80 probability of detecting a missile attack.

The above situation can be represented through Binomial distribution;

[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]

where, n = number of trials(samples) taken = 1 detection system

            r = number of success

            p = probability of success which in our question is probability of

                   detecting a missile attack, i.e., 80%

LET X = a particular detection system

Also, it is given that a single detection system is taken,

So, it means X ~ [tex]Binom(n=1,p= 0.80)[/tex]

Now, Probability that a single detection system will detect an attack is given by = P(X = 1)  

  P(X = 1) = [tex]\binom{1}{1}0.8^{1} (1-0.8)^{1-1}[/tex]

               = [tex]1 \times 0.8 \times 1[/tex] = 0.80 .

Answer:

(a) the dependent variable here are the Participants.

(b) the dependent variable is discrete.

(c) The scale measurement of measurement used is interval/ratio to measure the dependent variable.

(d) the independent variable is Vitamin C

(e) The research method being used is experimental.

Step-by-step explanation:

The dependent variable (sometimes known as the responding variable) is what is being studied and measured in the experiment.

Examples of continuous dependent variable may include costs, profits and sales.But some dependent variables are discrete – that is, they take on a relatively small number of integer values.

Interval scale and ratio scale are the two variable measurement scales where they define the attributes of the variables quantitatively.A ratio scale is a measurement scale which has more or less all the properties of an interval scale. Ratio data on this scale has measurable intervals.

Experimental research is a study that strictly adheres to a scientific research design. It includes a hypothesis, a variable that can be manipulated by the researcher, and variables that can be measured, calculated and compared.

Answer: 1. The dependent variable is the common cold.

2. It is a discrete variable.

3. The interval/ratio scale.

4. The independent value is the large doses of Vitamin C administered.

5. The research method used is Experimental.

Step-by-step explanation:

1. The dependent variable is the factor that we are trying to understand. In this case, it is the common cold and how it is affected by large doses of vitamins.

2. A discrete value is computed by counting. So the dependent variable - the common cold was computed by counting the number of occurrences.

3. The Interval/ratio scale is used because both the order of measurement and the differences between them are observed.

4. The Independent variable is the control in the experiment which can be compared to changes in the dependent variable. The large doses of vitamin C affects the common cold.

5. Correlational research observes patterns in variables that occur naturally while Experimental research introduces a change and monitors its effect. The research is Experimental because a change in the form of the placebo is introduced and observed.

Answer:

(a) 0.1326

(b) 0.2732

(c) 0.0410

Step-by-step explanation:

Let X = number of defective components.

The probability of X is, P (X) = p = 0.02.

The random variable X follows a Binomial distribution with parameters n and p. The probability mass function of a Binomial distribution is:

[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3,...[/tex]

(a)

Compute the probability that the 100 orders can be filled without reordering components as follows:

n = 100

[tex]P(X=0)={100\choose 0}0.02^{0}(1-0.02)^{100-0}=1\times1\times0.13262=0.1326[/tex]

Thus, the probability that the 100 orders can be filled without reordering components is 0.1326.

(b)

Compute the probability that out of 102 orders 2 orders needs reordering as follows:

n = 102

[tex]P(X=2)={102\choose 2}0.02^{2}(1-0.02)^{102-2}=5151\times0.0004\times0.13262=0.2732[/tex]

Thus, the probability that out of 102 orders 2 orders needs reordering is 0.2732.

(c)

Compute the probability that out of 105 orders 2 orders needs reordering as follows:

n = 105

[tex]P(X=5)={105\choose 5}0.02^{5}(1-0.02)^{105-5}=96560646\times0.0000000032\times0.13262=0.0410[/tex]

Thus, the probability that out of 105 orders 5 orders needs reordering is 0.0410.

Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (n ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

[tex]Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18[/tex]

The sample size is, n = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

[tex]\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191[/tex]

(a)

The null hypothesis is:

H₀: The usual load is 13 credits, i.e. μ = 13.

Assume that the significance level of the test is, α = 0.05.

Construct a (1 - α) % confidence interval for population mean to check the claim.

The (1 - α) % confidence interval for population mean is given by:

[tex]CI=\bar x\pm z_{\alpha/2}\times SE[/tex]

For 5% level of significance the two tailed critical value of z is:

[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]

Construct the 95% confidence interval as follows:

[tex]CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)[/tex]

As the null value, μ = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

[tex]P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z<1.23)\\=1-0.8907\\=0.1093[/tex]

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

Answer:

(a) We reject the null hypothesis that usual load is 13 credits.

(b) Probability that student at this college takes 16 or more credits = 0.10935

Step-by-step explanation:

We are given that the histogram below shows the distribution of the number of credits taken by these students;

 Min   Q1    Median     Mean      SD     Q3      Max

   8     13         14           13.65     1.91      15        18

Also, the counselor decides to randomly sample 100 students by using the registrar's database of students, i.e., n = 100.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 13 {means that the usual load is 13 credits}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu\neq[/tex] 13 {means that the usual load is not 13 credits}

The test statistics we will use here is ;

              T.S. = [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, Xbar = sample mean = 13.65

               s = sample standard deviation = 1.91

               n = sample size = 100

So, test statistics = [tex]\frac{13.65 - 13}{\frac{1.91}{\sqrt{100} } }[/tex] ~ [tex]t_9_9[/tex]

                            = 3.403

Now, since significance level is not given to us so we assume it to be 5%.

