explain how the is related to the standard error of the estimated difference in average birth weight for smoking and nonsmoking mothers.
Babies born to nonsmokers typically weighed between 1,000 and 5,000 grams. Babies of smokers typically ranged in weight from 500 to 4500 grams.
Given,
What is the weight at birth?
Birth weight is the term used to describe a newborn's body weight. Typically, newborns of European heritage weigh between 2.5 and 4.5 kilos, on average 3.5 kilograms, at delivery. Babies from South Asia and China typically weigh 3.26 kg.
Here,
Babies of nonsmokers frequently ranged in weight from 1,000 to 5,000 grams. Babies born to smokers frequently weighed between 500 and 4500 grams.
The usual ranges of birth weights between the two groups do show some notable differences, though. Nearly half of neonates among nonsmokers (56 out of 115, 48.7%) have birth weights between 3,000 and 4,000 grams since there are fewer babies in the lower weight ranges. Fewer babies are in the bigger weight groups, with 40 of 74, or 54%, of the newborns delivered to smokers having birth weights between 2,000 and 3,000 grams.
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plzz help me find the value of x im being timed plz !!
Answer:
I found it is beside of and next to when
Factor completely: 10a2 - 15a + 4ab - 6b.
A)(2a - 3)(5a + 2b)
B)(2a - 3b)(5a + 2)
C)(2a + 3)(5a - 2b)
D)(2a- 3) (5a²-2b)
Answer:
A
Step-by-step explanation:
\(10 {a }^{2} - 15a + 4ab - 6b \\ 5a(2a - 3) + 2b(2a - 3) \\ (5a + 2b)(2a - 3)\)
hopefully it makes sense
:)
(づ。◕‿‿◕。)づneed help asap!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
1) Ratio: 342:9 Rate: 38
2) Ratio: 152:38 Rate: 4
3) Ratio: 40:2 Rate: 20
Step-by-step explanation:
what is the mean of this data set 8 +16 +10+21 +6 +13+8 +26
How would you describe the following events, of randomly drawing a King OR a card
with an even number?
a) Mutually Exclusive
b)Conditional
c)Independent
d)Overlapping
Events, of randomly drawing a King OR a card with an even number describe by a) Mutually Exclusive.
The events of randomly drawing a King and drawing a card with an even number are mutually exclusive. This means that the two events cannot occur at the same time.
In a standard deck of 52 playing cards, there are no Kings that have an even number.
Therefore, if you draw a King, you cannot draw a card with an even number, and vice versa.
The occurrence of one event excludes the possibility of the other event happening.
It is important to note that mutually exclusive events cannot be both independent and conditional. If two events are mutually exclusive, they cannot occur together, making them dependent on each other in terms of their outcomes.
The correct option is (a) Mutually Exclusive.
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use integration by parts to evaluate the following integral. ∫−[infinity]−6θeθ dθ
The integral evaluates to [θeθ]∞−6 - e^(-6). To evaluate the integral using integration by parts, we first need to identify the parts of the integrand to be differentiated (u) and integrated (dv).
Let's choose:
u = -6θ
dv = e^θ dθ
Now, we need to differentiate u and integrate dv:
du = -6 dθ
v = ∫ e^θ dθ = e^θ
Integration by parts formula is given by:
∫u dv = uv - ∫v du
Applying this formula, we get:
∫(-6θ e^θ) dθ = (-6θ e^θ) - ∫(e^θ (-6)) dθ
Now, integrate the second term:
= -6θ e^θ + 6 ∫ e^θ dθ
Integrate e^θ:
= -6θ e^θ + 6 (e^θ) + C
Now, since the integral is from -∞ to a specific value, the integral is an improper integral. However, it's important to note that e^θ will go to 0 as θ approaches -∞, so we can evaluate the improper integral as:
∫[-∞, a] -6θ e^θ dθ = -6a e^a + 6 (e^a) - 6 (e^(-∞)) + C
So, the final answer is: -6θ e^θ + 6 e^θ + C
To use integration by parts to evaluate ∫−∞−6θeθ dθ, we need to choose two functions to differentiate and integrate. Let's choose u = θ and dv = eθ dθ. Then, du/dθ = 1 and v = eθ.
Using the integration by parts formula, we have:
∫−∞−6θeθ dθ = [θeθ]∞−6 - ∫−∞−6eθ dθ
Now, we need to evaluate the second integral. This is a straightforward integral, and we can evaluate it using the antiderivative of eθ:
∫−∞−6eθ dθ = [eθ]∞−6 = e^(-6)
Substituting this back into the original equation, we get:
∫−∞−6θeθ dθ = [θeθ]∞−6 - e^(-6)
Therefore, the integral evaluates to [θeθ]∞−6 - e^(-6).
