The mean of a normally distributed group of weekly incomes of a large group of executives is $1,000 and the standard deviation is $100. What is the z-score (value of z) for an income of $1,100

Answer:

[tex]33.6\ cm^{2}[/tex]

Step-by-step explanation:

Given that

The Base of a rhombus = 8 centimetres

Height of a rhombus = 4.2 centimetres

Based on the above information, the area of a rhombus is

[tex]Area\ of\ rhombus = Base \times height[/tex]

[tex]= 8 \times 4.2[/tex]

[tex]= 33.6\ cm^{2}[/tex]

By multiplying the base of a rhombus with the height of a rhombus we can get the area of the rhombus and the same is applied & shown in the computation part i.e to be shown above

Answer:

Please see attachment

Question:

The options are;

(a) (9 - 2s1, 3 + 3s1, s1)

solution

not a solution

(b) (-4 - 5s1, s1,  -(3 + s1)/2)

solution

not a solution

(c) (11 + 10s1, -3 - 2s1, s1)

solution

not a solution

(d) ((6 - 4s1)/3, s1, -(7 - s1)/4)

solution

not a solution

Answer:

The options that form a solution of the given system are;

(b) and (c)

Step-by-step explanation:

Here we have

3·x₁ + 8·x₂ − 14·x₃ = 9

x₁ + 3·x₂ − 4·x₃ = 1

(a) (9 - 2·s₁, 3 + 3·s₁,s₁)

3·(9 - 2·s₁) + 8·(3 + 3·s₁) − 14·s₁ = 4s₁+51

Not a solution

(b) (-4 - 5s₁, s₁, -(3 + s1)/2)

3·(-4 - 5s₁) + 8·(s₁) − 14·-(3 + s1)/2 = 9

(-4 - 5s₁) + 3·(s₁) − 4·-(3 + s1)/2 = 2

Solution

(c) (11 + 10s₁, -3 - 2s₁, s₁ )

3·(11 + 10s₁) + 8·(-3 - 2s₁) − 14·s₁  = 9

(11 + 10s₁) + 3·(-3 - 2s₁) − 4·s₁ = 2

Solution

(d) ((6 - 4s1)/3, s1, -(7 - s1)/4)

3·(6 - 4s1)/3+ 8·s1− 14·-(7 - s1)/4 = 0.5s₁ +30.5  

Not a solution

Answer:

Being p1 the proportion for people of ages 36-50 and p2 the proportion for people of ages 21-35, the null and alternative hypothesis will be:

[tex]H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0[/tex]

Step-by-step explanation:

A hypothesis test on the difference of proportions needs to be performed for this case.

We have two sample proportions and we want to test if the true population proportions differ from each other, usign the information given by the sample statistics.

The claim is that the proportion of people of ages 36-50 who own homes is significantly greater than the proportin of people age 21-35 who own homes.

The term "higher" will define the alternative hypothesis, that is the hypothesis that represents what is claimed. The null hypothesis always include the equal sign, and will state that both proportions do not differ.

Being p1 the proportion for people of ages 36-50 and p2 the proportion for people of ages 21-35, the null and alternative hypothesis will be:

[tex]H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0[/tex]

The null hypothesis (H0) is that the proportion of homeowners ages 36-50 is equal to the proportion of homeowners ages 21-35 (H0: P1 = P2), and the alternative hypothesis (Ha) is that the proportion of homeowners ages 36-50 is greater than that of ages 21-35 (Ha: P1 > P2).

To answer the question, the null hypothesis (H0) and the alternative hypothesis (Ha) must be formulated based on the given data about homeownership across different age groups. In this research, the null hypothesis would state that the proportion of homeowners who are ages 36-50 is equal to the proportion of homeowners who are age 21-35. Mathematically, this can be represented as H0: P1 = P2.

The alternative hypothesis is what the researchers are trying to support, which is that the proportion of homeowners who are ages 36-50 is significantly greater than the proportion of homeowners who are age 21-35, represented as Ha: P1 > P2.

It is important to note that a hypothesis test will be used to determine if there is enough statistical evidence to reject the null hypothesis in favor of the alternative hypothesis.

Answer:

2/4

Step-by-step explanation:

2+2=4 so 2/4

50%

Answer:

50% chance

Step-by-step explanation:

since half of the balls are red and half are black there will be a 50%chance of picking a black one from the bag

Answer:

its answer C

Step-by-step explanation:

x² - 12x + 4x + 48

Answer:

A.