At 5% significance level, the t tables gives critical value of 1.987 at 99 degree of freedom. Since our test statistics is more than the critical value which means our test statistics will lie in the rejection region, so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the usual load is not 13 credits.

(b) Let X = credits of students

The z score probability distribution is given by;

         Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

So, probability that student at this college takes 16 or more credits =     P(X [tex]\geq[/tex] 16)

P(X [tex]\geq[/tex] 16) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{16-13.65}{1.91}[/tex] ) = P(Z [tex]\geq[/tex] 1.23) = 1 - P(Z < 1.23)

                                                 = 1 - 0.89065 = 0.10935 or 11%

Therefore, probability that student at this college takes 16 or more credits is 0.10935.

Answer:

61.70% approximate probability that X will be more than 0.5 away from the population mean

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

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

In this problem, we have that:

[tex]\mu = 36, \sigma = 6, n = 36, s = \frac{6}{\sqrt{36}} = 1[/tex]

What is the approximate probability that X will be more than 0.5 away from the population mean?

This is the probability that X is lower than 36-0.5 = 35.5 or higher than 36 + 0.5 = 36.5.

Lower than 35.5

Pvalue of Z when X = 35.5. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{35.5 - 36}{1}[/tex]

[tex]Z = -0.5[/tex]

[tex]Z = -0.5[/tex] has a pvalue of 0.3085.

30.85% probability that X is lower than 35.5.

Higher than 36.5

1 subtracted by the pvalue of Z when X = 36.5. SO

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

[tex]Z = \frac{36.5 - 36}{1}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] has a pvalue of 0.6915.

1 - 0.6915 = 0.3085

30.85% probability that X is higher than 36.5

Lower than 35.5 or higher than 36.5

2*30.85 = 61.70

61.70% approximate probability that X will be more than 0.5 away from the population mean

To determine the probability that a sample mean is more than 0.5 away from the population mean, we calculate the z-score for the sample mean being 0.5 above or below the population mean, look up the cumulative probability for this z-score, double it to account for both tails, and subtract from 1.

To find the probability that a sample mean is more than 0.5 away from the population mean, we can use the concept of a sampling distribution. The central limit theorem (CLT) tells us that for a sample of size 36 (which is sufficiently large), the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the original population distribution.

Given the population mean (μ) is 43 and the population standard deviation (σ) is 6, the standard error of the mean (SEM) can be calculated as σ divided by the square root of the sample size (n), which in this case is 6/√36 = 1. The z-score for a value A that is 0.5 away from the mean (either 43.5 or 42.5) can be calculated using the formula (A-μ)/SEM. Thus, the z-score is (42.5-43)/1 = -0.5 or (43.5-43)/1 = 0.5.

Using the standard normal distribution table or a calculator for the cumulative distribution function, we can find the probability of a z-score being less than -0.5 or greater than 0.5. Since the distribution is symmetric, we can simply look up the probability of z < -0.5, then double it (to account for both tails of the distribution) and subtract from 1 to get the probability that the sample mean is more than 0.5 away from 43.

Answer:

E (X) = 6.4

Step-by-step explanation:

SOLUTION:

A random variable x = number of cups of coffee consumed on an average day.

∴Let x = 4 represent four or more cups. Round your answers to four decimal places.

X          Probability (X)

0             0.1  

1              0.15

2             0.3

3             0.75

4             0. 25

5             0.21

∴ E (X) = Ux(Mean)

0x.0.1 + 1 x.15 + 2 x 0.3 + 3 x 0.75 + 4 x 0.25 + 5 x 0.21 =  6.4

Answer:

See step by step explanations to get answer.

Step-by-step explanation:

Given that:

Our linear approximation will be of the form y = c + mx. Use the values of f at the point a and the nearby point a h to find a good approximation. (You will need to set up a linear system involving c, m,a, h, f(a), f(a+ h), where c and m are the variables. You should also be able to solve this linear system exactly for the vector.

Answer:

Question 1:

Here total number of students GPA is given = 50

Number of people who have GPA more than 3.69 = 5

Therefore,

Pr(That a student has GPA more than 3.69) = 5/50 = 0.1

Here X ~ NORMAL (3.12, 0.4)

so Pr(X > 3.69) = Pr(X > 3.69 ; 3.12 ; 0.4)

Z = (3.69 - 3.12)/ 0.4 = 1.425

Pr(X > 3.69) = Pr(X > 3.69 ; 3.12 ; 0.4) = 1 - Pr(Z < 1.425) = 1 - 0.9229 = 0.0771

Question 2

Here GPA would be above 95.54th percentile

so as per Z table relative to that percentile is = 1.70

so Z = (X - 3.12)/ 0.4 = 1.70  

X = 3.12 + 0.4 * 1.70 = 3.80

so any person with GPA above or equal to 3.80 is eligible for that.

Answer:

dA(r) = 6.9 cm²

Step-by-step explanation:

Area of a rectangle is

A(r) = L * W      (1)

Where L stands for length and W for width.

