Does anybody know both of these answere

Answer:

5 /36

Step-by-step explanation:

Sample space :

Number of faces^number of dice = 6^2 = 36

Sum of 8 on a roll of 2 dies :

Probability = required outcome / Total possible outcomes

Required outcome = number of times a sum of 8 is obtained on a roll of 2 dies = 5

Total possible outcomes = sample space = 36

P(obtaining a sum of 8) = 5 /36

Answer:

6, √38, √42, 7

Step-by-step explanation:

The numbers in order from least to greatest is 6,√38,√42,7

Algebra is the study of abstract symbols, while logic is the manipulation of all those ideas.

The acronym PEMDAS stands for Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction. This approach is used to answer the problem correctly and completely.

Finding the least common multiples of the denominator expressions can help. Then using the similar method as we use in sum of fractions would give the sum of algebraic fractions.

Given;

√42, 7, 6, √38

√42=6.4

√38=6.1

Therefore, the order of algebra will be 6,√38,√42,7

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Answer:

Angles of a triangle add to 180 degrees

42+58+x=180

x=80

Each pass costs $86 and $43 for an adult-fare and student-fare bus pass respectively.

Let the cost of an adult-fare bus pass be A student-fare bus pass costs half as much as an adult-fare pass, hence, the cost of a student pass will be $129 - A.

Mathematically, this is represented as:A = 2($129 - A) $A = $258 - 2A 3A = $258 A = $86Therefore, the cost of an adult-fare bus pass is $86.A student pass costs half as much as an adult-fare pass.

Since the cost of an adult-fare pass is $86, therefore the cost of a student pass will be half of $86. Mathematically, this can be represented as:Cost of student pass = 1/2 x $86 = $43

Therefore, each pass costs $86 and $43 for an adult-fare and student-fare bus pass respectively.

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Answer:

10/26

Step-by-step explanation:

out of 26 student 10 have a brother 8 people have only a brother and 2 people have both a sister and a brother. so you have 10/26

Answer:

r17 is that a mistake or actually the number?

Answer:

[tex]\boxed {\boxed {\sf sin(B)= \frac {21}{29}}}[/tex]

Step-by-step explanation:

Sine is the ratio of the opposite side to the hypotenuse.

[tex]sin \theta= \frac {opposite}{hypotenuse}[/tex]

We want to find the sine of angle B. The side AC, which measures 21, is opposite angle B.

The side AB, which measures 29, is the hypotenuse because it is the longest side and opposite the right angle.

[tex]opposite= 21\\hypotenuse=29[/tex]

Substitute the values into the formula.

[tex]sinB= \frac {21}{29}[/tex]

This ratio cannot be reduced further, so it is the final answer.

The sine of B is 21/29

Answer:

7,3 if rounded up...

not rounded up..... 7, 3.3????? I think???

Slope-intercept form: y = mx + b

First find the slope : [tex]\frac{rise}{run} = \frac{-1}{3}[/tex]

Slope = -1/3

Now we got  y = -1/3x + b

Next find y-intercept. This is where your line crosses the y line (vertical)

Y-intercept = 3

b = 3

Answer: y = -1/3x + 3

Answer:

2/5

Step-by-step explanation:

Add up everything which would be 15, then put the number of ovens their was as the numerator. Then you'd have 6/15 so then just simplify!

Answer:

x=23

Step-by-step explanation:

Hello There!

The relationship between chords can be found below in the image.

Pretty much the product of the segments in the same chord is equal to the product of the other segments in the other chord if that makes sense

So more specifically for this problem

10 * 18 = 9(x-3)

once we are able to create a formula/equation this problem is a lot easier to understand

Now we use basic algebra to solve for x

10 * 18 = 9(x-3)

step 1 combine like terms

10 * 18 = 180

now we have

180 = 9(x-3)

step 2 distribute the 9 to what's in the parenthesis (x and -3)

9*x=9x

9*-3=-27

now we have

180 = 9x - 27

step 3 add 27 to each side

-27 + 27 cancels out

180 + 27 = 207

now we have

207 = 9x

step 4 divide each side by 9

207/9 = 23

9x/9=x

we're left with x = 23

Now we want to check our answer

is 10 * 18 = 9*(23-3) then our answer is correct

10*18=180

23-3=20

20*9=180

180=180 is true hence the answer is 23

Answer: x = 33°

Step-by-step explanation: A triangle's angles should all equal 180, so we can set up an equation to help us.

