Austin is a human resources executive for a technology company. He is deciding between two types of plans for vacation allowance for the employees of the company: Unlimited and Traditional, Austin wants to deterrine, for workers in the tech Industry, if the yearly mean number of vacation days taken by workers with an Unlimited plan is greater than the yearly mean number of vacation days taken by workers with a Traditional plan Austin surveys a random sample of 17 workers who have the Unlimited plan and a random sample of 15 workers who have the Traditional plan. (These samples are chosen independently.) For each worker, he records the number of vacation days taken last year. For the workers with an Unlimited plan, the sample mean is 18.9 with a sample variance of 30.8. For the workers with a Traditional plan, the sample mean is 17.4 with a sample variance of 6.1 Assume that the two populations of vacation days taken are approximately normally distributed. Can Austin conclude, at the 0.10 level of significance, that the population mean of the yearly number of vacation days taken by workers with an Unlimited plan is greater than the population mean of the yearly number of vacation days taken by workers with a Traditional plan? Perform a one-talled test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, (If necessary, consult a list of formulas.) (*) State the null hypothesis and the alternate hypothesis (a) Determine the type of test statistic to use (a) state the null hypothesis H, and the alternate hypothesis (b) Determine the type of test statistic to use. (c) Find the value of the test statistic (Round to three or more decimal places.) (d) Find the critical value. (Round to three or more decimal places) (e) At the 0.10 level of significance, can Austin conclude that the yearly mean number of vacation days taken by workers with an Unlimited plan is greater than the yearly mean number of vacation days taken by workers with a Traditional plan?

[tex]1.2 \: degrees \: drop \: a \: day \\ 14 \: days \: in \: two \: weeks \\ net \: change = 14 \times 1.2 = 16.8 \: deg[/tex]

[tex]{\large{\textsf{\textbf{\underline{\underline{Question \: 1 :}}}}}}[/tex]

[tex]\star\:{\underline{\underline{\sf{\purple{Solution:}}}}}[/tex]

❍ Arrange the given data in order either in ascending order or descending order.

2, 3, 4, 7, 9, 11

❍ Number of terms in data [n] = 6 which is even.

As we know,

[tex]\star \: \sf Median_{(when \: n \: is \: even)} = {\underline{\boxed{\sf{\purple{ \dfrac{ { \bigg (\dfrac{n}{2} \bigg)}^{th}term +{ \bigg( \dfrac{n}{2} + 1 \bigg)}^{th} term } {2} }}}}}[/tex]

[tex]\\[/tex]

[tex] \sf Median_{(when \: n \: is \: even)} ={ \dfrac{ { \bigg (\dfrac{6}{2} \bigg)}^{th}term +{ \bigg( \dfrac{6}{2} + 1 \bigg)}^{th} term } {2} }[/tex]

[tex]\\[/tex]

[tex] \sf Median_{(when \: n \: is \: even)} ={ \dfrac{ {3}^{rd} term +{ \bigg( \dfrac{6 + 2}{2} \bigg)}^{th} term } {2} }[/tex]

[tex]\\[/tex]

[tex] \sf Median_{(when \: n \: is \: even)} ={ \dfrac{ {3}^{rd} term +{ \bigg( \cancel{ \dfrac{8}{2}} \bigg)}^{th} term } {2} }[/tex]

[tex]\\[/tex]

[tex] \sf Median_{(when \: n \: is \: even)} ={ \dfrac{ {3}^{rd} term +{ 4}^{th} term } {2} }[/tex]

• Putting,

3rd term as 4 and the 4th term as 7.

[tex]\longrightarrow \: \sf Median_{(when \: n \: is \: even)} ={ \dfrac{ 4 + 7 } {2} }[/tex]

[tex]\longrightarrow \: \sf Median_{(when \: n \: is \: even)} ={ \dfrac{ 11} {2} }[/tex]

[tex]\longrightarrow \: \sf Median_{(when \: n \: is \: even)} = \purple{5.5}[/tex]

[tex]\\[/tex]

[tex]{\large{\textsf{\textbf{\underline{\underline{Question \: 2 :}}}}}}[/tex]

[tex]\star\:{\underline{\underline{\sf{\red{Solution:}}}}}[/tex]

❍ Arrange the given data in order either in ascending order or descending order.

