In a standard Normal​ distribution, if the area to the left of a​ z-score is about 0.3500​, what is the approximate​ z-score? Draw a sketch of the Normal​ curve, showing the area and​ z-score.

Answer:

The cost function is [tex]C(x)=100000+x\cdot 28[/tex]

The revenue function is [tex]R(x)=x\cdot 74[/tex]

The profit function is [tex]P(x)=46x-100000[/tex]

Step-by-step explanation:

We have the following definitions:

The cost function consists of variable costs and fixed costs and is given by

[tex]C(x)=fixed\:costs+x\cdot variable\:costs[/tex]

The revenue function is given by

[tex]R(x)=x\cdot p(x)[/tex]

where x are the units sold and p(x) is the price per unit.

The profit function is given by

[tex]P(x)=R(x)-C(x)[/tex]

Given:

Fixed costs = $100,000

Variable costs = $28 per unit

Price per unit = $74 per unit

Applying the above definitions and the information given, we get that:

The cost function is [tex]C(x)=100000+x\cdot 28[/tex]

The revenue function is [tex]R(x)=x\cdot 74[/tex]

The profit function is [tex]P(x)=74x-(28x+100000)=46x-100000[/tex]

Final answer:

The total cost for producing 1,000 units of output, given average fixed costs of $100 and average variable costs of $50, is calculated to be $150,000.

Explanation:

The question asks us to calculate total cost of producing 1,000 units of output given the average fixed costs and average variable costs. To find the total cost, we need to add together the total fixed costs (average fixed cost × quantity) and the total variable costs (average variable cost × quantity).

The total fixed cost is $100,000 (since $100 × 1,000 units) and the total variable cost is $50,000 (since $50 × 1,000 units). Therefore, the total cost of producing 1,000 units of output is $150,000.

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.

Final answer:

The appropriate conclusion from the survey is that the true percentage of all people who regularly use sunscreen in the region is estimated to be between 18% and 20%, considering the margin of error of 1 percentage point.

Explanation:

Based on the estimate and the margin of error, an appropriate conclusion is that it can be said with a certain level of confidence, usually 95%, that the true percentage of all people living in that region who regularly use sunscreen when going outdoors is between 18% and 20%. The margin of error gives a range around the estimate to account for the sampling error, and it implies that the actual value in the population should fall within this range. This helps to understand the precision of the survey result and allows for the natural variability that comes from the fact that not every individual in the region was surveyed.

The variability also depends on the size of the sample, which relates to the confidence interval. For example, smaller samples tend to have larger margins of error, while larger samples tend to lead to smaller margins of error and thus, more precise estimates of the population parameter.

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:

The t-test statistic is -3.33.and the p-value is two times P(t_51 < -3.33).

Step-by-step explanation:

Data given and notation  

Assuming that the real sample mean is: [tex]\bar X=98.285[/tex] represent the sample mean

[tex]s=0.682[/tex] represent the sample standard deviation  

[tex]n=52[/tex] sample size  

[tex]\mu_o =98.6[/tex] represent the value that we want to test  

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to apply a two tailed test.  

What are H0 and Ha for this study?  

Null hypothesis: [tex]\mu = 98.6[/tex]  

Alternative hypothesis :[tex]\mu \neq 98.6[/tex]  

Compute the test statistic  

The statistic for this case is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]t=\frac{98.2-98.6}{\frac{0.682}{\sqrt{52}}}=-3.298[/tex]  

Now we can calculate the degrees of freedom and we got:

[tex] df = n-1 = 52-1 = 51[/tex]

P value

Since is a two tailed test the p value would be:  

[tex]p_v =2*P(t_{51}<-3.3)=0.0018[/tex]  

So the most appropiate conclusion for this case would be:

The t-test statistic is -3.33.and the p-value is two times P(t_51 < -3.33).

Answer:1.04 ml/g.

Step-by-step explanation:

The formula we use to find the density :

[tex]\text{Density}=\dfrac{\text{Volume}}{\text{Mass}}[/tex]

We are given that ,

Volume of a sample of water = 11.4 mL

Mass of sample of water = 10.98 g

Then , the experimental density of the water sample will be :-

[tex]\text{Density}=\dfrac{11.4}{10.98}=1.03825136\approx1.04[/tex]  

Hence, the experimental density of the water sample is 1.04 ml/g.