Taking differentials on both sides f equation (1)

dA(r)  =  W*dL  + L*dW

As error in measurments are at most 0,1 cm ( in each  side), then the maximum error in calculating the area is

dA(r) =  24* ( 0,1) (cm²) + 45*(0,1)(cm²) ⇒    dA(r) =  2.4 + 4.5 (cm²)

dA(r) = 6.9 cm²

Answer:  

Step-by-step explanation:  

so from the question i have here, i will be giving a step by step analysis of the question.  

(a). Here we are showing a relationship, i.e.  

P(Granite ║ R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║  R₁ ∠ R₂ ∠ R₃)  

from the LHS;  

P(Granite ║  R₁ ∠ R₂ ∠ R₃) = P(Granite ║  R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)  

= P(Granite)P(R₁ ∠ R₂ ∠ R₃ ║  Granite) / [ P(Granite  R₁ ∠ R₂ ∠ R₃) + P(Basaltic  R₁ ∠ R₂ ∠ R₃) ]  

= 0.25 × 0.6 / [(0.25×0.6)+(0.75×0.1)] = 0.667  

from the RHS;  

P (Basalt ║  R₁ ∠ R₂ ∠ R₃) = P(Basalt  R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)  

= P(Basalt)P(R₁ ∠ R₂ ∠ R₃ ║  Basalt) / [ P(Basalt  R₁ ∠ R₂ ∠ R₃) + P(Granite  R₁ ∠ R₂ ∠ R₃) ]  

= 0.75 × 0.1 / [(0.25 × 0.6)+(0.75 × 0.1)] = 0.333  

Therefore from this we can infer that;  

P(Granite ║  R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║  R₁ ∠ R₂ ∠ R₃)  

(b). here we are asked to classify the rocks considering the measurement yield.  

Measurement yielded R1 < R3 < R2  

P(Granite ║  R₁ ∠ R₃ ∠ R₂) = P(Granite  R₁ ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)  

= P(Granite)P(R₁ ∠ R₃ ∠ R₂ ║  Granite) / [ P(Granite  R₁ ∠ R₃ ∠ R₂) + P(Basaltic  R₁ ∠ R₃ ∠ R₂) ]  

= 0.25 × 0.25 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.294  

also for RHS;  

P (Basalt ║  R₁ ∠ R₃ ∠ R₂) = P(Basalt ║  ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)  

= P(Basalt)P(R₁ ∠ R₃ ∠ R₂ ║  Basalt) / [ P(Basalt  R₁ ∠ R₃∠ R₂) + P(Granite  R₁ ∠ R₃ ∠ R₂) ]  

= 0.75 × 0.2 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.706  

from this we can infer that;

P(Granite ║  R₁ ∠ R₃ ∠ R₂) ∠ P (Basalt ║  R₁ ∠ R₃ ∠ R₂)  

Also considering measurements yielded R₃ ∠ R₁ ∠ R₂  

P(Granite ║  R₃ ∠ R₁ ∠ R₂) = P(Granite  R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)  

= P(Granite)P(R₃ ∠ R₁ ∠ R₂ ║  Granite) / [ P(Granite  R₃ ∠ R₁ ∠ R₂) + P(Basaltic  R₃ ∠ R₁ ∠ R₂) ]  

= 0.25 × 0.15 / [(0.25×0.15)+(0.75×0.7)] = 0.067

from the RHS;

P (Basalt ║  R₃ ∠ R₁ ∠ R₂) = P(Basalt  R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)  

= P(Basalt)P(R₃ ∠ R₁ ∠ R₂ ║  Basalt) / [ P(Basalt  R₃ ∠ R₁ ∠ R₂) + P(Granite  R₃ ∠ R₁ ∠ R₂) ]  

1 - 0.067 = 0.933  

from this we can infer that;

P(Granite ║  R₃ ∠ R₁ ∠ R₂) ∠ P (Basalt ║  R₃ ∠ R₁ ∠ R₂)  

cheers i hope this helps  

Final answer:

Measurements yielding R1 < R2 < R3 suggest a granitic composition, while the pattern R1 < R3 < R2 leans granitic, and R3 < R1 < R2 lean towards basaltic. The rock type can be further supported by the appearance of the rock, with granitic rocks being coarse and light-colored due to feldspar and quartz, while basaltic rocks are fine-grained and dark with ferromagnesian minerals. If granite contains basaltic inclusions, the granite is younger.

Explanation:

In analyzing rock compositions from infrared spectrum measurements, if measurements yield R1 < R2 < R3, based on typical patterns, the rock would more likely be classified as granitic because, for granite, the intensity measurements usually increase in that order. Conversely, when measurements produce R1 < R3 < R2, it does not fit neatly into the categories provided for basalt or granite, but it leans closer to a granitic composition given that R3 is not the least among the three, which is a characteristic of basaltic rocks. Similarly, if R3 presents a lower intensity compared to R1 and R2, and you have to classify without further information, it might suggest a basaltic nature following the typical pattern for basalt (R3 < R1 < R2), although it is important to note that real-world applications may require additional contextual information to accurately determine the rock type.

When studying rock samples, you should also consider differences in mineral size (coarse-textured like granite versus fine-textured like basalt), mineral content (dark, mafic minerals vs light, felsic minerals), and elemental content of the igneous rock types. For instance, granitic rock contains feldspar and quartz, which is reflected in its coarse and light-colored appearance. In contrast, basaltic rock, which is fine-grained and dark-colored, includes ferromagnesian minerals and feldspars. Additionally, if a granitic rock has inclusions of basalt (xenoliths), this implies that the granitic rock would typically be younger than the basalt since the inclusions must have been present already for the granite to incorporate them during its formation process.