Its the simplkified version of the question and they both have the same out come which is 22.4

The equivalent expression for 10√5 is √500.

An equivalent expression is an expression that has the same value but does not look the same. When you simplify an expression, you're basically trying to write it in the simplest way possible.

For the given situation,

The expression is 10√5.

If a square number lies inside the square root, then we write that number outside the square root. So,

⇒ [tex]10\sqrt{5}[/tex]

⇒ [tex]\sqrt{10^{2}(5) }[/tex]

⇒ [tex]\sqrt{(100)(5)}[/tex]

⇒ [tex]\sqrt{500}[/tex]

Hence we can conclude that the equivalent expression for 10√5 is √500.

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Answer:

a) 10.34% probability​ that, in a​ year, there will be 4 hurricanes.

b) 3.62 years are expected to have 4 ​hurricanes

c) Either 3 or 4 hurricanes(discrete number) are close to the mean of 3.62, which means that the Poisson distribution works well in this case.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

6.7 per year.

This means that [tex]\mu = 6.7[/tex]

a. Find the probability​ that, in a​ year, there will be 4 hurricanes.

This is P(X = 4).

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 4) = \frac{e^{-6.7}*(6.7)^{4}}{(4)!} = 0.1034[/tex]

10.34% probability​ that, in a​ year, there will be 4 hurricanes.

b. In a 35​-year ​period, how many years are expected to have 4 ​hurricanes?

Each year, 0.1034 probability of 10 hurricanes.

In 35 years

35*0.1034 = 3.62

3.62 years are expected to have 4 ​hurricanes

c. How does the result from part​ (b) compare to a recent period of 35 years in which 3 years had 4 ​hurricanes? Does the Poisson distribution work well​ here?

Either 3 or 4 hurricanes(discrete number) are close to the mean of 3.62, which means that the Poisson distribution works well in this case.

Final answer:

Using the Poisson distribution with a mean of 6.7 hurricanes per year, the probability of exactly 4 hurricanes occurring in one year is calculated. Multiplying this probability by 35 provides the expected number of years with 4 hurricanes in a 35-year period, which is then compared to an actual historical period to evaluate the fit of the Poisson distribution to the data.

Explanation:

The student's question pertains to the application of the Poisson distribution to determine the probability of certain events. Given that the mean number of hurricanes in a certain area is 6.7 per year, we can use the Poisson formula to calculate the probabilities:

To find the probability that there will be 4 hurricanes in a year, we use the formula:

P(x; μ) = ( × e^-μ) / x!

To determine how many years are expected to have 4 hurricanes in a 35-year period, we multiply the probability found in part a by 35.

When comparing the expected number of years with 4 hurricanes to an actual 35-year period where 3 years had 4 hurricanes, it can be seen whether the Poisson distribution provides a good fit for the actual data.

Answer:

pls I can't see any diagram anywhere

Answer:

the term is 2/3(6)n-1

Step-by-step explanation:

Common ratio, r= 6

3rd term=24

Finding first term=

24=a 6^{2}  

24=36a

a=24/36

a=2/3

Answer:

F (x) = -3x+5x^2+8 has complex roots.

Step-by-step explanation:

number of roots?

ok.

f(x) =   5xx - 3x + 8  =   5xx + 5x  - 8x  + 8   ( I was guessing what numbers would sum to -3)

nope.

ok try discriminant:    (-3)^2 - 4*5 * 8 =   9 - 160  < 0

2 complex roots

The input units of the function f(t) are students multiplied by dollars per student.

The function[tex]\( f(t) = s(t) \cdot c(t) \)[/tex] represents the total cost to educate all the students in the elementary school t years after 2002. To determine the input units of this function, we need to analyze the units of its individual components.

s(t) is the number of students at time t. Its unit is "students."

c(t) is the cost to educate one student at time t. Its unit is "dollars per student."

When you multiply s(t)  by c(t), the units combine as follows:

[tex]\[ f(t) = s(t) \cdot c(t) = \text{"students"} \times \left(\frac{\text{"dollars"}}{\text{"student"}}\right) = \text{"students"} \cdot \text{"dollars per student"} \][/tex]

Therefore, the input units of the function f(t) are students multiplied by dollars per student.

Answer:

The upper limit for population proportion is 0.3666

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 219

Number of people who have ability to speak a language other than English, x = 63

[tex]\hat{p} = \dfrac{x}{n} = \dfrac{63}{219} = 0.2877[/tex]

99% Confidence interval:

[tex]\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 2.58[/tex]

Putting the values, we get:

[tex] 0.2877\pm 2.58(\sqrt{\dfrac{ 0.2877(1- 0.2877)}{219}})\\\\ = 0.2877\pm 0.0789\\\\=(0.2088,0.3666)[/tex]

is the required 99% confidence interval for population proportion.