1, 2, 3, 4, 5, 6, 7

❍ Number of terms in data [n] = 7 which is odd.

As we know,

[tex]\star \: \sf Median_{(when \: n \: is \: odd)} = {\underline{\boxed{\sf{\red{ { \bigg( \frac{n + 1}{2} \bigg)}^{th} term}}}}}[/tex]

[tex]\\[/tex]

[tex] \sf Median_{(when \: n \: is \: odd)} = {{ \bigg(\dfrac{ 7 + 1 } {2} \bigg) }}^{th} term[/tex]

[tex]\\[/tex]

[tex] \sf Median_{(when \: n \: is \: odd)} = { \bigg(\cancel{\dfrac{8}{2}} \bigg)}^{th} term[/tex]

[tex]\\[/tex]

[tex] \sf Median_{(when \: n \: is \: odd)} ={ 4}^{th} term[/tex]

• Putting,

4th term as 4.

[tex]\longrightarrow \: \sf Median_{(when \: n \: is \: odd)} = \red{ 4}[/tex]

[tex]\\[/tex]

[tex]{\large{\textsf{\textbf{\underline{\underline{Question \: 3 :}}}}}}[/tex]

[tex]\star\:{\underline{\underline{\sf{\green{Solution:}}}}}[/tex]

The frequency distribution table for calculations of mean :

[tex]\begin{gathered}\begin{array}{|c|c|c|c|c|c|c|} \hline \rm x_{i} &\rm 3&\rm 1&\rm 7&\rm 4&\rm 6&\rm 2 \rm \\ \hline\rm f_{i} &\rm 4&\rm 6&\rm 2&\rm 2 & \rm 1&\rm 1 \\ \hline \rm f_{i}x_{i} &\rm 12&\rm 6&\rm 14&\rm 8&\rm 6&\rm \rm 2 \\ \hline \end{array} \\ \end{gathered} [/tex]

☆ Calculating the [tex]\sum f_{i}[/tex]

[tex] \implies 4 + 6 + 2 + 2 + 1 + 1[/tex]

[tex] \implies 16[/tex]

☆ Calculating the [tex]\sum f_{i}x_{i}[/tex]

[tex] \implies 12 + 6 + 14 + 8 + 6 + 2[/tex]

[tex]\implies 48[/tex]

As we know,

Mean by direct method :

[tex] \: \: \boxed{\green{{ { \overline{x} \: = \sf \dfrac{ \sum \: f_{i}x_{i}}{ \sum \: f_{i}}}}}}[/tex]

here,

• [tex]\sum f_{i}[/tex] = 16

• [tex]\sum f_{i}x_{i}[/tex] = 48

By putting the values we get,

[tex]\sf \longrightarrow \overline{x} \: = \: \dfrac{48}{16}[/tex]

[tex]\sf \longrightarrow \overline{x} \: = \green{3}[/tex]

[tex]{\large{\textsf{\textbf{\underline{\underline{Note\: :}}}}}}[/tex]

• Swipe to see the full answer.

[tex]\begin{gathered} {\underline{\rule{290pt}{3pt}}} \end{gathered}[/tex]

Answer:

Hey! I'm gonna go ahead and give ya the "meaning" of the fundamental theorem of algebra but not the answer to the A. B. and C.

Step-by-step explanation:

fundamental theorem of algebra, theorem of equations proved by Carl Friedrich Gauss in 1799. It states that every polynomial equation of degree n with complex number coefficients has n roots, or solutions, in the complex numbers. The roots can have a multiplicity greater than zero.

Answer:

figure A = 3 units

figure B = 9 x 3 = 27 units

scale factor =

[tex] \frac{3}{27 } = \frac{1}{9} [/tex]

so the scale factor = 9

Answer: 40 inches

Step-by-step explanation: If the 20 points on the line were spaced 1 inch apart, the result would be 20 inches. Each point you're going to is 1 inch in spacing and there is only going to be 20 inches of spacing (1 inch per point.)

For 2 inches spaced apart, the result would be simply 40 inches. You just need to add the distance of 20 more to the already distance of 20 points, since if 2 were to be divided, it would be 1, which is 20 inches in distance as stated. Another 20 points would be another inch in distance, and adding them gives you 2 inches in distance or 40 inches all together.