106 + 41 + x = 180

Add 106 and 41

147 + x = 180

Subtract 147 on both sides.

x = 33

And there you go :)

Answer:

(-a, 0)

Step-by-step explanation:

By telling you that the quadrilaretal is a kite the problem is telling you that SQ is perpendicular to PR, and that PS=PQ (by simmetry, it follows that also RS=RQ). So S has to be the symmetric of Q to the center (intersection of the diagonals), which means that, since Py=0, its coordinates are (-a, 0)

Identification of the relevant MERS: Slow moving inventories (cost $350,000) held at West Co.'s warehouseTotal borrowing of $12 million for the construction of a building Non-current tangible assets held by Hazelnut CoThe accounting treatment for each of the above situations and relevant journal entries are as follows:

Accounting Treatment for Slow moving inventories held at West Co.'s warehouse. The slow-moving inventories held at West Co.'s warehouse will be valued at its cost of $350,000 at 31 December 2016. Since 70% of the inventory was sold, the remaining inventory will be valued at 30% of $350,000 = $105,000The accounting entry to record the sale of the inventory and the commission earned is as follows :Accounting Treatment for Slow moving inventories soldDateAccountTitlesDebitCredit4 April 2016Cost of Goods Sold 84,000Slow moving Inventories 84,000(70% x $350,000)Sales Commission 10,000Sales 10,000(70% x $120,000)Accounting Treatment for Non-current assets held by Hazelnut CoOn 31 December 2016, Hazelnut Co is undecided whether to sell or lease out the non-current assets, but it wants to recognize them as held for sale. The fair value less selling costs of the asset is $37 million, and the value in use is estimated at $38 million. Since the fair value less selling costs is lower than the value in use, the recoverable amount of the asset is lower than the carrying amount, and the asset will be written down.The accounting entry to write down the asset and recognize it as held for sale is as follows:Accounting Treatment for Non-current assets held by Hazelnut CoDateAccountTitlesDebitCredit31 December 2016Loss on Impairment 2,000,000Non-current Assets Held for Sale 2,000,000(Write down of non-current asset to fair value less selling costs of $37m)As of 31 December 2016, Hazelnut Co will recognize depreciation expense for the non-current assets for the year. The accounting entry to recognize the depreciation expense is as follows:Accounting Treatment for Depreciation of Non-current assets held by Hazelnut CoDateAccountTitlesDebitCredit31 December 2016Depreciation Expense 4,000,000Accumulated Depreciation - Non-current Assets 4,000,000(10% x $40,000,000)iii. Accounting Treatment for Total borrowing of $12 million for the construction of a buildingOn 1 January 2018, an entity borrowed $12 million for the construction of a building that is estimated to cost $30 million. Only $5 million was utilized during the year, and the balance of $7 million was invested temporarily in a fixed deposit which yielded an interest of 5% per annum. Total interest accrued on the borrowing of $12 million for the year 2016 was $0.84 million.On 31 December 2016, the entity will recognize the interest expense for the interest accrued on the borrowing of $12 million.The accounting entry to record the interest expense is as follows:Accounting Treatment for Total borrowing of $12 millionDateAccountTitlesDebitCredit31 December 2016Interest Expense 840,000Interest Payable 840,000(5% x $12,000,000)Thus, the accounting treatment for slow-moving inventories sold and non-current assets held by Hazelnut Co includes the recognition of losses due to impairment. The accounting treatment for total borrowing of $12 million includes the recognition of interest expense.

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The relevant MERS for each situation are as follows:

i. West Co.: Inventory

ii. The entity: Borrowing

iii. Hazelnut Co.: Assets

The accounting treatments and journal entries for each situation are given below:

i. West Co.: Inventory

The accounting treatment for slow-moving inventory valuation is to record the inventory’s cost of $350,000 on the balance sheet. When 70% of the inventory is sold for $120,000, the sales amount of $120,000 is recognized on the income statement, while the commission amount of $10,000 is recognized as selling expenses.

The journal entry is as follows:

Income statement: Dr. Cash $120,000Cr.

Revenue $120,000

Dr. Commission expense $10,000Cr.