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a laboratory scale is known to have a standard deviation of 0.005 in repeated weightings. scale readings in repeated weightings are normally distributed, with the mean of all possible measurements equal to the true weight of the specimen. a specimen weighted 4 times on this scale. the average weight is 4.53 grams. suppose that we needed a 95% confidence interval for this specimen using this data. what would the margin of error be?
The margin of error for a 95% confidence interval for this specimen is 0.008 grams. Then the 95% confident specimen's true weight is 0.008 grams of the sample mean of 4.53 grams.
To find the margin of error for a 95% confidence interval, we need to determine the standard error of the mean. The standard error of the mean formula can be written as:
SE = σ/√n
where:
SE = standard error of the mean,
σ = the standard deviation of the population
n = the sample size.
the standard deviation of the laboratory = 0.005
sample size = 4.
calculating the standard error of the mean:
SE = 0.005/√4
= 0.0025
Next, we need to find the critical value for a 95% confidence interval. the sample size is small (n < 30), then we need to use a t-distribution.
Critical value = 3.182.
The formula for calculating the margin of error is:
ME = t*SE
where :
ME = margin of error
t = critical value = 3.182.
SE = standard error of the mean = 0.0025
By substuting the values, we get:
ME = t*SE
= 3.182*0.0025
= 0.008 grams
Therefore, the margin of error for a 95% confidence interval for this specimen is 0.008 grams. Then the 95% confident specimen's true weight is 0.008 grams of the sample mean of 4.53 grams.
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At the beginning of the school year, Oak Hill Middle School has 480 students. There are 270 seventh graders and 210 eighth graders
At the beginning of the school year, Oak Hill Middle School has a total of 480 students. Out of these students, there are 270 seventh graders and 210 eighth graders.
To determine the total number of students in the school, we add the number of seventh graders and eighth graders:
270 seventh graders + 210 eighth graders = 480 students
So, the number of students matches the total given at the beginning, which is 480.
Additionally, we can verify the accuracy of the information by adding the number of seventh graders and eighth graders separately:
270 seventh graders + 210 eighth graders = 480 students
This confirms that the total number of students at Oak Hill Middle School is indeed 480.
Therefore, at the beginning of the school year, Oak Hill Middle School has 270 seventh graders, 210 eighth graders, and a total of 480 students.
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if p 62-5x and r =133 find qrs
The answer to this question is that we cannot find qrs without additional information about what q and s represent. We can only make educated guesses based on the context of the problem.
Sure, I'd be happy to help! To find qrs, we need to know what these variables represent. Based on the given information, we know that p=62-5x and r=133, but we don't know what q represents. Therefore, we cannot directly solve for qrs.
However, we can make an educated guess based on the context of the problem. It's possible that q is related to p and r in some way. For example, q could be the sum of p and r, or it could be the product of p and r. Without more information, we can't be sure.
Another possibility is that qrs is actually three separate variables, q, r, and s. In this case, we would need to know what s represents in order to solve for qrs.
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Planes X and Y intersect at a right angle. Line A B and Line C G lie in plane X and do not intersect. Line R S lies in plane Y.
Vertical plane x intersects horizontal plane y. Line R S is horizontal on plane y. Lines A B and C G are vertical and beside each other on plane x. A right angle is indicated between the lines on the y plane and x plane. Lines A B and D G are parallel.
Which statements are true? Select three options.
Based on the given information, the following statements are true:
1. Line R S is perpendicular to lines A B and C G on the y plane.
2. Line R S is parallel to line D F on the x plane.
3. Lines A B and C G are perpendicular to each other on the x plane.
What is Right angle ?
A right angle is an angle that measures exactly 90 degrees, or a quarter of a complete turn. It is often represented by a small square drawn at the vertex of the angle, as shown in the symbol "∟".
Explanation:
Statement 1 is true because line R S lies in plane Y and is perpendicular to the intersection of planes X and Y, which is represented by the right angle between lines A B and C G in the x plane.
Statement 2 is true because line R S is horizontal on plane Y, and line D G is parallel to line A B on the x plane, so line R S is also parallel to line D F on the x plane.
Statement 3 is true because lines A B and C G are beside each other and do not intersect on plane X, so they must be parallel to each other and perpendicular to line R S in the y plane.
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In the game of roulette, a steel ball is rolled onto a wheel that contains 18 red, 18 black, and 2 green slots. If the ball is rolled 25 times, find the probabilities of the following events. The ball falls into the green slots two or more times. The ball does not fall into the green slots. The ball falls into black slots 15 or more times. The ball falls into red slots 10 or fewer times.