Thus, the upper limit for population proportion is 0.3666

Answer:

The upper limit of the 99% confidence interval for the population proportion based on these statistics is 0.3665.

Step-by-step explanation:

We are given that of the 219 white GSS 2008 respondents in their 20's, 63 of them claim the ability to speak a language other than English.

So, the sample proportion is : [tex]\hat p[/tex]  = X/n = 63/219

Firstly, the pivotal quantity for 99% confidence interval for the population proportion  is given by;

     P.Q. =  [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion = [tex]\frac{63}{219}[/tex]

           n = sample of respondents = 219

           p = population proportion

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population​ proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at

                        0.5% level of significance are -2.5758 & 2.5758}

P(-2.5758 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 2.5758) = 0.99

P( [tex]-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99

P( [tex]\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99

99% confidence interval for p = [ [tex]\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]]

   = [ [tex]\frac{63}{219} -2.5758 \times {\sqrt{\frac{\frac{63}{219}(1-\frac{63}{219})}{219} } }[/tex] , [tex]\frac{63}{219} +2.5758 \times {\sqrt{\frac{\frac{63}{219}(1-\frac{63}{219})}{219} } }[/tex] ]

   = [0.2089 , 0.3665]

Therefore, 99% confidence interval for the population proportion based on these statistics is [0.2089 , 0.3665].

Hence, the upper limit of the population proportion based on these statistics is 0.3665.

Answer:

For a duration of 5 years, Monthly Payment =$600.42

For a duration of 6 years, Monthly Payment =$508.83

Step-by-step explanation:

[tex]P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\[/tex]

where a= Amount to Finance=$33,000

Annual interest rate = 3.5%=0.035

r=Monthly Interest Rate= 0.035 ÷ 12 =[tex]\frac{7}{2400}[/tex]

n=number of months to pay

For a duration of 5 years

n=5X12=60 months

[tex]P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\\\P=\dfrac{33000 X\frac{7}{2400} (1+\frac{7}{2400} )^{60}}{(1+\frac{7}{2400})^{60}-1} \\=\dfrac{96.25 (1.1909)}{1.1909-1}\\=\dfrac{96.25 (1.1909)}{0.1909}\\=\dfrac{114.62}{0.1909}=\$600.42[/tex]

For a duration of 6 years

n=6X12=72 months

[tex]P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\\\P=\dfrac{33000 X\frac{7}{2400} (1+\frac{7}{2400} )^{72}}{(1+\frac{7}{2400})^{72}-1} \\=\dfrac{96.25 (1.2333)}{1.2333-1}\\=\dfrac{96.25 (1.2333)}{0.2333}\\=\dfrac{118.71}{0.2333}=\$508.83[/tex]

Answer:

0.124 = 12.4% probability that exactly 9 small aircraft arrive during a 1-hour period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given time interval.

Rate of 8 per hour

This means that [tex]\mu = 8[/tex]

(a) What is the probability that exactly 9 small aircraft arrive during a 1-hour period?

This is P(X = 9).

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 9) = \frac{e^{-8}*8^{9}}{(9)!} = 0.124[/tex]

0.124 = 12.4% probability that exactly 9 small aircraft arrive during a 1-hour period.

Answer:

Step-by-step explanation:

Hi there,

To get started, recall the area of a bound circle formula:

[tex]A = \pi r^{2}[/tex]    where r is radius of the circle. However, Juan used an approximate value of π, 3.14. So for our purposes, the formula becomes:

[tex]A=[/tex] [tex](3.14)r^{2}[/tex]

Juan measured the circle's diameter, so we can find radius from diameter first. Radius is simply twice the length of the diameter; from one circle endpoint to the center, to the endpoint across, making a straight line:

[tex]d=2r[/tex] ⇒ [tex]r=\frac{d}{2} = \frac{5 \ cm}{2} = 2.5 \ cm[/tex]

Now plug in to obtain area:

[tex]A = (3.14)(2.5 cm)^{2} =19.625 \ cm^{2}[/tex]

The area is 19.265 centimetres squared.

Cross-sections are the area shapes when you cut through a 3D volume; if you cut through a pipe perpendicular to where it flows, you can see it is a circle! If you cut straight through a cube, it would be a square, etc.