Answer:

1. Let amount deposited now be P.

So P*(1.1106^44) = 1,000,000, which means P=9896.06

2. Let amount deposited now be P.

So P*(1.0553^44) = 1,000,000, which means P=93639.42

Step-by-step explanation:

Answer:

Step-by-step explanation:

Let P represent the amount that you need to invest today. Thus, principal = $P

The return would be compounded annually. So

n = 1

The rate at which the principal would be compounded is 11.06%. So

r = 11.06/100 = 0.1106

The investment would be for 44 years. So

t = 44

The formula for compound interest is

A = P(1+r/n)^nt

A = total amount in the account at the end of t years.

A = $1000000

Therefore

1000000 = P(1+0.1106/1)^1×44

1000000 = P(1.1106)^44

P = 1000000/101.05

P = $9896.1

2) if you can earn an annual return of 5.53 percent, then

r = 5.53/100 = 0.053

1000000 = P(1+0.53/1)^1×44

1000000 = P(1.053)^44

P = 1000000/9.7

P = $103093

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:

There are 320 milliliters of orange juice in the glass.

Step-by-step explanation:

Unit conversion problems can be solved by rules of three.

We have that each dL is worth 100 mL(milliliters).

How many milliliters of orange juice are in the glass?

How many ml are 3.2dL of juice?

1 dL - 100 mL

3.2 dL - x mL

[tex]x = 100*3.2[/tex]

[tex]x = 320[/tex]

There are 320 milliliters of orange juice in the glass.

The measure of juice is 320 milliliters.

Given that

A glass of orange juice contains 3.2 dL of juice.

We have to determine

How many milliliters of orange juice is in the glass?

According to the question

To determine the measure of juice in milliliters following all the steps given below.

A glass of orange juice contains 3.2 dL of juice.

Here 1 dL = 100 milliliters

Then,

The measure of juice in milliliters is,

[tex]\rm = 3.2 \times 100\\\\= 320 \ milliliters[/tex]

Hence, The measure of juice is 320 milliliters.

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The expression 15t – 2t not equivalent to 2t – 15t because the negative sign in 15t – 2t belongs to the term 2t.

Solution:

Given expressions are 15t – 2t and 2t – 15t.

To determine 15t – 2t is equivalent to 2t – 15t or not.

Substitute t = 2 in above two expressions.

15t – 2t = 15(2) – 2(2)

            = 30 – 4

            = 26

2t – 15t = 2(2) – 15(2)

            = 4 – 30

            = –26

The values of the expressions are different when t = 2.

So, 15t – 2t is not equivalent to 2t – 15t.

Hence the expression 15t – 2t not equivalent to 2t – 15t because the negative sign in 15t – 2t belongs to the term 2t.

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:

a) There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b) 0.0064% probability that all of the selected adults believe in reincarnation.

c) There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d) Since [tex]P(X \geq 5) < 0.05[/tex], 5 is a significantly high number of adults who believe in reincarnation in this sample.

Step-by-step explanation:

For each of the adults selected, there are only two possible outcomes. Either they believe in reincarnation, or they do not. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

In this problem we have that:

[tex]n = 6, p = 0.2[/tex]

a. What is the probability that exactly five of the selected adults believe in reincarnation?

This is P(X = 5).

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

[tex]P(X = 5) = C_{6,5}.(0.2)^{5}.(0.8)^{1} = 0.0015[/tex]

There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b. What is the probability that all of the selected adults believe in reincarnation?

This is P(X = 6).

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

[tex]P(X = 6) = C_{6,6}.(0.2)^{6}.(0.8)^{0} = 0.000064[/tex]

There is a 0.0064% probability that all of the selected adults believe in reincarnation.

c. What is the probability that at least five of the selected adults believe in reincarnation?

This is

[tex]P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564[/tex]

There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation?