The required probabilities for the events in question are:
0.000800.9992\(2.21\times10^{-8}\)\(1.63\times10^{-8}\)Based on the parameters given :
Number of red slots = 18
Number of black slots = 18
Number of green slots = 2
Total number of slots = 18+18+2 = 38
A.)
Probability of green slots 2 or more times :
(2/38)² × (36/38)²³ = 0.00080
B.)
Probability of ball not entering the green slots
1 - P(green slots)
P(not entering green slot ) = 1 - 0.0008 = 0.9992
C.)
probability that ball falls into black slot 15 or more times :
(18/38)¹⁵ * (20/38)¹⁰ = \(2.21\times10^{-8}\)
D.)
Probability that ball falls into red slot 10 or less times :
(10/38)¹⁰ * (28/38)¹⁵ = \(1.63\times10^{-8}\)
Therefore, the required probabilities are :
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Please help I will mark BRAINLIST
Answer:
Mean:
10.5+9.2+0+9.3+10.4+9.2=48.6
48.6 divded by 6 =8.1
Median is 9.3
Mode is 9.2
Step-by-step explanation:
Rafael runs 4 miles in 30 minutes. At the same rate, how many minutes would he take to run 10 miles?
Rafael will take 75 minutes to run 10 miles at the same rate.
Solution:
We know that Rafael can run 4 miles in 30 minutes.
So, we can find his pace, as follows:
30 minutes/ 4 miles = 7.5 minutes per mile.
So, for 10 miles he will take 7.5* 10 = 75 minutes.
Hence, it will take Rafael 75 minutes to run 10 miles.
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hElP mE plS anD ty I hAd tO rEpoSt thiS :')
Answer:
1: below its initial height.
2: if Jimmy has 20 balls and Chris take 15,how many balls does Jimmy have?
3: I would want Mrbeast to be the President because he would help everyone.
find the volume of this composite object!
(use 3 for pi)
The volume of the composite object can be calculated as approximately: 285 ft³.
How to Find the Volume of the Composite Object?The composite object is composed of a cylinder and a rectangular prism, therefore:
The volume of the composite object = volume of cylinder + volume of rectangular prism
Volume of cylinder = πr²h
radius (r) = 1/2(4) = 2 ft
height (h) = 5 ft
Volume of cylinder = 3 * 2² * 5 = 60 ft³
Volume of rectangular prism = length * width * height
= 9 * 5 * 5
= 225 ft³
Volume of the composite object = 60 + 225 = 285 ft³.
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On July 8 , Jones Inc. issued an $74,200,6%,120-day note payable to Miller Company. Assume that the fiscal year of Jones ends July 31 . Using a 360 -day year, what is the amount of interest expense recognized by Jones in the current fiscal year? When required, round your answer to the nearest dollar. a. $4,452 b. $568 c. $284 d. $852
The amount of interest expense recognized by Jones Inc. in the current fiscal year for the $74,200, 6%, 120-day note payable is $852.
To calculate the interest expense, we need to determine the interest for the 120-day period using a 360-day year. The formula for calculating interest is Principal × Rate × Time.
In this case, the Principal is $74,200, the Rate is 6% (0.06 as a decimal), and the Time is 120 days divided by 360 days.
Interest = $74,200 × 0.06 × (120/360) = $852.
Therefore, the correct answer is option d. $852, which represents the amount of interest expense recognized by Jones Inc. in the current fiscal year.
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what year are we in??
Lucy has $7 less than Kristine and $5 more than Nina together,the three have $35 how much does Lucy have?
Lucy has $7 less than Kristine and $5 more than Nina together, the three have $35. Lucy has $11.
Let's denote the amount of money that Kristine has as K, the amount of money that Lucy has as L, and the amount of money that Nina has as N.
According to the given information, we can form two equations:
Lucy has $7 less than Kristine: L = K - 7
Lucy has $5 more than Nina: L = N + 5
We also know that the three of them have a total of $35: K + L + N = 35
We can solve this system of equations to find the values of K, L, and N.
Substituting equation 1 into equation 3, we get:
K + (K - 7) + N = 35
2K - 7 + N = 35
Substituting equation 2 into the above equation, we get:
2K - 7 + (L - 5) = 35
2K + L - 12 = 35
Since Lucy has $7 less than Kristine (equation 1), we can substitute K - 7 for L in the above equation:
2K + (K - 7) - 12 = 35
3K - 19 = 35
Adding 19 to both sides:
3K = 54
Dividing both sides by 3:
K = 18
Now we can substitute the value of K into equation 1 to find L:
L = K - 7
L = 18 - 7
L = 11
Finally, we can find the value of N by substituting the values of K and L into equation 3:
K + L + N = 35
18 + 11 + N = 35
N = 35 - 18 - 11
N = 6
Therefore, Lucy has $11.
