Using area under normal curve and z-score, approximately 21.32% of shafts have a diameter between 19.987 mm and 20.000 mm.
What is the proportion of shafts with a diameter between 19.987mm and 20.000mm?To find the proportion of shafts with a diameter between 19.987 mm and 20.000 mm, we need to calculate the probability that a randomly selected shaft falls within this range.
Given that the diameters of the shafts are normally distributed with a mean of 20.003 mm and a standard deviation of 0.005 mm, we can use the properties of the normal distribution to determine the desired proportion.
To calculate this proportion, we need to find the area under the normal curve between the values of 19.987 mm and 20.000 mm.
Let's denote the random variable X as the diameter of the shafts. We want to find P(19.987 ≤ X ≤ 20.000).
To do this, we can standardize the values by converting them to z-scores using the formula:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
For 19.987 mm:
z₁ = (19.987 - 20.003) / 0.005
For 20.000 mm:
z₂ = (20.000 - 20.003) / 0.005
We can then use a standard normal distribution table or calculator to find the corresponding probabilities associated with these z-scores.
Using a standard normal distribution table, we find that P(Z ≤ z₁) ≈ 0.2119 and P(Z ≤ z₂) ≈ 0.4251.
To find the proportion of shafts between 19.987 mm and 20.000 mm, we subtract the probabilities:
P(19.987 ≤ X ≤ 20.000) = P(Z ≤ z₂) - P(Z ≤ z₁) ≈ 0.4251 - 0.2119
P(19.987 ≤ X ≤ 20.000) ≈ 0.2132
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what equation matches the graph?
Answer:
b
Step-by-step explanation:
If arc ED=(9x-3) , arc BF=(15x-39) and angle BCF=(11x-9) find arc ED
Answer:
ED = 105
Step-by-step explanation:
Answer:
ED=105 and x=12
Step-by-step explanation:
The largest U.S flag in the world is 225 feet by 505 feet.
Is the ratio of the length to the width equivalent to 1:19,
the ratio for official government flags?
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.
Find the value of X.
Answer:
Angles of a triangle add to 180 degrees
42+58+x=180
x=80
Can someone help me with this problem
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 :)
d) Vegetable ghee is stored in a rectangular vessel of internal dimensions
60 cm x 10 cm x 45 cm. It is transferred into the identical cubical vessels. If the
internal length of each cubical vessel is 10 cm. how many vessels are required to
empty the rectangular vessel?
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
Please help!!! I’ll do anything!!
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
5. Let T1 and T2 be two stop times with respect to the same filtration. Prove that me (T1, T2) and T₁ +T2 are also stopping times.
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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Consider this right triangle.
Enter the ratio equivalent to sin (B)
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
Does anyone know the answer to this and if you do pls give it to me I need it ASAP
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
y = 100(1.25)^t
1. What is your initial value?
A student-fare bus pass costs half as much as an adult-fare pass. Together, one student pass and one adult pass cost $129. How much does each pass cost?
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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Prince Ivan rides Grey Wolf at a constant speed from King's Castle to the Magic Apple Garden in 5 hours. On their return trip to King's Castle, Grey Wolf runs at that original constant speed for the first 36 km. Then he runs the rest of the way 3 km/h faster. What was Grey Wolf's original speed if the return trip took 15 minutes less than the trip from King's Castle to the Magic Apple Garden?
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.
Im giving brainliest
Answer:
> there you go
Step-by-step explanation:
Answer:
>
Step-by-step explanation:
Is the spinner shown a fair spinner? Explain
why or why not.
solve the equation
13. 9^3c + 1 = 27^3c - 1
give the slope of the line with equation 17x = -34; then graph the line
please help me out! I don’t understand
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
A system of equations is created by using the line that is created by the equation 3x-2y=-4 and the line that is created
by the data in the table below.
Х
y
-9
-3
-1
-5
3
3
5
7
What is the y-value of the solution to the system?