5 is significantly high if [tex]P(X \geq 5) < 0.05[/tex]

We have that

[tex]P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564 < 0.05[/tex]

Since [tex]P(X \geq 5) < 0.05[/tex], 5 is a significantly high number of adults who believe in reincarnation in this sample.

a. The probability that exactly five of the selected adults believe in reincarnation is approximately 0.00256.

b. The probability that all of the selected adults believe in reincarnation is approximately 0.000064.

c. The probability that at least five of the selected adults believe in reincarnation is approximately 0.002624.

d. To determine if five is significantly high, we need a significance level for comparison, which isn't provided in the question.

To solve this problem, we can use the binomial probability formula, where "n" is the number of trials, "p" is the probability of success (believing in reincarnation in this case), and "x" is the number of successes.

a. The probability that exactly five of the selected adults believe in reincarnation is calculated as follows:

P(X = 5) = C(6, 5) * (0.20)^5 * (0.80)^(6-5),

where C(6, 5) is the number of ways to choose 5 out of 6 adults, which equals 6.

P(X = 5) = 6 * (0.20)^5 * (0.80)^1 ≈ 0.00256

b. The probability that all of the selected adults believe in reincarnation is:

P(X = 6) = (0.20)^6 ≈ 0.000064

c. The probability that at least five of the selected adults believe in reincarnation is the sum of the probabilities from parts (a) and (b):

P(X ≥ 5) = P(X = 5) + P(X = 6) ≈ 0.00256 + 0.000064 ≈ 0.002624

d. To determine if five is a significantly high number who believe in reincarnation, we can compare the probability of getting at least five believers (from part c) to a significance level. If this probability is less than the significance level, it would be considered significant. The significance level would depend on the context and what is considered "significant" in the specific analysis.

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complete question should be :

See You Later Based on a Harris Interactive poll, 20% of adults believe in reincarnation. Assume that six adults are randomly selected, and find the indicated probability. a. What is the probability that exactly five of the selected adults believe in reincarnation? b. What is the probability that all of the selected adults believe in reincarnation? c. What is the probability that at least five of the selected adults believe in reincarnation? d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation .

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:

a) 0.9,b) 1.5,c) 0.3hrs, d) 0.5hrs,e)  60%

Step-by-step explanation:

Given Data:

rate of arrival   = 3customers/hr ;

rate of service = 5 haircuts/hr    ;

a)

Average number of customers = La = λ²/[μ(μ-λ)]

                                                           = 3²/[(5(5-3)]

Average number of customers = La = 0.9

b)

Number of customers in system = Ls = λ/(μ-λ)

                                                             = 3/(5-3)

Number of customers in system = Ls = 1.5

c)

Average waiting time = Ta = λ/[μ(μ-λ)]

                                             = 3/[(5(5-3)]

Average waiting time = Ta =0.3hrs or 18mins

d)

Average time spent by customer = Ts = 1/(μ-λ)

                                                               = 1/(5-3)

Average time spent by customer = Ts = 0.5hrs or 30mins

e)

% of time  = Tr = λ/μ

                        = 3/5

% of time  = Tr = 0.6 or 60%

The arrival of customers follows a Poisson distribution

The given parameters are:

[tex]\mathbf{\lambda = 3}[/tex] --- rate of arrival

[tex]\mathbf{\mu= 5}[/tex] ---- rate of service

(a) Average number of customers waiting

This is calculated using:

[tex]\mathbf{L_a = \frac{\lambda^2}{\mu(\mu - \lambda)}}[/tex]

So, we have:

[tex]\mathbf{L_a = \frac{3^2}{5(5 - 3)}}[/tex]

[tex]\mathbf{L_a = \frac{9}{5 \times 2}}[/tex]

[tex]\mathbf{L_a = \frac{9}{10}}[/tex]

[tex]\mathbf{L_a = 0.9}[/tex]

Hence, the average number of customers waiting for haircut is 0.9

(b) Average number of customers in the shop

This is calculated using:

[tex]\mathbf{L_s = \frac{\lambda}{\mu - \lambda}}[/tex]

So, we have:

[tex]\mathbf{L_s = \frac{3}{5 - 3}}[/tex]

[tex]\mathbf{L_s = \frac{3}{2}}[/tex]

[tex]\mathbf{L_s = 1.5}[/tex]

Hence, the average number of customers in the shop is 1.5

(c) Average time of waiting

This is calculated using:

[tex]\mathbf{T_a = \frac{\lambda}{\mu(\mu - \lambda)}}[/tex]

So, we have:

[tex]\mathbf{T_a = \frac{3}{5(5 - 3)}}[/tex]

[tex]\mathbf{T_a = \frac{3}{5 \times 2}}[/tex]

[tex]\mathbf{T_a = \frac{3}{10}}[/tex]

[tex]\mathbf{T_a = 0.3}[/tex]

Hence, the average time of waiting for haircut is 0.3 hour

(d) Average time spent in the shop

This is calculated using:

[tex]\mathbf{T_s = \frac{1}{\mu - \lambda}}[/tex]

So, we have:

[tex]\mathbf{T_s = \frac{1}{5 - 3}}[/tex]

[tex]\mathbf{T_s = \frac{1}{2}}[/tex]

[tex]\mathbf{T_s = 0.5}[/tex]

Hence, the average time spent in the shop is 0.5 hour

(e) Percentage of time Rich is busy

This is calculated as:

[tex]\mathbf{T = \frac{\lambda}{\mu}}[/tex]

So, we have:

[tex]\mathbf{T = \frac{3}{5}}[/tex]

Divide

[tex]\mathbf{T = 0.6}[/tex]

Express as percentage

[tex]\mathbf{T = 60\%}[/tex]

Hence, Rich is busy 60% of the time

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Answer: the perimeter of the shape is 19.1

Step-by-step explanation:

To determine the length of each side of the quadrilateral, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side

For line AD,

AD² = 2² + 4² = 4 + 16 = 20

AD = √20 = 4.47

For line AB,

AB² = 1² + 5² = 1 + 25 = 26

AB = √26 = 5.1

For line BC,

BC² = 2² + 4² = 4 + 16 = 20

BC = √20 = 4.47

For line CD,

CD² = 1² + 5² = 1 + 25 = 26

CD = √26 = 5.1

The perimeter of a plane figure is the distance around the figure. Therefore

Perimeter = AB + AB + BC + CD

Perimeter = 4.47 + 5.1 + 4.47 + 5.1 =

19.1

Answer:

19.1

Step-by-step explanation:

Got it right on the test! <3

Answer:

Step-by-step explanation:

we know

[tex]\vec{i}\times \vec{j}=\vec{k}[/tex]

[tex]\vec{j}\times \vec{k}=\vec{i}[/tex]

[tex]\vec{k}\times \vec{i}=\vec{j}[/tex]

(a)[tex]\left [ \left ( \hat{i}+\hat{j}\right )\times \hat{i}\right ]\times \hat{j}[/tex]

[tex]=\left [ \hat{i}\times \hat{i}+\hat{j}\times \hat{i}\right ]\times \hat{j}[/tex]

[tex]=\left [ 0-\hat{k}\right ]\times \hat{j}[/tex]

[tex]=\hat{i}[/tex]

(b)[tex]\left ( \hat{j}+\hat{k}\right )\times \left ( \hat{j}\times \hat{k}\right )[/tex]

[tex]=\left ( \hat{j}+\hat{k}\right )\times \left ( \hat{i}\right )[/tex]

[tex]=\hat{k}+\hat{j}[/tex]

(c)[tex]4\hat{i}\times \left ( \hat{i}+\hat{j}\right )[/tex]

[tex]=4\hat{i}\times \hat{i}+4\hat{i}\times \hat{j}[/tex]

[tex]=0+4\hat{k}[/tex]

(d)[tex]\left ( \hat{k}+\hat{j}\right )\times \left ( \hat{k}-\hat{j}\right )[/tex]

[tex]=\hat{k}\times \hat{k}-\hat{k}\times \hat{j}+\hat{j}\times \hat{k}-\hat{j}\times \hat{j}[/tex]

[tex]=0+\hat{i}+\hat{i}-0[/tex]

[tex]=2\hat{i}[/tex]

The direction of the cross product depends on the right-hand rule and the resulting cross-product is located on the plane that is perpendicular to the vectors undergoing the cross product.

Taking i, j, k as the unit vector along x, y, z-direction.