I apologize if I wasn't clear enough, I couldn't find a way to explain it as good as I thought it was in my mind.

Hope it helped though!

The corresponding angles are ∠2 and ∠6 i.e., ∠2≅∠6 according to the corresponding angles theorem. So, option 2 is correct.

The theorem states that,

The lines 'AB' and 'CD' are parallel and cut by a transverse 'EF' if and only if their corresponding angles are congruent.

The corresponding angles are those which lie on the same side of the transverse one is interior and the other is an exterior angle.

From the diagram,

Lines 'AB' and 'CD' are parallel and line 'EF' cuts these parallel lines.

So, the corresponding angles formed are:

(∠1, ∠5), (∠2, ∠6), (∠3, ∠7), and (∠4, ∠8)

According to the theorem, they are congruent. I.e.,

∠1 ≅ ∠5

∠2 ≅ ∠6

∠3 ≅ ∠7

∠4 ≅ ∠8

So, option 2 is correct.

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Disclaimer: The given question is incorrect. Here is the correct question.

Question: Lines 'c' and 'd' are parallel lines cut by the transversal 'p'. Which must be true by the corresponding angles theorem?

1) ∠1 = ∠7

2) ∠2 = ∠6

3) ∠3 = ∠5

4) ∠5 = ∠7

Answer:

$40/unit?

Step-by-step explanation:

I'll assume the the fixed and variable costs are for 2000 units.

Fixed Costs = $35,000

Variable Costs = $45,000

Tota Costs = $80,000

The selling price can be anything. Free and we lose $80,000; $40 each, net profit of $0; $100 each for a net profit of $120,000.

In the absence of any infoprmation regarding selling price (e.g., 50% profit margin), I'd select the $40/unit selling price since that is breakeven.

Hi there,

please see below for solution steps :

‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗

First let us revise the rounding rules before getting down to the process of rounding :

[tex]\sf{763.9 \ to \ the \ nearest \ whole \ number :}\\\hfill\stackrel{\small\text{add 1 }}{3^{+1}}.\\\hfill\stackrel{\small\text{drop it}}{9}}[/tex]

[tex]\sf{763.9\approx764}[/tex]

‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗

Answer:

6.5%

Step-by-step explanation:

I = prt

5616 = 7200 × r × 12

r = 5616/(7200 × 12)

r = 0.065

r = 6.5%

The sum of the equation is  = 5494000.

The outcome of adding numbers or quantities mathematically is a summation, often known as a sum. A summation always has an even number of terms in it. There may be just two terms, or there may be 100, 1000, or even a million. Some summations include an infinite number of terms.

Distribute 2j to (j+3).

Rewrite the summation as the sum of two individual summations.

Evaluate each summation using properties or formulas from the lesson.

The lower index is 1, so any properties can be used.

The sum is 5,494,000.

[tex]\sum_{j=1}^{200} 2 j(j+3)[/tex]

simplifying them we get:

[tex]\sum_{j=1}^{200} 2 j^{2}+6 j[/tex]

Split the summation into smaller summations that fit the summation rules.

[tex]\sum_{j=1}^{200} 2 j^{2}+6 j=2 \sum_{j=1}^{200} j^{2}+6 \sum_{j=1}^{200} j[/tex]

[tex]\text { Evaluate } 2 \sum_{j=1}^{200} j^{2}[/tex]

The formula for the summation of a polynomial with degree 2

is:

[tex]\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}[/tex]

Substitute the values into the formula and make sure to multiply by the front term.

[tex](2)$$\left(\frac{200(200+1)(2 \cdot 200+1)}{6}\right)$$[/tex]

we get: 5373400

Evaluating same as above : [tex]6 \sum_{j=1}^{200} j[/tex]

we get: 120600

Add the results of the summations.

5373400 + 120600

= 5494000

The sum of the equation is  = 5494000.

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Answer: A. V: x = –4; H: y = 4

Step-by-step explanation:

Answer:

3

Step-by-step explanation:

[tex]6=2 \times 3 \\ \\ 12=3 \times 2^2 \\ \\ 21=3 \times 7[/tex]

72 grams tea is needed to prepare 288 fluid ounces of tea.

According to the statement

we have given that the yvette uses 6 grams of tea leaves to make 24 fluid ounces of tea.