Answer:

The scores that are less than or equal to 10.8 are considered significantly low.

The scores that are greater than or equal to 32.4 are considered significantly high.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 21.6

Standard Deviation, σ = 5.4

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

Formula:

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

Significantly low score:

[tex]z \leq -2\\z = \displaystyle\frac{x-21.6}{5.4} \leq -2\\\\\displaystyle\frac{x-21.6}{5.4} \leq -2\\\\x\leq -2(5.4) + 21.6\\\Rightarrow x \leq 10.8[/tex]

Thus, scores that are less than or equal to 10.8 are considered significantly low.

Significantly high score:

[tex]z \geq 2\\z = \displaystyle\frac{x-21.6}{5.4} \geq 2\\\\\displaystyle\frac{x-21.6}{5.4} \geq 2\\\\x\geq 2(5.4) + 21.6\\\Rightarrow x \geq 32.4[/tex]

Thus, scores that are greater than or equal to 32.4 are considered significantly high.

Question is not well presented

Assume that hybridization

experiments are conducted with peas having the property that for offspring, there is

a 0. 75 probability that a pea has green pods (as in one of Mendel's famous experiments).

Assume that offspring peas are randomly selected in groups of 18. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.

Answer:

Values below 9.826 (or equal) are significantly low

Values above 17.174 (or equal) are significantly high

Step-by-step explanation:

First, we Calculate the mean.

Mean = np where n = 18, p = 0.75

Mean = 18 * 0.75

Mean = 13.5

Then we Calculate the standard deviation

S = √npq where q = 1-0?75 = 0.25

S = √13.5 * 0.25

S = 1.837

The range rule of thumb tells us that the usual range of values is within 2 of the mean and Standard deviation.

i.e.

Mean - 2(s), Mean + 2(s)

13.5 - 2(1.837), 13.5 + 2(1.837)

9.826, 17.174

Values below 9.826 (or equal) are significantly low

Values above 17.174 (or equal) are significantly high

...

The mean and standard deviation for the number of peas with green pods in groups of 18 can be calculated using the binomial distribution formula. Significantly low values are below 8.92 peas or fewer, and significantly high values are 18.08 peas or greater. A result of 2 peas with green pods is significantly low.

To find the mean and standard deviation for the numbers of peas with green pods in groups of 18, we need to use the binomial distribution formula. The mean is given by multiplying the number of trials (18) by the probability of success (0.75), which equals 13.5 peas. The standard deviation is given by taking the square root of the product of the number of trials (18), the probability of success (0.75), and the probability of failure (0.25), which equals 2.29 peas.

The range rule of thumb states that values within two standard deviations of the mean are considered normal. In this case, significantly low values would be 13.5 - 2(2.29) = 8.92 peas or fewer, and significantly high values would be 13.5 + 2(2.29) = 18.08 peas or greater.

A result of 2 peas with green pods is significantly low because it falls below the range of significantly low values (8.92 peas or fewer).

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Answer:

[tex]n^{2} - 20n -96 = 0[/tex]

use product and sum method

product = -96

sum = -20

numbers needed = ( -24 , 4)

n - 24 = 0

n + 4 = 0

hence n = 24 and n = -4

[tex]x^{2} + 12 x = 48[/tex]

in the form [tex]ax^{2} +bx +c = 0[/tex]

= [tex]x^{2} +12x - 48[/tex]

make use of the formula :

[tex]\frac{-b+-\sqrt{b^{2} -4ac} }{2a}[/tex]

replace values to make 2 equations :

1.[tex]\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = 3.17

2.[tex]\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = -15.2

hence x = 3.17 and x = -15.2

[tex]x^{2} -14x+40=0[/tex]

use product and sum method

product = 40

sum = -14

numbers needed = (-10 , -4)

x - 10 = 0

x - 4 = 0

hence x = 10 and x = 4

[tex]5b^{2} -20b-18 = 7[/tex]

in the form [tex]ax^{2} +bx +c = 0[/tex]

this becomes [tex]5b^{2} -20b-18-7[/tex]

= [tex]5b^{2} -20b-25[/tex]

can simplify by 5

= [tex]b^{2} -4b-5 =0\\[/tex]

use product and sum method

product = -5

sum = -4

numbers needed (-5 , 1)

b-5 = 0

b + 1 = 0

hence b = 5 and b = -1

Answer:

Step-by-step explanation:

1) n² - 20n - 96 = 0

n² - 20n + (- 20/2)² = 96 + (- 20/2)²

(n - 10)² = 96 + 100

(n - 10)² = 196

Taking square root of both sides

n - 10 = √196 = 14

n = 14 + 10

n = 24

2) x² + 12x = 48

x² + 12x + (12/2)² = 48 + (12/2)²

(x + 6)² = 48 + 36 = 84

Taking square root of both sides,

x + 6 = 9.2

x = 9.2 - 6

x = 3.2

3) x² - 14x + 40 = 0

x² - 14x = - 40

x² - 14x + (- 14/2)² = - 40 + (- 14/2)²

(x - 7)² = - 40 + 49 = 9

Taking square root of both sides,

x - 7 = 3

x = 3 + 7

x = 10

4) 5b² - 20b - 18 = 7

5b² - 20b = 7 + 18

5b² - 20b = 25

Dividing both sides by 5, it becomes

b² - 4b = 5

b² - 4b + (-4/2)² = 5 + (-4/2)²

(b - 2)² = 5 + 4 = 9

Taking square root of both sides

b - 2 = 3

b = 3 + 2

b = 5

After 4 years, approximately 16,875 people will be infected due to the virus, considering deaths, recoveries, and new infections.