If you liked this solution, hit Thanks or give a Rating!

thanks,

Answer: I think it’s B and D on eduinuity 2021

Step-by-step explanation:

The required 5m²2n² is the greatest common factor of (b) 5m⁴n₃ and (d) 15m²n².

The greatest common factor (GCF) is the largest positive integer that divides two or more numbers without leaving a remainder. In other words, it is the largest number that is a factor of all the given numbers.

For example, the GCF of 12 and 18 is 6, because 6 is the largest positive integer that divides both 12 and 18 without leaving a remainder.

here,
The terms that could have the greatest common factor of 5m²n² are:
5m⁴n₃, since 5m²n² is a factor of both 5m⁴n³ and 5m²n².
15m²n², since 5m²n² is a factor of both 15m²n² and 5m²2n².

Therefore, the correct options are (b) 5m⁴n₃ and (d) 15m²n².

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The complete question is given in the attachment below,

Answer:

[tex]t=\frac{423-433}{\frac{2.6}{\sqrt{15}}}=-14.896[/tex]    

[tex]df=n-1=15-1=14[/tex]  

We need to find in the t distribution with df=14 a value who accumulates 0.1 of the area in the left and we got [tex]t_{crit}= -1.345[/tex].

Since our calculated value for the statistic is is so much lower than the critical value we have enough evidence to reject the null hypothesis, and we can conclude that the true mean for this case is significantly less than 433 and then the machine is underfilling.

Step-by-step explanation:

Data given

[tex]\bar X=423[/tex] represent the sample mean

[tex]s=26[/tex] represent the sample standard deviation

[tex]n=15[/tex] sample size  

[tex]\mu_o =433[/tex] represent the value that we want to test

[tex]\alpha=0.1[/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

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 433 (underfilling), the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 433[/tex]  

Alternative hypothesis:[tex]\mu < 433[/tex]  

The statistic is:

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

Calculate the statistic

[tex]t=\frac{423-433}{\frac{2.6}{\sqrt{15}}}=-14.896[/tex]    

Decision rule

The degrees of freedom are:

[tex]df=n-1=15-1=14[/tex]  

We need to find in the t distribution with df=14 a value who accumulates 0.1 of the area in the left and we got [tex]t_{crit}= -1.345[/tex]

Since our calculated value for the statistic is is so much lower than the critical value we have enough evidence to reject the null hypothesis, and we can conclude that the true mean for this case is significantly less than 433 and then the machine is underfilling.

Answer:

97.73% of bags weigh more than 14.8 ounces.

Step-by-step explanation:

We are given that the weight of potato chip bags filled by a machine at a packaging plant is normally distributed with a mean of 15.0 ounces and a standard deviation of 0.1 ounces.

Let X = weight of potato chip bags filled by a machine

So, X ~ N([tex]\mu=15.0,\sigma^{2} =0.1^{2}[/tex])

The z-score probability distribution for normal distribution is given by;

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

where, [tex]\mu[/tex] = mean weight = 15.0 ounces

            [tex]\sigma[/tex] = standard deviation = 0.1 ounces

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

So, percentage of bags that weigh more than 14.8 ounces is given by = P(X > 14.8 ounces)

  P(X > 14.8 ounces) = P( [tex]\frac{ X -\mu}{\sigma}[/tex] > [tex]\frac{14.8-15.0}{0.1}[/tex] ) = P(Z > -2) = P(Z < 2)

                                                                   = 0.9773 or 97.73%

Now, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 in the z table which has an area of 0.9773.

Hence, 97.73% of bags weigh more than 14.8 ounces.

To find the percentage of bags weighing more than 14.8 ounces, we calculate the z-score for 14.8 ounces and use the normal distribution properties. A z-score of -2 corresponds to 2.28% of bags weighing less, so about 97.72% weigh more.

To determine what percentage of potato chip bags weigh more than 14.8 ounces when the mean weight is 15.0 ounces with a standard deviation of 0.1 ounces, we use the properties of the normal distribution. First, we calculate the z-score for 14.8 ounces:

Z = (X - μ) / σ = (14.8 - 15.0) / 0.1 = -2

The z-score tells us how many standard deviations away 14.8 ounces is from the mean. A z-score of -2 indicates that 14.8 ounces is 2 standard deviations below the mean. Using the z-score table or a calculator with normal distribution functions, we find the area to the left of z = -2. This area represents the percentage of bags weighing less than 14.8 ounces. Since we're interested in bags weighing more than 14.8 ounces, we subtract this value from 1 (or 100% if we're working with percentages).