Cash $10,000

Balance sheet:

Dr. Inventory $245,000Cr.

Cost of goods sold $245,000

ii. The entity: Borrowing

When an entity borrows funds, it records the principal amount on the balance sheet as a liability and recognizes interest expenses as the cost of borrowing. The interest amount for the year 2016 was $0.84 million. Therefore, the journal entry is as follows:

Balance sheet:

Dr. Cash $5,000,000

Dr. Fixed deposit $7,000,000 Cr.

Borrowing $12,000,000

Income statement:

Dr. Interest expense $0.84 million Cr.

Cash $0.84 million

iii. Hazelnut Co.: Assets

When an asset is recognized as held for sale, it is carried at the lower of the fair value less selling costs or value in use. The carrying amount of the asset is reduced to its fair value less selling costs.

The journal entry is as follows:

Balance sheet:

Dr. Provision for impairment $1 million Cr.

Tangible non-current assets $1 million

Balance sheet:

Dr. Tangible non-current assets $3 million Cr.

Provision for impairment $3 million

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Answer:

y=6x-8

Step-by-step explanation:

3x-2y=-4

3x^2-2y^2=-4^2

6x-y=8

y=6x-8

T1 and T2 are stop times, both of these events belong to Ft, their union also belongs to Ft. Hence, we can conclude that T1 + T2 is also a stop time.

We are given two stop times T1 and T2 with respect to the same filtration.

We are to prove that the maximum and the sum of T1 and T2, i.e., max(T1, T2) and T1 + T2 are also stop times.  

Let us consider the stop time T1.

This means that the event {T1 ≤ t} belongs to the sigma-algebra Ft, for all t≥0.

Similarly, let us consider the stop time T2.

This means that the event {T2 ≤ t} belongs to the sigma-algebra Ft, for all t≥0.

We are to prove that max(T1, T2) is also a stop time.

We can do so by considering the following event:{max(T1, T2) ≤ t}.

If T1 ≤ T2, then this event reduces to {T2 ≤ t} which belongs to Ft.

Similarly, if T2 ≤ T1, then this event reduces to {T1 ≤ t} which also belongs to Ft.

Thus, we can conclude that max(T1, T2) is a stop time.

We are to prove that T1 + T2 is also a stop time.

We can do so by considering the following event:{T1 + T2 ≤ t}.

This event can be expressed as:{T1 ≤ t − T2} ∪ {T2 ≤ t − T1}.

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Answer:

b

Step-by-step explanation:

The net financing need considering the three accounts is approximately $185,753.42.

To determine the average investment in accounts receivable, inventories, and accounts payable, we need to calculate the Inventory Period, Receivables Period, and Payment Period.

Inventory Period:

The Inventory Period is the average number of days it takes for finished goods to be sold. In this case, the average production process time is given as 40 days, and finished goods are kept on hand for an average of 15 days before they are sold. Therefore, the Inventory Period is the sum of these two periods:

Inventory Period = Production Process Time + Days Kept on Hand

Inventory Period = 40 days + 15 days

Inventory Period = 55 days

Receivables Period:

The Receivables Period is the average number of days it takes for accounts receivable to be collected. It is given that accounts receivable are outstanding for an average of 40 days. Therefore, the Receivables Period is 40 days.

Payment Period:

The Payment Period is the number of days the firm receives credit on its purchases from suppliers. It is given that the firm receives 40 days of credit. Therefore, the Payment Period is 40 days.

Now, we can calculate the average investment in accounts receivable, inventories, and accounts payable.

Average Investment in Accounts Receivable:

Average Investment in Accounts Receivable = (Net Sales / 365) * Receivables Period

Average Investment in Accounts Receivable = ($1,200,000 / 365) * 40

Average Investment in Accounts Receivable ≈ $131,506.85

Average Investment in Inventories:

Average Investment in Inventories = (Cost of Goods Sold / 365) * Inventory Period

Average Investment in Inventories = ($900,000 / 365) * 55

Average Investment in Inventories ≈ $142,191.78

Average Investment in Accounts Payable:

Average Investment in Accounts Payable = (Cost of Goods Sold / 365) * Payment Period

Average Investment in Accounts Payable = ($900,000 / 365) * 40

Average Investment in Accounts Payable ≈ $87,945.21

Finally, to calculate the net financing need, we subtract the average investment in accounts payable from the sum of the average investment in accounts receivable and inventories:

Net Financing Need = Average Investment in Accounts Receivable + Average Investment in Inventories - Average Investment in Accounts Payable

Net Financing Need = $131,506.85 + $142,191.78 - $87,945.21

Net Financing Need ≈ $185,753.42

Therefore, the net financing need considering the three accounts is approximately $185,753.42.

