Item 5 Item 5

You are earning an average of $47,400 and will retire in 10 years. If you put 20% of your gross average income in an ordinary annuity compounded at 7% annually, what will be the value of the annuity when you retire?

Answer:

746 mi^2

Step-by-step explanation:

The top rectangle has an area of

A = 22*23 =506

The bottom rectangle has an area of

A =10 *24 = 240

Add the areas together

506+ 240 =746

Answer:

746

Step-by-step explanation:

22*23= 506

24*10= 240

506+240= 746

plz mark brainliest

Answer:

  x = -1/5, x = 1

Step-by-step explanation:

Maybe you want to find x.

Subtract the right side and collect terms.

  6x^2 +4x^2 -6x -4 -(2x -2) = 0

  10x^2 -8x -2 = 0

  5x^2 -4x -1 = 0 . . . . . . divide by 2

  (5x +1)(x -1) = 0 . . . . . . factor

Solutions are the values of x that make these factors zero:

  5x +1 = 0   ⇒   x = -1/5

  x -1 = 0   ⇒   x = 1

Solutions are x = -1/5, x = 1.

Answer:

For the given data, the mean is 50.8, the median is 51, and the mode is 52.

For the sample with this error, the mean is 48.7, the median is 51, and the mode is 52.

For the sample with this value removed, the mean is 43.2, the median is 50, and the mode is 52.

Step-by-step explanation:

We are given the following set of sample data below;

18, 26, 30, 42, 50, 52, 52, 76, 78, 84.

The formula for calculating mean is given by;

         Mean  =  [tex]\frac{\text{Sum of all data values}}{\text{Total number of observations}}[/tex]

                     =  [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 76+ 78+ 84}{10}[/tex]  

                     =  [tex]\frac{508}{10}[/tex]  =  50.8

For calculating median, we have to observe that the number of observations (n) in our data is even or odd, i.e;

                    Median  =  [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]

                    Median  =  [tex]\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}[/tex]

Now, here in our data the number of observations is even, i.e. n = 10.

So, Median  =  [tex]\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}[/tex]

                    =  [tex]\frac{(\frac{10}{2})^{th}\text{ obs.} +(\frac{10}{2}+1)^{th}\text{ obs.} }{2}[/tex]

                    =  [tex]\frac{5^{th}\text{ obs.} +6^{th}\text{ obs.} }{2}[/tex]

                    =  [tex]\frac{50+52 }{2}[/tex]  =  51

A Mode is a value that appears the maximum number of times in our data.

In our data, the value 52 is appera]ing maximum number of times, i.e. 2 times which means that mode of our data is 52.

Now, suppose the value 76 in the data is mistakenly recorded as 55 instead of 76. For the sample with this error,

            New Mean  =   [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 55+ 78+ 84}{10}[/tex]  

                                =  [tex]\frac{487}{10}[/tex]  =  48.7

Now, suppose the value 76 in the original sample is inadvertently removed from the sample. For the sample with this value removed,

            New Mean  =   [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 78+ 84}{9}[/tex]  

                                =  [tex]\frac{432}{10}[/tex]  =  43.2

            So, Median  =  [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]

                                 =  [tex](\frac{9+1}{2})^{th} \text{ obs.}[/tex]

                                 =  [tex]5^{th} \text{ obs.}[/tex] = 50

Answer:

480

Step-by-step explanation:

Answer:

4, 8, 16,64 and 4096.

Step-by-step explanation:

We are already given: [tex]4096=2^{12}[/tex]

To determine other whole numbers that can be raised to a power to obtain 4096, we apply the product rule of indices.

Product Rule of Indices: [tex]a^{xy}=(a^x)^y[/tex]

Now 12 can be factored in the following ways where one of the terms must be a perfect square:

[tex]2^{12}=(2^2)^6=4^6\\\\2^{12}=(2^6)^2=64^2\\\\2^{12}=(2^4)^3=16^3\\\\2^{12}=(2^3)^4=(8^2)^2=8^{2*2}=8^4\\\\2^{12}=(2^{12})^1=4096^1 $(This is the trivial case)[/tex]

Therefore, the other whole numbers that can be raised tp a power to obtain 4096 are: 4, 8, 16, 64 and 4096.