Since [tex]i \times i = 0[/tex] it implies the angle∠ between them is 0;

Then: sin(0) = 0

Also;

[tex]j \times j = 0 \\ \\ k\times k = 0[/tex]

Similarly, [tex]i \times j = k[/tex] which implies that the angle ∠ between them = 90°;

Then: sin (90°) = 1

Also;

[tex]j \times k= i \\ \\ k \times i = j[/tex]

[tex]j \times i = - k[/tex] which implies that the angle ∠ between them = -90° or 270°

Then; sin (-90) or Sin ( 270) = -1

Also;

[tex]i \times k = -j \\ \\ j \times i = -k[/tex]

As such i,  j, k are unit vectors for x, y, and z-axis.

To determine the following calculations, we have;

(a)

[tex]\Big ( ( i^{\to }+ j^{\to })\times i^{\to }\Big)\times j^{\to }[/tex]

[tex]= ( i^{\to } \times i^{\to } + j^{\to }\times i^{\to }) \times j^{\to }[/tex]

[tex]=(0 - k^{\to })\times j^{\to }[/tex]

[tex]= - k^{\to } \times j^{\to }\\\\= -(-i^{\to }) \\ \\ \mathbf{= i^{\to }}[/tex]

(b)

[tex](j^{\to }+ k^{\to }) \times (j^{\to } + k^{\to })[/tex]

[tex]= j^{\to } \times (j^{\to } \times k ) + k^{\to } \times (j \times k)\\ \\ = j^{\to } \times i^{\to } + k^{\to } \times i^{\to } \\ \\ \mathbf{ = -k^{\to } + j^{\to }}[/tex]

(c)

[tex]4i^{\to } \times (j \times k^{\to }) \\ \\ = 4 (i^{\to } \times i^{\to }) \\ \\ \mathbf{= 0}[/tex]

(d)

[tex](k^{\to} + j^{\to}) \times (k^{\to} - j^{\to})) \\ \\ =(k^{\to} \times k^{\to}) - (k^{\to} \times j^{\to})+(j^{\to} + k^{\to}) - (j^{\to} \times j^{\to}) \\ \\ = 0 + i^{\to} + i^{\to} -0 \\ \\ \mathbf{ = 2 i^{\to}}[/tex]

Therefore, we can conclude that the calculations of the cross-product are well defined from the above explanations.

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Answer:

240,000 AUD to USD = 161,924.79 US Dollars or 150,000 USD to AUD = 222,316.63 Australian Dollars

Step-by-step explanation:

Answer:

There were 28.75 large boxes and 96.25 small boxes.

Step-by-step explanation:

Create a system of equations to solve.

State your variables

let x be the number of small boxes

let y be the number of large boxes

35x + 55y = 4950               Equation for weight

x + y = 125                          Equation for number of boxes

Rearrange equation for number of boxes to isolate "x".

x = 125 - y

Substitute the new equation into the equation for weight

35x + 55y = 4950

35(125 - y) + 55y = 4950

Expand the brackets

4375 - 35y + 55y = 4950       Combine like terms

4375 + 20y = 4950

Start isolating "y"

4375 + 20y = 4950

4375 - 4375 + 20y = 4950 - 4375         subtract 4375 from both sides

20y = 575

20y/20 = 575/20                         divide both sides by 20

y = 28.75                   number of large boxes

Calculate "x". Rearrange the equation for number of boxes to isolate "y".

y = 125 - x                    Substitute this expression

35x + 55y = 4950                          equation for weight

35x + 55(125 - x) = 4950                 expand the brackets

35x + 6875 - 55x = 4950                     combine like terms

6875 - 20x = 4950                              start isolating "x"

6875 - 6875 - 20x = 4950 - 6875            subtract 6875 on both sides

-20x = -1925

-20x/-20 = -1925/-20                     divide both sides by -20

x = 96.25                    number of small boxes

There is a mistake with this question because you should not be able to have partial boxes. However, the answer is:

There were 28.75 large boxes and 96.25 small boxes.

Answer:

P(20≤x≤35) = 0.75 .

Step-by-step explanation:

We know that the probability distribution function of Uniform Distribution is represented as :

     If x follows Uniform(c,d) then,

          f(x) = [tex]\frac{1}{d-c}[/tex] where c < x < d

To fond the given probability it is better to first calculate the Cumulative Distribution Function(CDF) of Uniform Distribution.