Answer:
y=6x-8
Step-by-step explanation:
3x-2y=-4
3x^2-2y^2=-4^2
6x-y=8
y=6x-8
If a and b are the legs of a right triangle, and c is the hypotenuse, what is
the length of b if a = 6 and c = 18.5? (If necessary, round to the nearest
tenth)
West Co.'s 2016 financial statements were authorized for issue by its directors on 28 April 2017 and the annual general meeting will be held on 1 June 2017. Slow moving inventories held at one of its warehouses were valued at its cost of $350.000 at 31 December 2016.On 4 April 2016,70% of this inventory was sold for $120,000 on which West Co.'s sales staff earned a commission of $10,000. On 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 utilised 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 Hazelnut Co leased out tangible non-current assets as operating leases. At 1 January 2016, the carrying amount of such assets was $40 million while the remaining useful life was 10 years. These assets were leased out on operating leases that have recently expired. On 31 December 2016, the company is undecided to whether sell or lease the assets to customers under operating leases but wants to recognise it 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. For each of the above situation: i. Identify the relevant MERS i. Explain the accounting treatment Prepare relevant journal entries (note: for the financial year-end is 31 December 2016)
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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The tangent lines of a simple curve have azimuths 300 and bearing N 04° E, respectively. A third tangent line AB intersects the two tangent lines at bearing S 34 E. Stationing of the Pl of the curve is 16 + 464.35 and the distance from point B to the Pl of the curve is 277.6 m (ie BV = 277.6 m). Determine the following: a. Radius of the simple curve that shall be tangent to these three lines. b. Stationing of the PC Stationing of the PT
The radius of the simple curve is not provided and b. The stationing of the PC and PT is not provided.
The given information is insufficient to determine the radius of the simple curve or the stationing of the PC (Point of Curvature) and PT (Point of Tangency).
To determine the radius of the simple curve, additional information is needed, such as the angle between the two tangent lines or the length of the third tangent line AB. Without this information, we cannot calculate the radius.
Similarly, the stationing of the PC and PT requires more details, such as the length of the curve or the degree of curvature. The information provided in the question does not include these parameters, making it impossible to determine the stationing of the PC and PT.
Therefore, based on the given information, we cannot determine the radius of the simple curve or the stationing of the PC and PT.
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please help!
PQRS is a kite. Enter coordinates
for point S.
P(0, b)
S
Q(a, 0)
R(0, -c)
S([ ? ] ,[ ? ])
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)
Need help with this
Answer:
search it up
Step-by-step explanation:
help with segment relationships in circles...picture attatched.
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
It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet. A transportation researcher wants to determine if the statement made in the advertisement is false. She randomly test drives 34 small cars at 65 miles per hour and records the braking distance. The sample average braking distance is computed as 115 feet. Assume that the population standard deviation is 20 feet. Use Table 1.
Use α = 0.01 to determine if the average braking distance differs from 120 feet. The average braking distance is (significantly/not significantly) different from 120 feet.
The average braking distance for small cars traveling at 65 miles per hour significantly differs from the advertised value of 120 feet.
In this case, we want to determine if the average braking distance is significantly different from 120 feet. Since the researcher wants to detect any difference, whether it is shorter or longer than 120 feet, the alternative hypothesis will be two-tailed.
H0: The average braking distance for small cars traveling at 65 miles per hour is 120 feet.
Ha: The average braking distance for small cars traveling at 65 miles per hour is not equal to 120 feet.
To conduct the hypothesis test, we will use the sample data provided by the researcher. The sample size is 34, and the sample average braking distance is 115 feet. The population standard deviation is given as 20 feet.
The formula for the test statistic (z-score) is:
z = (sample average - hypothesized population average) / (population standard deviation / √sample size)
Plugging in the values from the problem:
z = (115 - 120) / (20 / √34)
z = -5 / (20 / √34)
Using Table 1 or a statistical calculator, we can determine the critical z-value corresponding to a significance level of 0.01. Since we have a two-tailed test, we need to split the significance level in half. Each tail will have an alpha of 0.005 (0.01/2).
Looking up the z-value for α/2 = 0.005, we find it to be approximately 2.576.
Now we compare the calculated z-value to the critical z-value:
If the calculated z-value falls outside the range defined by the critical z-values, we reject the null hypothesis. Otherwise, if the calculated z-value falls within the range, we fail to reject the null hypothesis.
In our case, the calculated z-value is -5 / (20 / √34), which we need to compare to -2.576 and +2.576.
If the calculated z-value is less than -2.576 or greater than +2.576, we reject the null hypothesis. Otherwise, if the calculated z-value is between -2.576 and +2.576, we fail to reject the null hypothesis.
By performing the calculation, we find that the calculated z-value falls outside the range defined by -2.576 and +2.576. Therefore, we can reject the null hypothesis.
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Put the following numbers in order from least to greatest: √42, 7, 6, √38.
Answer:
6, √38, √42, 7
Step-by-step explanation:
The numbers in order from least to greatest is 6,√38,√42,7
What is Algebra?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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0548 f(x) = 2100 The mean age of a woman in a certain country when her child is born can be approximated by the function 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, The mean age of a woman at the birth of her first child in 2015 is
(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.)
Need help with this question
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
r17 is greater than or equal to 545.7
Answer:
r17 is that a mistake or actually the number?