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Answer:

> there you go

Step-by-step explanation:

Answer:

>

Step-by-step explanation:

Answer:

Function A rate = 5/1

Function B rate = -4.5

Function A because its rate is positive while Function B's rate is negative

Answer:

Step-by-step explanation:

Remark

The ratio should be 10 : 19

Given

Flag Ratio: 225 : 505

Break the dimensions into  prime factors.

225: 15 * 15 = 3*5 * 3*5

505: 5 * 101

Conclusion

101 is prime so this dimension cannot be broken down any further

The fives cancel out. The dimensions of this flag are in the ratio of 45/101 which is 0.4455

10/19 = 0.5263

I would say this is reasonably close.

Answer:

search it up

Step-by-step explanation:

(a) Mean age in 2010 ≈ 23.9 years

(b) Mean age in 2013 ≈ 24.4 years

(c) Mean age in 2016 ≈ 24.9 years

To estimate the mean age of a woman at the birth of her first child in the given years, we can substitute the corresponding values of x into the function f(x) = 21 × [tex]x^{0.0521[/tex].

(a) For the year 2010 (x = 10):

f(10) = 21 × [tex](10)^{0.0521[/tex] ≈ 21 × 1.136 ≈ 23.856

The mean age of a woman at the birth of her first child in 2010 is approximately 23.9 years.

(b) For the year 2013 (x = 13):

f(13) = 21 × [tex](13)^{0.0521[/tex] ≈ 21 × 1.161 ≈ 24.381

The mean age of a woman at the birth of her first child in 2013 is approximately 24.4 years.

(c) For the year 2016 (x = 16):

f(16) = 21 × [tex](16)^{0.0521[/tex] ≈ 21 × 1.185 ≈ 24.885

The mean age of a woman at the birth of her first child in 2016 is approximately 24.9 years.

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The question is -

The mean age of a woman in a certain country when her child is born can be approximated by the function

f(x)=21x0.0521​,

where

x=10

corresponds to the year 2010. Estimate the mean age of the woman at the birth of her first child in the following years.

​(a) 2010

​(b) 2013

​(c) 2016

​(a) The mean age of a woman at the birth of her first child in 2010 is?

​(b) The mean age of a woman at the birth of her first child in 2013 is?

​(c) The mean age of a woman at the birth of her first child in 2016 is?

​(Type an integer or decimal rounded to one decimal place as​ needed.)

27 vessels

Step-by-step explanation:

get the volume of the rectangular vessel 60cm ×10cm ×45cm giving you 27000cm3.Find volume of one of the cube 10cm×10cm×10cm giving 1000cm3.

if 1 cube =1000cm3

? = 27000cm3

27000cm3× 1cube

1000cm3

di

vide then get your answer as 27 vessels

Answer:

48 and 9

Step-by-step explanation:

36/x + 5x-36/x+3 = 5-0.25

Answer:

9 and 48 km/hr

Step-by-step explanation:

The Formula [tex]S=\frac{D}{T}[/tex] along with its variations are used

Let's say for the initial ride, the speed was x. Because the time was 5 hours, the distance was 5x km. Now for the return. For the first part, the speed remains x but our distance is 36km. So, our time is 36/x hours. For the second part, the speed is x+3 but our distance is 5x-36 (The total minus the distance of the first part.) So, our time is [tex]\frac{5x-36}{x+3}[/tex]. Now we have the equation [tex]\frac{36}{x} + \frac{5x-36}{x+3}=5-\frac{1}{4}[/tex] as our times add up to 5 hours minus 1/4 of and hour. Solving, we get x=9, x=48.

Answer:

I believe the answer would be m = -28

Step-by-step explanation:

To remove the demoninater, multiple both sides by 7

12*7 = 84 so now you have -3 = 84

Divide 84 by -3, giving you -28