Answer:

The second choice.

Step-by-step explanation:

Answer:

The amount of heat lost by the aluminum is 31,500 J

Step-by-step explanation:

Given;

mass of aluminum, m =  500 g

initial temperature of the aluminum, θ₁ = 100° C

final temperature of the aluminum, θ₂ = 30°C

specific heat capacity of water, C = 4.18 J/g °C

specific heat capacity of aluminum , C = 0.90 J/g

Heat lost by the aluminum is equal to heat gained by the water.

The amount of heat lost by the aluminum, is calculated as;

Q = MCΔθ

Q = 500 x 0.9 (100 - 30)

Q = 500 x 0.9 x 70

Q = 31,500 J

Therefore, the amount of heat lost by the aluminum is 31,500 J

Answer:

there are 86,400 seconds in one day

Answer:

[tex]2043.67[/tex]

Step-by-step explanation:

Hundredths is at 2 decimal places.

The thousandths place is higher than 5, so add 1 to the hundredths place.

Answer:

2043.67

Step-by-step explanation:

If you’ve ever rounded a number, you would know that if it’s 5 or higher, round it up, and if it’s 4 or lower, round it down. In this case, the second decimal place reads ’6’ which is higher that 5, so we round up. The rest of the numbers stay the same

2043.67

Answer:

the third answer i am pretty sure because my brother is learning this and he told me and he has As

Step-by-step explanation:

if yo need any more help please tell me please mark as brainliest i only got it twice

Answer: 20% of 500= 100

So 500-100 = 400

4x100= 400

Step-by-step explanation:

Answer:

Step-by-step explanation:

Hello!

So in the refrigerator factory there are three shifts. Each shift records their quality based on the quantity of defective and working parts assembled.

Using a Chi-Square test of independence you have to test the claim that quality and shifts are independent.

The hypotheses are:

H₀: The variables are independent.

H₁: The variables are not independent.

α: 0.05

[tex]X^2= sum\frac{(O_{ij}-E_{ij})^2}{E_{ij}} ~X_{(r-1)(c-1)}[/tex]

r= total number of rows

c= total number of columns

i= 1, 2 (categories in rows)

j=1, 2, 3 (categories in columns)

To calculate the statistic you have to calculate the expected frequencies for each category:

[tex]E_{ij}= \frac{O_{i.}*O_{.j}}{n}[/tex]

[tex]O_{i.}[/tex] Represents the marginal value of the i-row

[tex]O_{.j}[/tex] Represents the marginal value of the j-column

[tex]E_{11}= \frac{O_{1.}*O_{.1}}{n}= \frac{21*40}{120}= 7[/tex]

[tex]E_{12}= \frac{O_{1.}*O_{.2}}{n}= \frac{21*40}{120}= 7[/tex]

[tex]E_{13}= \frac{O_{1.}*O_{.3}}{n}= \frac{21*40}{120}= 7[/tex]

[tex]E_{21}= \frac{O_{2.}*O_{.1}}{n}= \frac{99*40}{120}= 33[/tex]

[tex]E_{22}= \frac{O_{2.}*O_{.2}}{n}= \frac{99*40}{120}= 33[/tex]

[tex]E_{23}= \frac{O_{2.}*O_{.3}}{n}= \frac{99*40}{120}= 33[/tex]

[tex]X^2_{H_0}= \frac{(7-7)^2}{7} + \frac{(5-7)^2}{7} + \frac{(9-7)^2}{7} + \frac{(33-33)^2}{33} + \frac{(35-33)^2}{33} + \frac{(31-33)^2}{33} = 1.385= 1.34[/tex]

Using the critical value approach, the rejection region for this test is one-tailed to the right, the critical value is:

[tex]X^2_{(c-1)(r-1);1-\alpha }= X^2_{2; 0.95}= 5.991[/tex]

Decision rule:

If [tex]X^2_{H_0}[/tex] ≥ 5.991, reject the null hypothesis.

If [tex]X^2_{H_0}[/tex] < 5.991, do not reject the null hypothesis.