For z = -2, the area to the left is approximately 0.0228 (or 2.28%). Therefore, the percentage of bags that weigh more than 14.8 ounces is about 100% - 2.28% = 97.72%.

The equation of a parabola that separates the blue points from the red points can be written in the form y = ax² + bx + c, where a, b, and c are constants. The specific coefficients will determine the direction, width, and position of the parabola.

To create an equation for a parabola that separates the blue and red points, we need more information about the specific characteristics desired for the parabola. The general form of the equation y = ax² + bx + c allows us to customize the parabola's shape. The coefficient a determines the direction of the parabola (opening upwards or downwards), while b and c influence its horizontal shift and vertical position.

For example, if we want a parabola that opens upwards with its vertex at the origin (0,0) and separates the points above from those below, the equation could be y = ax², where a is a positive constant. If a horizontal shift or translation is needed, adjustments to b and c can be made accordingly.

In summary, the equation of the parabola depends on the specific requirements for separating the blue and red points, and the general form y = ax² + bx + c provides the flexibility to tailor the parabola's characteristics to meet those needs.

Final answer:

The probability that all three students were born on Wednesday is 1/343.

Explanation:

To find the probability that all three students were born on Wednesday, we need to consider the total number of possible outcomes and the number of favorable outcomes.

There are 7 days in a week, so each student has a 1/7 chance of being born on Wednesday. Since the students are chosen at random and the choices are independent, we can multiply the probabilities together to find the probability that all three students were born on Wednesday:

P(all three born on Wednesday) = (1/7) * (1/7) * (1/7) = 1/343.

Answer:

1/2

Step-by-step explanation:

Answer:

11/12

Step-by-step explanation:

If you convert 3/4 and 1/6,

you will get 9/12+2/12.

If you add both of them, you will get your answer.

In the plot, I have graphed two functions on the same set of axes.

The first function, y(x) = e^(-0.5x) * sin(2x), is represented by a solid blue line with a 2-point line width.

The second function, y(x) = e^(-0.5x) * cos(2x), is shown with a dashed red line with a 3-point line width. Both functions are evaluated for 100 values of x ranging from 0 to 10.

The solid blue line represents the sine function, and the dashed red line represents the cosine function.

The legend, title, axis labels, and grid have been included to make the plot more informative and visually appealing.

In this plot, you can observe the oscillatory behavior of both functions as they decay exponentially with decreasing x.

The legend distinguishes between the two functions, and the grid helps in reading the values accurately.

The choice of line widths and colors enhances the visibility of the two functions, making it easier to compare their behavior over the specified range of x values.

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Answer:

it's B) 7

Step-by-step explanation:

Answer:

The answer is B

Step-by-step explanation:

So if Lisa scored less than 13 points then the answer has to lie in the intervals 1-12. You had 2 from intervals 1-6 and 5 from intervals 7-12 and you get 7

Answer:

  10.2

Step-by-step explanation:

Apparently, you want to find the solution for ...

  132.2 = 23.2√(5 +2.7t)

  132.2/23.2 = √(5 +2.7t) . . . . divide by 23.2

  (132.2/23.2)² = 5 +2.7t . . . . . square both sides

  (132.2/23.2)² -5 = 2.7t . . . . . subtract 5

  ((132.2/23.2)² -5)/2.7 = t . . . . divide by the coefficient of t

  10.2 ≈ t

It will take about 10.2 years for the demand for computers to reach 132.2 million.

Answer:

  about 100 cm²

Step-by-step explanation:

The side length of the rhombus is 1/4 of the perimeter so is 10 cm. The length of half of the other diagonal will be the length of the leg of a right triangle with hypotenuse 10 and leg 7 (half the given diagonal).

  d= √(10² -7²) = √51

Then the area of the rhombus is the product of this and the given diagonal:

  A = (14 cm)(√51 cm) ≈ 99.98 cm²

The area of the rhombus is about 100 cm².

This problem can be approached as a binomial distribution. The probability of a particular number of flights on time is calculated using the binomial probability formula. Determining 'unusual' can be subjective but normally a probability less than 0.05 is considered unusual.

This problem is a binomial probability problem because we have a binary circumstance (flight is either on time or it isn't) and a fixed number of trials (14 flights). The binomial probability formula is P(X=k) = C(n, k) * (p^k) * ((1 - p)^(n - k)) where n is the number of trials, k is the number of successful trials, p is the probability of success on a single trial, and C(n, k) represents the number of combinations of n items taken k at a time.