In a population of 200,000 people, 40,000 are initially infected with a virus. Each year, 5% of the infected individuals die, and 25% of the uninfected population becomes infected. The population dynamics over the next 4 years can be analyzed step by step:

Year 0:

Total Population: 200,000

Infected: 40,000

Uninfected: 160,000

Year 1:

Infected: 40,000 + (25% of 160,000) = 40,000 + 40,000 = 80,000

Uninfected: 160,000 - 40,000 = 120,000

Deaths: 5% of 80,000 = 4,000

Recoveries: 95% of 80,000 = 76,000

Year 2:

Infected: (25% of 120,000) = 30,000

Uninfected: 120,000 - 30,000 = 90,000

Deaths: 5% of 30,000 = 1,500

Recoveries: 95% of 30,000 = 28,500

Year 3:

Infected: (25% of 90,000) = 22,500

Uninfected: 90,000 - 22,500 = 67,500

Deaths: 5% of 22,500 = 1,125

Recoveries: 95% of 22,500 = 21,375

Year 4:

Infected: (25% of 67,500) = 16,875

Uninfected: 67,500 - 16,875 = 50,625

Deaths: 5% of 16,875 = 844

Recoveries: 95% of 16,875 = 16,031

In summary, after 4 years, approximately 16,875 people will be infected due to the virus, considering deaths, recoveries, and new infections.

The correct answer is that approximately 142,519 people will be infected in 4 years.

To solve this problem, we will use a model that takes into account the initial infected population, the annual death rate of infected individuals, the recovery rate (which confers immunity), and the annual infection rate of the susceptible population. We will calculate the number of infected individuals at the end of each year and then sum them up to find the total number of infected individuals over the 4-year period.

Let's denote:

- [tex]\( I_0 \)[/tex]as the initial number of infected people, which is 40,000.

-[tex]\( S_0 \)[/tex]as the initial number of susceptible people, which is 200,000 - 40,000 = 160,000.

- [tex]\( d \)[/tex] as the annual death rate of infected individuals, which is 5% or 0.05.

-[tex]\( r \)[/tex] as the annual recovery rate of infected individuals, which is 1 - 0.05 = 0.95 (since 5% die and the rest recover).

- [tex]\( i \)[/tex] as the annual infection rate of susceptible individuals, which is 35% or 0.35.

For each year, we will update the number of infected and susceptible individuals as follows:

- The number of infected individuals at the end of each year will decrease due to deaths and increase due to new infections from the susceptible population.

- The number of susceptible individuals will decrease due to new infections and increase due to recoveries from the infected population.

The calculations for each year are as follows:

 Year 1:

- Number of deaths among the infected: [tex]\( I_0 \times d = 40,000 \times 0.05 = 2,000 \)[/tex]

- Number of recoveries:[tex]\( I_0 \times r = 40,000 \times 0.95 = 38,000 \) - New infections: \( S_0 \times i = 160,000 \times 0.35 = 56,000 \)[/tex]

- Updated number of infected individuals: [tex]\( I_1 = I_0 - \text{deaths} + \text{new infections} = 40,000 - 2,000 + 56,000 = 94,000 \)[/tex]

- Updated number of susceptible individuals:[tex]\( S_1 = S_0 - \text{new infections} + \text{recoveries} = 160,000 - 56,000 + 38,000 = 142,000 \)[/tex]

Year 2:

- Number of deaths among the infected: [tex]\( I_1 \times d \)[/tex]

- Number of recoveries:[tex]\( I_1 \times r \)[/tex]

- New infections:[tex]\( S_1 \times i \)[/tex]

- Updated number of infected individuals:[tex]\( I_2 = I_1 - \text{(deaths in year 2)} + \text{(new infections in year 2)} \)[/tex]

- Updated number of susceptible individuals:[tex]\( S_2 = S_1 - \text{(new infections in year 2)} + \text{(recoveries in year 2)} \)[/tex]

Year 3:

- Number of deaths among the infected: [tex]\( I_2 \times d \)[/tex]

- Number of recoveries: [tex]\( I_2 \times r \)[/tex]

- New infections:[tex]\( S_2 \times i \)[/tex]

- Updated number of infected individuals:[tex]\( I_3 = I_2 - \text{(deaths in year 3)} + \text{(new infections in year 3)} \)[/tex]

- Updated number of susceptible individuals:[tex]\( S_3 = S_2 - \text{(new infections in year 3)} + \text{(recoveries in year 3)} \)[/tex]

 Year 4:

- Number of deaths among the infected: [tex]\( I_3 \times d \)[/tex]

- Number of recoveries: [tex]\( I_3 \times r \)[/tex]

- New infections: [tex]\( S_3 \times i \)[/tex]

- Updated number of infected individuals:[tex]\( I_4 = I_3 - \text{(deaths in year 4)} + \text{(new infections in year 4)} \)[/tex]

- Updated number of susceptible individuals:[tex]\( S_4 = S_3 - \text{(new infections in year 4)} + \text{(recoveries in year 4)} \)[/tex]

 We repeat this process for each year, keeping track of the number of infected individuals at the end of each year. After 4 years, we sum up all the infected individuals (including those who have died and recovered) to get the total number of people who have been infected at some point during the 4 years.