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Please help! the photo is below
Answer:
8 Green beads
Step-by-step explanation:
12 divided by 3 = 4
4 times 2 = 8
A rectangular board is 104cm wide and 156. cm long Equal squares, os large as possible,are ruled of this board find the size of the squares, how many squares are there?
Answer:
52 cm on a side
6 squares
Step-by-step explanation:
The side length of each square will be the greatest common factor (GCF) of 104 cm and 156 cm.
Square sizeEuclid's algorithm is useful for finding the GCF of two numbers whose factors you may not immediately know.
156/104 = 1 r 52 . . . . . . divide and keep the remainder
Replace the smaller number with the remainder and repeat:
104/52 = 2 r 0 . . . . . . . when the remainder is 0, the divisor is the GCF
The greatest common factor, and the side length of the square, is 52 cm.
Board sizeIn terms of the size of the square, the board is ...
104 cm / (52 cm) = 2 squares wide
156 cm / (52 cm) = 3 squares long
The number of squares that can be ruled on the board is 2×3 = 6.
Have to find x,y and z
Answer:
x=2, y=5
Step-by-step explanation:
the triangle that x is in is a right triangle (don't remember the theorem sorry)
use pythagorean theorem (a^2 + b^2 = c^2)
x is the hypotenuse(c)
1 squared plus /3 squared =c squared
1+3= c squared
4= c squared
c=2
x=2
find y using pythagorean theorem as well
/12 squared plus 2 squared = y squared
y is the hypotenuse
12+4=y squared
16= y squared
4=y
add one to account for that extra little piece
y=5
... I don't see z
Answer:
y= if it is the hypoteneuse for the whole triangle it √16 or 4, if it is the side of the smaller triangle it is 3
x=√4 which is 2
z= I don't think ther's a z
PLEASE HELP!!:)Rewrite the given equation without logarithms. DO NOT SOLVE.
2 log 4(x + 11) = log 416 + 2
Rewrite the given equation without logarithms. Do not solve for x.
Answer:
2 log4 (x + 11) = log4 16 + 22 log4 (x + 11) = 2 + 2log4 (x + 11) =2x + 11 = 4²x + 11 = 16From a full deck of 52 bridge cards you receive 6 cards. (a) What is the probability that they are all of the same suit? (b) What is the probability that they contain only one pair?
The probability of drawing 6 cards of the same suit is 0.002% while the probability of drawing a hand containing only one pair is 42.3%.
How to Solve the Probability?(a) The probability of drawing 6 cards of the same suit can be calculated as follows:
First, choose one of the four suits. There are 4 ways to do this.
Next, choose 6 cards from the chosen suit. There are 13 cards in each suit, so there are C(13,6) ways to do this.
Finally, choose any 6 cards from the deck. There are C(52,6) ways to do this.
Therefore, the probability of drawing 6 cards of the same suit is:
P(6 cards of the same suit) = (4 * C(13,6)) / C(52,6) ≈ 0.002%
(b) The probability of drawing a hand containing only one pair can be calculated as follows:
First, choose a rank for the pair. There are 13 ranks to choose from.
Next, choose 2 cards of the chosen rank. There are C(4,2) ways to do this.
Then, choose 4 cards from the remaining 48 cards. There are C(48,4) ways to do this.
Therefore, the probability of drawing a hand containing only one pair is:
P(only one pair) = (13 * C(4,2) * C(48,4)) / C(52,6) ≈ 42.3%
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Logic/Venn Diagram1. No environmentally produced diseases are inherited afflictions. Some psychological disorders are not inherited afflictions. Therefore, some psychological disorders are environmentally produced diseases.2. All watercolors are paintings. Some watercolors are masterpieces. Therefore, some paintings are masterpieces3. No illegal aliens are persons who have a right to welfare payments, and some migrant workers are illegal aliens. Thus, some persons who have a right to welfare payments are migrant workers.4. No insects that eat mosquitoes are insects that should be killed. All dragonflies are insects that eat mosquitoes. Therefore, no dragonflies are insects that should be killed.5. All corporations that overcharge their customers are unethical businesses. Some unethical businesses are investor-owned utilities. Therefore, some investor-owned utilities are corporations that overcharge their customers.
1. No environmentally produced diseases are inherited afflictions. Some psychological disorders are not inherited afflictions. Therefore, some psychological disorders are environmentally produced diseases. This statement invalid.