And we have to find that the how much grams of tea uses to make 288 fluid ounces of tea.

So, For this purpose,

the given terms are:

6 grams tea = 24 fluid ounces of tea

Now, we use division method

1 gram of tea = 4 fluid ounces of tea

It means

4 fluid ounces of tea prepared by 1 gram of tea.

then

288 fluid tea prepared = ?

So,

288 fluid tea prepared by grams of tea = 288/4

288 fluid tea prepared by grams of tea = 72.

So, 72 grams tea is needed.

So, 72 grams tea is needed to prepare 288 fluid ounces of tea.

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we can conclude that the last digit of the product of all the numbers between 11 and 29 is 0.

Here we want to find the last digit of the product between all the whole numbers larger than 11 and smaller than 29.

Then we have the product:

P = 12*13*14*15*16*17*18*19*20*21*22*23*24*25*26*27*28

Now, notice that there is a 20 there.

Any number times 20 will end with a zero, then:

P = 20*(12*13*14*15*16*17*18*19*21*22*23*24*25*26*27*28)

Only with that, we can conclude that the last digit of the product of all the numbers between 11 and 29 is 0.

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The amount of money she should deposit is 6131.14 dollars.

The amount she should deposit can be found as follows:

[tex]p = \frac{A}{(1+\frac{r}{n} )^{nt} }[/tex]

where

Therefore,

p = [tex]\frac{20000}{(1+\frac{0.12}{4}{} )^{40} }[/tex]

[tex]p=\frac{20000}{(1+0.03)^{40} }[/tex]

[tex]p = \frac{20000}{1.03^{40} }[/tex]

[tex]p=\frac{20000}{3.262037792}[/tex]

Hence,

p = $6,131.14

Therefore, she should deposit 6131.14 dollars.

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Answer:

30%

Step-by-step explanation:

1) Based on the given conditions, formulate:

30 meters/100 meters

2) Calculate 30 meters/100 meters:

3/10

3) Convert to percentage:

- 3/10 x 10/10

- 3x10/10x10

30/100

30/100 = 30%

You’re done!

Also, if you could label this brainliest that would be a great help!

Thanks xx

-Dante

[tex] \sqrt[4]{ - 8} [/tex]

Three students want to estimate the mean backpack weight of their schoolmates. To do this, they each randomly chose 8 schoolmates and weighed their backpacks. Then as per the given sample data,

(a) The sample means of the backpacks are: 6.375,6.375,6.625

(b) Range of sample means: 0.25

(c)The true statement is: The closer the range of the sample means is to 0, the less confident they can be in their estimate.

For the first sample, mean= 6.375

For the second sample, mean= 6.375

For the third sample, mean= 6.625

Range of sample means=Maximum Mean- Minimum Mean

                                        = 6.625 - 6.375

                                        = 0.25

The students will estimate the average backpack weight of their classmates using sample means, the true statement is:

The closer the range of the sample means is to 0, the more confident they can be in their estimate.

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Answer:

315 students

Step-by-step explanation:

Based on the question, we can deduce:

[tex] \frac{4}{5} = 80percent \: passing \: rate[/tex]

Next we can find out the total number of students who sat for the examination.

80% = 252

[tex]1percent \: = \frac{252}{80} \\ 100 \: percent = \frac{252}{80} \times 100 = 315 \: students [/tex]

Answer: $4.50

Step-by-step explanation:

Original Price: $6

Discount: 25%

Discounted Price: 6*25%= $1.5

                              $6-$1.5= $4.5

Answer:

75%

Step-by-step explanation:

3/4 ie 75/100. .........edge

The measurements which would most likely to have a negative exponent in scientific notation is the length of an amoeba in meters.

Given four measurements:

We are required to choose one measurement which would most likely to have a negative exponent in scientific notation.

A negative exponent is defined as the multiplicative inverse of the base,raised to the power which is of the opposite sign of tthe given power.It is expressed as [tex]e^{-x}[/tex].

We know that exponent shows continuous growth or continuous decay.

Among all the measurement the measurement which is most likely to have a negative exponent in scientific notation is the length of an amoeba in meters because among all the option amoeba can grow continously or decay continuously.

Hence the measurements which would most likely to have a negative exponent in scientific notation is the length of an amoeba in meters.