(a) For all 12 flights on time, it seems there's a typo; there are 14 flights in the sample. We can't calculate for 12 out of 14 flights without the rest of the information.

(b) For exactly 10 flights, we use n=14, k=10, p=0.85: P(X=10) = C(14, 10) * (0.85^10) * ((1 - 0.85)^(14 - 10)).

(c) For 10 or more flights on time, it's the sum of the probabilities for 10, 11, 12, 13, and 14 flights on time.

(d) For determining whether 11 or more on-time flights is unusual, it depends on the specific context, but we could consider it unusual if the probability is less than 0.05.

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Analyzing a sample of 14 flights at Denver International Airport, the probability of 10 or more flights arriving on time is 0.3783, and the probability of 11 or more flights arriving on time is 0.2142, which is not considered unusual.

(a) All 12 of the flights were on time.

(b) Exactly 10 of the flights were on time.

(c) 10 or more of the flights were on time.

(d) Would it be unusual for 11 or more of the flights to be on time?

We can use the binomial probability formula to solve this problem. The binomial probability formula is:

P(k successes in n trials) = (n choose k) *[tex](p)^k[/tex] * [tex](q)^(^n^-^k^)[/tex]

where:

n is the number of trials

k is the number of successes

p is the probability of success

q is the probability of failure

In this case, n = 14, p = 0.85, and q = 0.15.

(a) To find the probability that all 12 of the flights were on time, we can plug k = 12 into the binomial probability formula:

P(12 successes in 14 trials) = (14 choose 12) * [tex](0.85)^1^2[/tex]*[tex](0.15)^2[/tex]

Using a calculator, we can find that this probability is approximately 0.0032.

(b) To find the probability that exactly 10 of the flights were on time, we can plug k = 10 into the binomial probability formula:

P(10 successes in 14 trials) = (14 choose 10) *[tex](0.85)^1^0[/tex] *[tex](0.15)^4[/tex]

Using a calculator, we can find that this probability is approximately 0.1022.

(c) To find the probability that 10 or more of the flights were on time, we can add up the probabilities of 10, 11, 12, 13, and 14 successes:

P(10 or more successes) = P(10 successes) + P(11 successes) + P(12 successes) + P(13 successes) + P(14 successes)

Using a calculator, we can find that this probability is approximately 0.3783.

(d) To determine whether it would be unusual for 11 or more of the flights to be on time, we can find the probability of this event and compare it to a common threshold for unusualness, such as 0.05.

P(11 or more successes) = P(11 successes) + P(12 successes) + P(13 successes) + P(14 successes)

Using a calculator, we can find that this probability is approximately 0.2142. This probability is greater than 0.05, so it would not be considered unusual for 11 or more of the flights to be on time.

Answer:

2^3 n^2 or 8n^2

Step-by-step explanation:

Answer:

6 minutes

Step-by-step explanation:

Since it's you have 10 dollars in one hour, and you want to find 1/10th of the dollar value, you'd multiply 1/10th, or 0.1 by the time as well. I converted it into minutes because we're finding a smaller unit of time than 1 hour.

60 min x 1/10 (or 60 min/10)=  6 minutes

I hope this helped!

6 minutes

If it takes 1 hour to earn $10, we convert that 1 hour to 60 minutes. Now we will divide 60 by 10, and that gives you 6. Therefore every 6 minutes you earn $1.

Hope this helps!

Answer:

The standard deviation of the number of defective bulbs produced in an hour is 6.615

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

In this problem, we have that:

[tex]p = 0.0383, n = 1188[/tex]

What is the standard deviation of the number of defective bulbs produced in an hour

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1188*0.0383*(1-0.0383)} = 6.615[/tex]

The standard deviation of the number of defective bulbs produced in an hour is 6.615

Answer:

295

Step-by-step explanation:

multiply 410 x .72=295.2

you can not have .2 of a person, so you must round. Normally the question states what to round to, whether up or down, but generally .2 rounds down so: 295

Final answer:

295 students voted for Ben.

Explanation:

To calculate how many students voted for Ben in the class president election, we need to use the percentage of votes he received. Ben received 72% of the total votes from a class of 410 students.

First, convert the percentage to a decimal by dividing by 100:

72% = 72 ÷ 100 = 0.72

Then, multiply this decimal by the total number of students to find out how many voted for Ben:

Number of votes for Ben = 0.72 × 410

Now, we calculate the multiplication:

Number of votes for Ben = 295.2

Since we can't have a fraction of a vote, we'll round down to the nearest whole number. Thus, 295 students voted for Ben.