The CDF of Uniform Distribution is P(X<=x) = [tex]\frac{x-c}{d-c}[/tex] where d > c .

Therefore, P(20<=x<=35) = P(x<=35) - P(x<20)

  P(x<=35) =  [tex]\frac{x-20}{40-20}[/tex]  because we are given c = 20 and d = 40.

                 = [tex]\frac{35-20}{40-20}[/tex] = 0.75

 P(x<20) =  [tex]\frac{x-20}{40-20}[/tex]  = [tex]\frac{20-20}{40-20}[/tex] = 0

Hence, P(20<=x<=35) = 0.75 - 0 = 0.75.

The probability of x being between 20 and 35 is 0.75 or 75%.

The uniform probability distribution is defined as:

P(x) = 1/(b-a) for a ≤ x ≤ b and 0 otherwise.

In this case, a=20 and b=40, so the probability P(20≤x≤35) can be calculated as:

P(20≤x≤35) = (35-20)/(40-20) = 15/20 = 0.75.

Therefore, the probability of x being between 20 and 35 is 0.75 or 75%.

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Part(a):[tex]f(r)=f(4)[/tex]

Part(b):[tex]f(10.91)-f(10.9)[/tex]

Part(c):[tex]5 f(r)=5 f(12.7)[/tex]

Part(d):[tex]28+59 =f(28)+59[/tex]

Area of the circle:

The area of a circle is the region occupied by the circle in a two-dimensional plane. It can be determined easily using a formula,

[tex]A= \pi r^2[/tex]

where [tex]r[/tex] is the radius of the circle

The formula for the area of the circle is,

[tex]A=\pi r^2[/tex]

Part(a):

Given,

Radius([tex]r[/tex])=4 cm

So, the area is [tex]f(r)=f(4)[/tex]

Part(b):

Given,

[tex]r=10.91\\r=10.9[/tex]

The difference in area is,

[tex]f(10.91)-f(10.9)[/tex]

Part(c):

Area of 5 circles are,

[tex]5 f(r)=5 f(12.7)[/tex]

Part(d):

The area of the larger circle is,

Area of the circle of radius [tex]28+59 =f(28)+59[/tex]

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Answer:

The answer is shown in the attachment

Step-by-step explanation:

The detailed step by step and appropriate mathematical manipulation for the expression for the steady state capital stock per worker is as shown in the attachment.

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:

Height of Lhotse mountain is greater than Ngadi Chuli mountain.

Height of Lhotse mountain is 632.0432 meter more than height of Ngadi Chuli mountain.

Step-by-step explanation:

Height of Lhotse mountain = 8516 meters

Height of Ngadi Chuli mountain = 25,866 feet

1 ft = 30.48 cm

[tex]25,866 feet =25,866\times 30.48 m=788,395.68 cm[/tex]

1 cm = 0.01 m

[tex]788,395.68 cm=788,395.68\times 0.01 m=7883.9568 m[/tex]

8516 meters > 7883.9568 m

Height of Lhotse mountain is greater than Ngadi Chuli mountain.

Difference in their height :

8516 meters -  7883.9568 m = 632.0432 m

Height of Lhotse mountain is 632.0432 meter more than height of Ngadi Chuli mountain.

The height of Ngadi Chuli is approximately 7871 meters, while the height of Lhotse is 8516 meters. Therefore, Lhotse is the taller mountain and is about 645 meters taller than Ngadi Chuli.

The height of Ngadi Chuli is given in feet, so we must first convert that to meters using the conversion factor 1 ft = 30.48 cm = 0.3048 m. Multiplying 25,866 feet by 0.3048 m/ft gives us an approximate height of 7871 meters for Ngadi Chuli.

Lhotse is given to be 8516 meters tall. Therefore, by comparing these two heights, we can determine that Lhotse is the taller mountain.

The difference in height is found by subtracting the height of Ngadi Chuli from that of Lhotse: 8516 m - 7871 m = 645 m. So, Lhotse is approximately 645 meters taller than Ngadi Chuli.

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Answer:

a) 91% of the scores were at or below her score.

b) 9% of the scores were above her score.

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:

Angela scored on the 91st percentile.