The value of the statistic is less than the critical value, the decision is to not reject the null hypothesis.

At 5% significance level, you can conclude that the shift the pieces were assembled and the quality of said pieces are independent.

I hope this helps!

Answer:

[tex]P(210<X<290)=P(\frac{210-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{290-\mu}{\sigma})=P(\frac{210-202}{70}<Z<\frac{290-202}{70})=P(0.114<z<1.26)[/tex]

And we can find this probability with this difference

[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)[/tex]

And we can find the difference with the normal standard distirbution or excel:

[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351[/tex]

Step-by-step explanation:

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

[tex]X \sim N(202,70)[/tex]  

Where [tex]\mu=202[/tex] and [tex]\sigma=70[/tex]

We are interested on this probability

[tex]P(210<X<290)[/tex]

The z score formula is given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Using the formula we got:

[tex]P(210<X<290)=P(\frac{210-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{290-\mu}{\sigma})=P(\frac{210-202}{70}<Z<\frac{290-202}{70})=P(0.114<z<1.26)[/tex]

And we can find this probability with this difference

[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)[/tex]

And we can find the difference with the normal standard distirbution or excel:

[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351[/tex]

Answer: Option D.

Step-by-step explanation:

Here we see the cross-section of a cone when it is cut by a plane that is parallel to the base of the cone.

As the plane is parallel to the base, we expect to see a figure that has the same shape as the base ( a circle) (you can think that over the plane we have a smaller cone, and the base of that cone also must be circular)

So the correct option is D.

Complete question is:

Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of 32 grams of fat per pound, with a standard deviation of 7 grams of fat per pound. A random sample of 34 farm-raised trout is selected. The mean fat content for the sample is 29.7 grams per pound. Find the probability of observing a sample mean of 29.7 grams of fat per pound or less in a random sample of 34 farm-raised trout. Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.

Answer:

Probability = 0.0277

Step-by-step explanation:

We are given;

Mean: μ = 32

Standard deviation;σ = 7

Random sample number; n = 34

To solve this question, we would use the equation z = (x - μ)/(σ/√n) to find the z value that corresponds to 29.7 grams of fat.

Thus;

z = (29.7 - 32)/(7/√34)

Thus, z = -2.3/1.200490096

z = -1.9159

From the standard z table and confirming with z-calculator, the probability is 0.0277

Thus, the probability to select 34 fish whose average grams of fat per pound is less than 29.7 = 0.0277

Answer: 7/6

Step-by-step explanation:

If he is writing sixths, then we have multiples of 1/6.

this is:

0*(1/6) = 0

1*(1/6) = 1/6

.

.

.

5*(1/6) = 5/6 (the numer he wrote at the left of 1)

6*(1/6) =  6/6 = 1

7*(1/6) = 7/6

So the number next to 1, (at the right of 1) must be 7/6.

You also can find it by adding 1/6 to 1.

1/6 + 1 = 1/6 + 6/6 = 7/6.

Answer:

[tex]7/6[/tex]

Explanation:

[tex]1/6 \approx 0.16666[/tex]

[tex]2/6 \approx 0.33333[/tex]

[tex]3/6 = 0.5[/tex]

[tex]4/6 \approx 0.66666[/tex]

[tex]5/6 \approx 0.83333[/tex]

[tex]6/6=1[/tex]

[tex]7/6 \approx 1.16666[/tex]

Answer:

  x = 8

Step-by-step explanation:

The angle with the expression in it is complementary to the 30° angle, so is 60°. Then we have ...

  4 +7x = 60

  7x = 56 . . . . . . subtact 4

  x = 8 . . . . . . . . .divide by 7

Answer:

0.62% probability that the mean of our sample is greater than $70.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

[tex]\mu = 65, \sigma = 20, n = 100, s = \frac{20}{\sqrt{100}} = 2[/tex]

What is the probability mean of our sample is greater than $70.

This is 1 subtracted by the pvalue of Z when X = 70. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{70 - 65}{2}[/tex]

[tex]Z = 2.5[/tex]

[tex]Z = 2.5[/tex] has a pvalue of 0.9938

1 - 0.9938 = 0.0062

0.62% probability that the mean of our sample is greater than $70.