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The approximate solution of this system of equations is

(x,y)=(-0.25,1.25)

Equation is defined as the state of being equal and is often shown as a math expression with equal values on either side, or refers to a problem where many things need to be taken into account.

Given that,

y = |x − 1| and y = 3x+2

 y = |x − 1|

    = (x-1) ,    for  (x-1)>0 or x>1

and

y = -(x-1)  ,    for(x-1)<0  or x<1

y  = -x+1      

Now,   For x>1

y = x-1    ......(1)

y = 3x+2   ......(2)

Solving for x by equaliting's method:

 3x+2 = x-1

→ 3x-x = -1-2

→ 2x = -3

→ x = -3/2

→ x = -1.5 which is <1.

For x<1

y = -x+1    ......(1)

y = 3x+2   ......(2)

Solving for x by equaliting's method:

 3x+2 = -x+1

→ 3x+x = 1-2

→ 4x = -1

→ x = -1/4

→ x= -0.25 which is <1.

substitute the value of x = -0.25 in equation 1 for x<1.

y = -x+1

  = -(-0.25)+1

y  = 1.25

Hence, The approximate solution of this system of equations is

(x,y)=(-1/4,5/4)=(-0.25,1.25).

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Answer:

1.187991

Step-by-step explanation:

sin^2(225)−(cos(330))(cos(240))

=(−0.930095)2−(cos(330))(cos(240))

=0.865076−(cos(330))(cos(240))

=0.865076−−0.991199(cos(240))

=0.865076−(−0.991199)(0.325781)

=0.865076−(−0.322914)

=1.187991

Using the formula for a z-distribution confidence interval of proportions, he also needs to know the confidence level.

A confidence interval of proportions is given by:

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which:

Of these 3 itens, the problem states that he knows the sample size and the sample proportion, hence he needs to know z to determine the confidence interval. z depends on the confidence level, which is the remaining parameter.

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The value of each variable in the two-way table is:

a = 7

b = 31

c = 28

d = 13

e = 65

A Venn diagram uses circles that overlap each other to show logical relationships between two or more sets of items. A two-way table represents the frequency of dataset by arranging the dataset into a table made up of rows and columns.

The following have to be established:

a player can either be male or female

a player favors either the left or the right

a = females that favor the left. Looking at the Venn diagram, this number is 7.

b = total number of females : 24 + 7 = 31

c = males that favor the right = total number of males - males that favor the left

34 - 6 = 28

d = Total number of people who favor the left = 7 + 6 = 13

Total number of baseball players = 65

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The probability of correctly picking the two numbers is 2/10.

Probability determines the chances that an event would happen. The probability the event occurs is 1 and the probability that the event does not occur is 0.

The more likely the event is to happen, the closer the probability value would be to 1. The less likely it is for the event not to happen, the closer the probability value would be to zero.

The probability of correctly picking the two numbers = (numbers that have to be selected to win / total numbers)

2/10

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Answer:

Tools used to construct a bisector are 1》pencil ,2》Eraser ,3》protector ,4》Compass ,5》Sharpner etc ....

The measures of the two different angles are 70° and 110°.

What is angle in parallelogram ?

Thus from the given question, we have:

(6n − 70)° + (2n + 10)° =  180°

8n - 60 =  180° + 60

8n = 240

n = 240/8

n = 30°

So that,

i. (6n − 70)° = (6[30] − 70)°

                    = 180 - 70

                     = 110°

ii. (2n + 10)° = (2[30] + 10)°

                   = 60 + 10

                   = 70°

Therefore, the measures of the two different angles are 70° and 110°.

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The statement that is true about the box plots is: D. The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers.

For a data distribution that is displayed by a box plot, the measure of variation that can be used to describe the data distribution is the interquartile range.

Interquartile range, which is a measure of variation can be determined from a box plot by finding the value of the third quartile and first quartile, the finding their difference. The difference is the interquartile range.

Interquartile range = third quartile - first quartile.

Interquartile range for crackers = 85 - 75

Interquartile range for crackers = 10

Interquartile range for trail mix = 105 - 90

Interquartile range for trail mix = 15.

The bigger the value of the interquartile range, the more variation there is that exist in a data distribution.

Interquartile range for trial mix is the largest, therefore, the statement that is true about the box plots is: D. The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers.

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