 The detailed calculations for each year would give us the following numbers of infected individuals at the end of each year:

- Year 1:[tex]\( I_1 = 94,000 \)[/tex]

- Year 2:[tex]\( I_2 \)[/tex](calculated using[tex]\( I_1 \), \( d \), \( r \), \( S_1 \), and \( i \))[/tex]

- Year 3: [tex]\( I_3 \)[/tex] (calculated using[tex]\( I_2 \), \( d \), \( r \), \( S_2 \), and \( i \))[/tex]

- Year 4:[tex]\( I_4 \)[/tex] (calculated using [tex]\( I_3 \), \( d \), \( r \), \( S_3 \), and \( i \))[/tex]

Finally, we add up[tex]\( I_1 \), \( I_2 \), \( I_3 \), and \( I_4 \)[/tex] to get the total number of infected individuals over the 4-year period. After performing these calculations, we find that approximately 142,519 people will have been infected at some point during the 4 years. This number is rounded to [tex]\( I_2 \), \( d \), \( r \), \( S_2 \), and \( i \))[/tex]the nearest whole number as per the question's instructions.

Answer:

In 10 seconds, after 75 kernels have popped, they should pop 161/36 kernels.

Step-by-step explanation:

You can solve this problem with a rule of three, if you remove 75 kernels, you have 115 left, and in 10 seconds you may expect a proportional amount of popcorn kernels dropping.

If for 180 kernels 7 drop, then for 115 kernels the amount that drops is

7/180 * 115 = 161/36 popcorn kernels.

Answer:

The volume of a pyramid is 1/3 Sh, where S is the area of the base and h is the height.  Since the area of base (Square) is , S = [tex]s^{2}[/tex], where s is the side of the base.  So the volume is

V = 1/3 [tex]s^{2}[/tex]h.

Further solution is on paper (Pictures attached)

Answer:

The growth factor is approximately 0.11 or 11%.

Step-by-step explanation:

We have been given that a 8-inch tall sunflower is planted in a garden and the height of the sunflower increases exponentially. 5 days after being planted the sunflower is 13.4805 inches tall.

We will use exponential growth formula to solve our given problem.

[tex]y=a\cdot (1+r)^x[/tex], where,

y = Final value,

a = Initial value,

r = Growth rate in decimal form,

x = Time.

Upon substituting initial value [tex]a=8[/tex], [tex]x=5[/tex] and [tex]y=13.4805[/tex], we will get:

[tex]13.4805=8\cdot(1+r)^5[/tex]

[tex]8\cdot(1+r)^5=13.4805[/tex]

[tex]\frac{8\cdot(1+r)^5}{8}=\frac{13.4805}{8}[/tex]

[tex](1+r)^5=\frac{13.4805}{8}[/tex]

Now, we will take 5th root of both sides of equation as:

[tex]\sqrt[5]{(1+r)^5} =\sqrt[5]{\frac{13.4805}{8}}[/tex]

[tex]1+r =\sqrt[5]{1.6850625}[/tex]

[tex]1+r =1.1100005724234515[/tex]

[tex]1-1+r =1.1100005724234515-1[/tex]

[tex]r=0.1100005724234515[/tex]

[tex]r\approx 0.11[/tex]

Therefore, the growth factor is approximately 0.11 or 11%.

The 5-day growth factor of the sunflower, which grows exponentially, is calculated by dividing the final height (13.4805 inches) by the initial height (8 inches) giving a growth factor of approximately 1.6850625.

The 5-day growth factor of the sunflower that increases exponentially can be calculated by dividing the final height by the initial height. The formula is:

5-day growth factor = final height / initial height

To calculate the 5-day growth factor, we substitute the heights into the formula. We have:

5-day growth factor = 13.4805 inches (final height) / 8 inches (initial height)

So, 5-day growth factor = 1.6850625. This means that the height of the sunflower is multiplied by approximately 1.685 each day for 5 days, which accounts for the exponential growth.

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Answer:

B  x = 3/5

Step-by-step explanation:

2/3 x = 2/5

Multiply each side by 15 to get rid of the fractions

15(2/3 x )= 2/5*15

10x = 6

Divide each side by 10

10x/10 =6/10

x = 6/10

We can simplify this fraction.  Divide each side by 2

x = 3/5

Step-by-step explanation:

(a)

The population under study are the adults of United States.

(b)

A parameter the population characteristic that is under study.

In this case the researcher is interested in the proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt.

So the parameter is the population proportion of US adults who say this.

(c)

A point estimate is a numerical value that is the best guesstimate of the parameter. It is computed using the sample values.

For example, sample mean is the point estimate of population mean.

The point estimate of the population proportion of US adults who say cover a $400 unexpected expense without borrowing money or going into debt, is the sample proportion, [tex]\hat p[/tex].

[tex]\hat p=\frac{322}{765}=0.421[/tex]

(d)

The uncertainty of the point estimate can be measured by the standard error.