Answer:

Sqrt(13)

Step-by-step explanation:

d = sqrt(3^2 + 2^2) = sqrt (13)

uhh there is no such thing because -1 isn't a perfect square.

Answer:

c) descriptive statistics

Correct. When we present information without any type of hypothesis we are describing the information

Step-by-step explanation:

We know that we are presenting historical information without any hypothesis and we need to find the right term, let's analyze one by one

a) predictive statistics

False. We can't predict if we are using historical information because predict is for the future and that not applied here.

b) prescriptive statistics

False. This term not exists in reality the most similar term is prescriptive analytic who analyze a series of scenarios fr an information given

c) descriptive statistics

Correct. When we present information without any type of hypothesis we are describing the information

d) inferential statistics

False. If we don't have any hypothesis we can't apply any inferential study and for this case is not the correct option

Answer:

[tex]105.13ft^2[/tex]

Step-by-step explanation:

Rectangle

[tex]A=lw\\=10*8\\=80ft^2\\[/tex]

Semicircle

[tex]A=\frac{1}{2} \pi r^2\\=\frac{1}{2}* \pi *4^2\\=25.13ft^2[/tex]

Add both values together

[tex]80+25.13\\=105.13ft^2[/tex]

Answer: 105.13

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]y = f(x) =\frac{1}{\sqrt{3 \pi} } e^{-x^{2/3}}[/tex]

y = 0, x = 0 and x = 3

Consider an element of thickness dx at a distance x from the origin. By Cylindirical Shell Method, the volume of the element is given by

[tex]dV=(2\pi rdr)h=(2\pi xdx)f(x) => dV=(2\pi xdx) \frac{1}{\sqrt{3\pi}}e^{-x^{\frac{2}{3}}}[/tex]

[tex]dV=2\sqrt{\frac{\pi}{3}}xe^{-x^{\frac{2}{3}}}dx[/tex]

Integrate the above integral over the limits x=0 to x=3 which implies

[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}xe^{-x^{\frac{2}{3}}}dx[/tex]

Solve by subsititution

[tex]Let,\\ -x^{\frac{2}{3}}=y => \frac{-2}{3}x^{\frac{-1}{3}}dx=dy => x^{\frac{-1}{3}}dx=\frac{-3}{2}dy[/tex]

Also, apply the new limits

[tex]At,\\\\ x=0, y=0 \ and \ At, x=3, y=-\sqrt[3]{9}[/tex]

This implies,

[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}x^{\frac{4}{3}}e^{-x^{\frac{2}{3}}}x^{\frac{-1}{3}}dx=2\sqrt{\frac{\pi}{3}}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}(\frac{-3}{2})dy[/tex]

[tex]V=-\sqrt{3\pi}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]

Let,

[tex]I=\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]

Integrate by parts the above integral

[tex]u=y^2 \ and \ dv=e^ydy => du=2y \ and \ v=e^y[/tex]

Integrate by parts formula

[tex]\int udv=uv-\int vdu => y^2e^y-\int 2ye^ydy[/tex]

Again integrate by parts

[tex]u=y \ and \ dv=e^ydy => du=1 \ and \ v=e^y[/tex]

Integrate by parts formula

[tex]\int udv=uv-\int vdu => y^2e^y-2[ye^y-e^y]=e^y[y^2-2y+2][/tex]

Therefore,

[tex]I=[e^y(y^2-2y+2)]_{0}^{-\sqrt[3]{9}}\\\\=e^{-2.0802}[(2.0802)^2+2(2.0802)+2]-e^{0}[0-0+2]\\\\\frac{(4.3272+4.1604+2)}{8.0061}-2\\\\=\frac{10.4876}{8.0061}-2\\\\=1.3099-2\\\\=-0.6901[/tex]

This implies, the volume is

[tex]V=-\sqrt{3\pi}I\\\\=-\sqrt{3\times 3.142} \times (-0.6901)\\\\=3.0701 \times 0.6901\\\\=2.1186[/tex]

That is, up to three decimal places

[tex]V\approx 2.118[/tex]