(a) What percentage of the scores were at or below her score?

Since Angela scored on the 91st percentile, 91% of the scores were at or below her score.

(b) What percentage were above?

Since Angela scored on the 91st percentile, 9% of the scores were above her score.

Angela scored higher than or equal to 91% of her peers, meaning 9% scored above her.

If Angela scored in the 91st percentile on her general aptitude test, that means she scored higher than or equal to 91% of the other scores. Therefore, (a) 91% of scores were at or below Angela's score. As for (b), if you subtract her percentile from 100%, you can determine that 9% of scores were above hers.

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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:

Height(h) = 24in

Area = 204in²

Step-by-step explanation:

we would use the heron formula because we have all the sides to calulate the area

S  = (A+B+C)/2 = (25 + 17 + 26)/2 = 68/2 = 34

area = √s(s-a)(s-b)(s-c)  = √34(34-25)(34-17)(34-26) = √34(9)(17)(8) = √34x9x17x8 = √41616 = 204

area= (1/2)bh

    204 = (1/2)bh b= 17, h =?

204 = (1/2)17h

204 x 2 = 17h

408 = 17h

h = 408/17 = 24in

Height(h) = 24in

Area = 204in²

Answer:

0.42

Step-by-step explanation:

Let the events be given as:

For twp mutually exclusive events, the probability of A1 is given as follows:

P (B1A1) = [tex]\frac{ P(B1) x P (BA1)}{P(A1)}[/tex]

             = 0.6

Solving the equation above  to get B1:

P (B1) = [tex]\frac{(0.8)x (0.6)}{(0.6)}[/tex]

         = 0.8

Therefore, computing P (A1B1) gives P(A1) × P (B1)

                                                          =  (0.8) × (0.6)

                                                          =  0.42 Ans

Step-by-step explanation:

[tex] \because \triangle \: DEF \sim \triangle NPQ \\ \\ \therefore \: \frac{EF}{PQ} = \frac{DF}{NQ}... (csst) \\ \\ \therefore \: \frac{d}{ \frac{11}{2} } = \frac{7}{9} \\ \\ \therefore \: \frac{2d}{ 11 } = \frac{7}{9} \\ \\ \therefore \: d = \frac{11 \times 7}{2 \times 9} \\ \\ \therefore \: d = \frac{77}{18} \\ [/tex]

Answer:

77/18

Step-by-step explanation:

Similar figures have sides in the same ratio.

Ratio can be obtained using the ratio of DF/NQ

DF/NQ = 7/9

EF/PQ = 7/9

d/(11/2) = 7/9

d = 7/9 × 11/2

d = 77/18

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:

The additional amount of area covered between 4 to 9 days is 23.71 cm2

Step-by-step explanation:

As the relation is given as a combination of two functions so integration by parts is carried out thus

[tex]\int\limits^9_4 {\sqrt{t}\, ln t} \, dt[/tex]

In order to solve this integral, integration by parts is to be carried out which is given as

[tex]\int u v dx=u \int v dx -\int u'(\int vdx) dx[/tex]

Where

u(x) is a function of x

v(x) is a function of x

u' is the derivative of u wrt to x

Also u and v are defined on using the following sequence  ILATE RULE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)

As here Logarithmic function is present which is taken as u and the algebraic function is taken as v so

[tex]u= ln t\\v=\sqrt{t}\\u'=\frac{1}{t}[/tex]

[tex]\int v dt =\int t^{1/2} dt =\frac{2}{3}t^{3/2}[/tex]

Substituting the values in equation gives

[tex]\int u v dt=u \int v dt -\int u'(\int vdt) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\int(\frac{1}{t} )(\frac{2}{3}t^{3/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\int(\frac{2}{3}t^{1/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{2}{3}\int(t^{1/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{2}{3}(\frac{2}{3}t^{3/2}) +C\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2}) +C[/tex]

Now solving the definite integral

[tex]\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2}) +C\\\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=[ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2})]_{9} -[ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2})]_{4}\\\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=18 ln (9)-\frac{16}{3} ln 4 -\frac{76}{9}\\A=\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=23.71 cm^2\\[/tex]

So the additional amount of area covered between 4 to 9 days is 23.71 cm2