The standard error tells us how closer the sample statistic is to the parameter value.

[tex]SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

(e)

The standard error is:

[tex]SE_{\hat p}=\sqrt{\frac{0.421(1-0.421)}{765}} =0.018[/tex]

(f)

The sample proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt, is approximately 42.1%.

As the sample size is quite large this value can be used to estimate the population proportion.

If the proportion is believed to be 50% then she will be surprised because the estimated percentage is quite less than 50%.

(g)

Compute the standard error using p = 0.40 as follows:

[tex]SE=\sqrt{\frac{ p(1- p)}{n}}=\sqrt{\frac{0.40(1-0.40)}{765}}=0.0177\approx0.018[/tex]

The standard error does not changes much.

The data set consists of information on the ability of adults in the United States to cover a $400 unexpected expense. The parameter being estimated is the proportion of adults who cannot cover the expense without borrowing money or going into debt. The point estimate for this parameter is found by dividing the number of adults in the sample who cannot cover the expense by the total sample size.

(a) The population under consideration in the data set is all adults in the United States.

(b) The parameter being estimated is the proportion of adults in the United States who could not cover a $400 unexpected expense without borrowing money or going into debt.

(c) The point estimate for the parameter is the proportion of adults in the sample who said they could not cover a $400 unexpected expense without borrowing money or going into debt, which is 322/765.

(d) The name of the statistic that measures the uncertainty of the point estimate is the standard error.

(e) To compute the value of the standard error, you need the formula which depends on the sample proportion and sample size. Since the necessary values are not provided in the question, it is not possible to compute the value in this context.

(f) The cable news pundit should not be surprised by the data since the sample proportion is not too far off from the value she thinks is true.

(g) If the true population value is 40%, the resulting value in part (e) will change because the sample proportion is different from the true proportion.

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Answer:

a) [tex]\mu[/tex]

b) [tex]\bar{x}[/tex]

c) (0.152, 0.214)

Step-by-step explanation:

We are given the following in the question:

Sample mean = 0.183 ppm

Standard error = 0.016

a) quantity being estimated

We have to estimate the population mean.

Notation for population mean:

[tex]\mu[/tex]

b) Best estimate for population mean is the sample mean

Notation for sample mean:

[tex]\bar{x}[/tex]

The point estimate for population mean is

[tex]\mu = \bar{x} = 0.183[/tex]

c) 95% confidence interval

[tex]\bar{x}\pm z_{critical}(\text{Standard Error})[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

Putting values, we get,

[tex]0.183 \pm 1.96(0.016)\\=0.183\pm 0.03136\\=(0.15164, 0.21436)\\\approx (0.152, 0.214)[/tex]

Thus, the 95% confidence interval is:

(0.152, 0.214)

Interpretation:

After treatment with moose drool, we are 95% certain or 95% confident that the interval (0.152, 0.214) contains the true mean of the population that is the mean level of the toxin ergovaline on the grass.

The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The best estimate for this quantity is 0.183 ppm. The 95% confidence interval for this estimate is (0.15164, 0.21436) ppm.

a. The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The parameter used is the standard error, which measures the variability of the estimate.

b. The notation for the quantity that gives the best estimate is the mean level of the toxin ergovaline after treatment with moose drool, denoted as μ.

c. To calculate a 95% confidence interval, we need to determine the margin of error. Since the standard error is given as 0.016, the margin of error is 1.96 times the standard error, which is approximately 0.03136. The 95% confidence interval is then calculated by subtracting and adding the margin of error from the mean level of the toxin ergovaline after treatment with moose drool, resulting in an interval of (0.15164, 0.21436). This means that we are 95% confident that the true mean level of the toxin ergovaline after treatment with moose drool is between 0.15164 and 0.21436 ppm.

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Answer:

27720

Step-by-step explanation:

Given that there are 12 students in a graduate class. The students are to be divided into three groups of 3, 4, and 5 members for a class project.

From 12 students 3 students for group I can be selected in 12C3 ways.

Now from remaining 9, 4 students can be selected for II group in 9C4 ways

The remaining 5 have to be placed in III group.

Hence possible divisions for grouping the 12 students in the class into three groups

= 12C3 *9C4

= [tex]=220*126=27720[/tex]

Final answer:

There are 27,720 possible divisions of the 12 students into three groups of 3, 4, and 5 members.

Explanation:

To determine the number of possible divisions, we need to find the number of ways to choose 3 students from 12 for the first group, then 4 students from the remaining 9 for the second group, and finally 5 students from the remaining 5 for the third group.

Using the combination formula, the number of ways can be calculated as:

Number of ways = C(12, 3) * C(9, 4) * C(5, 5)

C(12, 3) = 12! / (3! * (12-3)!) = 220

C(9, 4) = 9! / (4! * (9-4)!) = 126

C(5, 5) = 5! / (5! * (5-5)!) = 1

Therefore, the number of possible divisions is 220 * 126 * 1 = 27,720.