Answer:

618

Step-by-step explanation:

l x w

34x5

14x27

5x14

618

Answer:

y = 12 + 6x

y = 12x

Step-by-step explanation:

From the information provided, the following equations are derived:

y = 12 + 6x    ------- Eqn 1

y = 12x          ------- Eqn 2

Since Eqns 1 and 2 have the same subject, we equate them to solve for x. We have:

12x = 12 + 6x

Putting like terms together, we have:

12x - 6x = 12 ⇒ (12 - 6)x = 12

6x = 12 ⇒ x = 2

x = 2

Substitute x into Eqn 1 or 2

Eqn 1

y = 12 + 6x

y = 12 + 6(2) = 12 + 12

y = 24

Eqn 2

y = 12x

y = 12(2)

y = 24

It means that it will take Ms. Ironperson and Mr. Thoro 2 days apiece to produce the same number of posters at the current rate (which is 24 posters). Both Ms. Ironperson and Mr. Thoro will individually take 2 days to produce 24 Avenger posters apiece.

Answer:

[tex]9\dfrac{5}{6}[/tex]

Step-by-step explanation:

[tex]5\dfrac{1}{6}+4\dfrac{2}{3}=\\\\5\dfrac{1}{6}+4\dfrac{4}{6}=\\\\9\dfrac{5}{6}[/tex]

Hope this helps!

Answer:

Step-by-step explanation:

Area = area of rectangle 1 + area of rectangle 2 + area of rectangle 3 +area of triangle

= 8*12 +  12*9 + 19 *5 +  (1/2) * 4 *12

= 96 + 108 + 95 + 24

= 323 sq. cm

Answer:

Step-by-step explanation:

Perhaps you're factoring ...

  2x² +3x -54

  = 2x² +12x -9x -54 . . . . rewrite 3x appropriately

  = 2x(x +6) -9(x +6) . . . . factor pairs of terms

  = (2x -9)(x +6) . . . . . . . . finish the factoring

The factors are (2x -9) and (x +6).

Answer:

The width or range of the confidence interval with sample size 200 will be about half of that of the confidence interval with sample 50.

Step-by-step explanation:

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)

- For the two random samples, of sizes 50 and 200, the Central limit theorem allows us to say that the sample mean is approximately equal to the population mean as this random sample satisfies the condition of being a simple random sample and a distribution obtained from a normal distribution.

- Making the right assumption that population standard deviation is known and z-distribution is used to find the critical value

Critical value for 95% = 1.96

The critical value for both samples are the same then.

- Standard Error of the mean = σₓ = (σ/√n)

where σ = population standard deviation

n = sample size

For the two distributions

Confidence Interval = (Sample mean) ± [(Critical value) × (Standard Error of the mean)

(Sample mean)₅₀ = (Sample mean)₂₀₀

(Critical value)₅₀ = (Critical value)₂₀₀

(Standard Error of the mean)₅₀ = (σ/√50) = 0.1414σ

(Standard Error of the mean)₂₀₀ = (σ/√200) = 0.0707σ

0.1414σ = 2 × 0.0707σ

(Standard Error of the mean)₅₀ = 2 × (Standard Error of the mean)₂₀₀

(Standard Error of the mean)₅₀ > (Standard Error of the mean)₂₀₀

Hence,

(Margin of Error)₅₀ > (Margin of Error)₂₀₀

(Margin of Error)₅₀ = 2 × (Margin of Error)₂₀₀

Confidence Interval = (Sample mean) ± (Margin of error)

Hence, the width or range of the confidence interval with sample size 50 will be about two times larger than the confidence interval with sample 200.

Hope this Helps!!!

[tex]answer \\ = 2 , 4 , 5 \\ additional \: information \\ let \: r \: be \: a \: relation \: a \: to \: b. \: then \: the \: set \\ of \: first \: components \: or \: the \: set \: of \: \\ elements \: of \: a \: are \: called \: domain \\ and \: the \: set \: of \: second \: components \\ or \: the \: set \: of \: elements \: of \: b \: are \: called \: the \: range. \\ hope \: it \: helps[/tex]