2. All watercolors are paintings. Some watercolors are masterpieces. Therefore, some paintings are masterpieces. This statement is invalid.
3. No illegal aliens are persons who have a right to welfare payments, and some migrant workers are illegal aliens. Thus, some persons who have a right to welfare payments are migrant workers. This statement is valid.
4. No insects that eat mosquitoes are insects that should be killed. All dragonflies are insects that eat mosquitoes. Therefore, no dragonflies are insects that should be killed. This statement is invalid.
5. All corporations that overcharge their customers are unethical businesses. Some unethical businesses are investor-owned utilities. Therefore, some investor-owned utilities are corporations that overcharge their customers. This statement is valid.
A Venn diagram uses overlapping circles and other shapes to show logical relationships between two or more groups of elements. It often helps to organize things graphically by highlighting similarities and differences between elements.
Venn diagrams, also known as set diagrams or logic diagrams, are widely used in mathematics, statistics, logic, education, linguistics, computer science, and economics. Venn diagrams became part of his "new mathematics" curriculum in the 1960s, so many people encounter them for the first time when learning mathematics or logic in school. These can be simple diagrams spanning 2 or 3 sets of a few items, or can be very sophisticated including 3D presentations as you progress to 6 or 7 sets or more. There is also They are used to reason about and represent how objects are related to each other within a particular "universe" or segment. Venn diagrams are widely used in presentations and reports as they allow users to visualize data in a clear and powerful way. These are closely related to Euler diagrams, differing in that the set is omitted if it has no elements. Venn diagrams show relationships even when the set is empty.
Suppose our universe is made up of pets and we want to compare the types of pets that families agree on.
Set A contains my favorites: dogs, birds, hamsters.
Set B contains Family B's preferences: dog, cat, fish.
Set C contains family C's favorites: dog, cat, turtle, and snake.
Overlapping or intersecting 3 sets contains only Dog. it's like having a dog
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Which of the following is not & characteristic of the t test? Choose the correct answer below: The Student t distribution has mean of t = 0 and a standard deviation of 5 = 1. The Student t distribution has the same general bell shape as the standard normal distribution_ The " Student t distribution is different for different sample sizes. The test is robust against - departure from normality:
The option that is not characteristic of the t-test is "The Student t distribution has mean of t = 0 and a standard deviation of 5 = 1". The correct answer is option A.
The t-test is a statistical test that uses the Student t-distribution. The Student t-distribution has a mean of 0, but its standard deviation is not equal to 1. The standard deviation of the t-distribution varies depending on the degrees of freedom, which is determined by the sample size. Therefore, the statement in option A is not true. The correct answer is option A.
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Can you help me with my work don’t send no link
Answer:
A and C!
Step-by-step explanation:
a: for a just move all the terms not containing x to the right side of the equation!
c: divide each term by 4 and simplify!
-
hope this helped <3
a.) How many ways are there to pack eight indistinguishable copies of the same book into five indistinguishable boxes, assuming each box can contain as many as eight books?
b.) How many ways are there to pack seven indistinguishable copies of the same book into four indistinguishable boxes, assuming each box can contain as many as seven books?
a.) To solve this problem, we can use a stars and bars approach. We need to distribute 8 books into 5 boxes, so we can imagine having 8 stars representing the books and 4 bars representing the boundaries between the boxes.
For example, one possible arrangement could be:
* | * * * | * | * *
This represents 1 book in the first box, 3 books in the second box, 1 book in the third box, and 3 books in the fourth box. Notice that we can have empty boxes as well.
The total number of ways to arrange the stars and bars is the same as the number of ways to choose 4 out of 12 positions (8 stars and 4 bars), which is:
Combination: C(12,4) = 495
Therefore, there are 495 ways to pack eight indistinguishable copies of the same book into five indistinguishable boxes.
b.) Using the same approach, we can distribute 7 books into 4 boxes using 6 stars and 3 bars.
For example:
* | * | * * | *
This represents 1 book in the first box, 1 book in the second box, 2 books in the third box, and 3 books in the fourth box.
The total number of ways to arrange the stars and bars is the same as the number of ways to choose 3 out of 9 positions, which is:
Combination: C(9,3) = 84
Therefore, there are 84 ways to pack seven indistinguishable copies of the same book into four indistinguishable boxes.
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A group of workers can plant 18 acres in 6 days.
What is their rate in acres per day?
Divide total acres by total days:
18 acres / 6 days = 3 acres per day.
The rate is 3 acres per day.
The Evaluation of -4 (4) (-1) (2)
Answer:
32
Step-by-step explanation:
calculator
Answer:
32
Step-by-step explanation:
-4*4=-16
-16*-1=16
16*2=32