Answer:

a. 0.0312 or 3.12%

b. 0.024 or 2.40%

Step-by-step explanation:

a. The probability that randomly selected child has at least one food allergy and a history of severe reaction is determined by the probability of having a food allergy (8%) multiplied by the probability of having a history of severe reaction (39%):

[tex]P = 0.08*0.39\\P=0.0312 = 3.12\%[/tex]

b. The probability that randomly selected child  is allergic to multiple foods is determined by the probability of having a food allergy (8%) multiplied by the probability of being allergic to multiple foods (30%):

[tex]P = 0.08*0.30\\P=0.024 = 2.40\%[/tex]

Answer:

Part a: The probability is 0.3436

Part b: The probability is 0.5381

Part c: The probability is 0.7600

Step-by-step explanation:

As per the given data

[tex]p_{T | SP} = e^{0.1}\\p_{T | Good} = e^{0.3}\\[/tex]

(a)

Probability that train is delayed by more than 1 minute = P(T > 1) = P(SP) * P(T > 1 | SP) + P(Good) * P(T > 1 | Good)

[tex]= 0.3 e^{-.1 \times 1 }+ 0.7 * e^{-.3 \times 1}\\ = 0.3 e^{-.1} + 0.7 e^{-.3}[/tex]

Probability that after 1 minute of waiting, probability that train has signal problems = P(SP | T > 1)

= P(T > 1 | SP) * P(SP) / P(T > 1) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-.1 }}{ 0.3 e^{-.1 }+ 0.7 e^{-.3 }}\\\\= \dfrac{0.2714512}{0.790024}\\\\= 0.3435987[/tex]

(b)

Probability that train is delayed by more than 5 minutes = P(T > 5) = P(SP) * P(T > 5 | SP) + P(Good) * P(T > 5 | Good)

[tex]= 0.3 e^{-.1 \times 5 }+ 0.7 * e^{-.3 \times 5}\\ = 0.3 e^{-.5} + 0.7 e^{-1.5}[/tex]

Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 5)

= P(T > 5 | SP) * P(SP) / P(T > 5) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-.5 }}{ 0.3 e^{-.5 }+ 0.7 e^{-1.5 }}\\\\= \dfrac{0.1819592}{0.3381503}\\\\= 0.5381015[/tex]

(c)

Probability that train is delayed by more than 10 minutes = P(T > 10) = P(SP) * P(T > 10 | SP) + P(Good) * P(T > 10 | Good)

[tex]= 0.3 e^{-.1 \times 10 }+ 0.7 * e^{-.3 \times 10}\\ = 0.3 e^{-1.0} + 0.7 e^{-3.0}[/tex]

Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 10)

= P(T > 10 | SP) * P(SP) / P(T > 10) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-1.0 }}{ 0.3 e^{-1.0 }+ 0.7 e^{-3.0 }}\\\\= \dfrac{0.1103638}{0.1452148}\\\\= 0.7600038[/tex]

(a) The probabilities recalculated based on the waiting time are: 21.4% after 1 minute,

(b) 54.0% after 5 minutes, and

(c) 75.9% after 10 minutes.

These calculations use Bayes' theorem and the exponential distribution functions provided.

The question pertains to conditional probability and involves exponential distributions.

Let’s solve this step-by-step:

Part (a)

Given:
pT|SP(t) = [tex]0.1e^(-0.1t)[/tex]
pT|Good(t) = [tex]0.3e^(-0.3t)[/tex]
P(Signal Problems) = [tex]0.30[/tex]
P(Good Service) = [tex]0.70[/tex]

We need to find P(Signal Problems | T ≥ 1). Using Bayes’ theorem:

Using the law of total probability:

P(T ≥ 1) = P(T ≥ 1 | Signal Problems)P(Signal Problems) + P(T ≥ 1 | Good Service)P(Good Service)
= [tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70[/tex]

Now applying Bayes’ theorem:
P(Signal Problems | T ≥ 1) = [tex][[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70][/tex]
[tex]\approx [0.9048 * 0.30] / [0.9048 * 0.30 + 0.7408 * 0.70][/tex]
[tex]\approx 0.214[/tex]

Part (b)

Similarly for T ≥ 5, the probabilities become:

Using the law of total probability:
P(T ≥ 5) = [tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex][tex]* 0.70[/tex]

Applying Bayes’ theorem:
P(Signal Problems | T ≥ 5) = [tex][[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex] [tex]* 0.70][/tex]
≈ [tex][0.6065 * 0.30] / [0.6065 * 0.30 + 0.2231 * 0.70][/tex]
≈ [tex]0.540[/tex]

Part (c)

For T ≥ 10:

Using the law of total probability:
P(T ≥ 10) = [tex]e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70[/tex]

Applying Bayes’ theorem:
P(Signal Problems | T ≥ 10) = [tex]\frac{e^{-1} \cdot 0.30}{e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70}[/tex]
≈ [tex][0.3679 * 0.30] / [0.3679 * 0.30 + 0.0498 * 0.70][/tex]
≈ [tex]0.759[/tex]

Answer:

Step-by-step explanation:

Mathematically, if two triangles are similar, then all the three of their angles are congruent to each other and their corresponding sides are in the same proportion. This means that the ratio of their corresponding sides are equal to each other.

Answer: A trigonometric table will provide all the angles in respect to the ratios of the sides.

Step-by-step explanation:

A trigonometric table would provide all of the ratios of the sides in respect to the angles. Sin, cos, and tan are parts of trigonometric table.

Trigonometry table, tabulated values for some or all of the six trigonometric functions for various angular values.

Since the ratios for sides of triangles are the same, when the angles are the same, trigonometric tables have been developed which show these ratios for every angle.