Answer:
X = 9
Step-by-step explanation:
please help me .......
Answer:
A
Step-by-step explanation:
4
Find the area of the figure and type your result in the empty box provided.
13 m
6 m
8 m
7 m
Answer:
Answer:
I need a picture of the figure to do it
T or F:
When finding the percentage of each section, you have to divide the part by the whole or total amount.
Answer:
true because you can do that in a amound of the total
El usuario se registra medio de una temperatura de 8 grados centígrados Sin embargo a las 22 horas la temperatura y ha bajado unos 10 grados centígrados Cuál es la temperatura en este momento
Answer:
La temperatura en estos momentos es -2 grados centígrados.
Step-by-step explanation:
Dado que conoces que había una temperatura de 8°C y que a las 22 horas esta ha bajado unos 10 grados, tienes que restar estos 10 grados de la temperatura inicial para saber cuál es la temperatura actual:
8-10=-2
De acuerdo a esto, la respuesta es que la temperatura en estos momentos es -2 grados centígrados.
A chi-squared test for homogeneity of proportions requires that
A. ni, n2, n3, ... > 30
B. all expected counts are > 5
C. nipi, n2p2, n3p3, ... > 10
Which of the following statement(s) with regard to a data set's measures of Central Tendency, Dispersion, and Paired Observations is(are) TRUE?
In a very small set of data, i.e. small n, the Sample Standard Deviation is generally smaller than its Population Standard Deviation.
It is possible for the value of the Correlation between a set of paired observations to be greater than 1.
Unlike STDEV.P Excel function for calculating a Population Standard Deviation, Excel has no direct functions for calculating the Range and Midrange values of a data set.
Mode and Range are both measures of central tendency.
In a set of paired observations of X and Y, if the Correlation is 0.75, then as the values of X increases (decreases), the values of Y also generally increases (decreases) in the same direction.
In a very large set of data, i.e. large n, the Standard Deviation is a better measurement than Range.
It is possible for Median of a data set to have a value that is not equal to any of the values in the data set.
The statements that are TRUE with regard to a data set's measures of Central Tendency, Dispersion, and Paired Observations are the following:In a very small set of data, i.e. small n, the Sample Standard Deviation is generally smaller than its Population Standard Deviation.
It is possible for the value of the Correlation between a set of paired observations to be greater than 1.In a set of paired observations of X and Y, if the Correlation is 0.75, then as the values of X increases (decreases), the values of Y also generally increases (decreases) in the same direction.In a very large set of data, i.e. large n, the Standard Deviation is a better measurement than Range.Why these statements are true?In a very small set of data, the Sample Standard Deviation is generally smaller than its Population Standard Deviation because when the sample size is smaller, there is less dispersion and thus the value of the sample standard deviation is generally smaller than that of the population standard deviation.It is possible for the value of the Correlation between a set of paired observations to be greater than 1 because the correlation coefficient r ranges from -1 to 1, inclusive of both endpoints.
However, it is practically impossible to get a value of r outside this range in a real dataset.In a set of paired observations of X and Y, if the Correlation is 0.75, then as the values of X increases (decreases), the values of Y also generally increases (decreases) in the same direction. This is because 0.75 is a strong positive correlation indicating that as the value of one variable increases, the value of the other variable also increases.In a very large set of data, i.e. large n, the Standard Deviation is a better measurement than Range because the standard deviation takes into account all values in the dataset and is less sensitive to outliers as compared to the range. On the other hand, the range only considers the minimum and maximum values of the dataset and thus is less informative.
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a. Determine whether the Mean Value Theorem applies to the function f(x) 4x^(1/7) on the interval [-128,128). b. If so, find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem
To determine if the Mean Value Theorem (MVT) applies to the function f(x) = 4x^(1/7) on the interval [-128, 128), we need to check if the function satisfies the conditions of the MVT.
The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
In this case, the interval is [-128, 128), which is a closed interval. To check if the MVT applies, we need to ensure that the function is continuous on [-128, 128] and differentiable on (-128, 128).
Continuity: The function f(x) = [tex]4x^(1/7)[/tex] is a power function and is continuous for all x values, including the interval [-128, 128]. Therefore, the function is continuous on the interval.
Differentiability: The function f(x) = [tex]4x^(1/7)[/tex]is differentiable for all x values except at x = 0. Since the interval (-128, 128) does not include x = 0, the function is differentiable on the interval.
Therefore, both conditions for the MVT are satisfied, and we can conclude that the Mean Value Theorem applies to the function f(x) = [tex]4x^(1/7)[/tex] on the interval [-128, 128).
Next, we can find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem.
The Mean Value Theorem guarantees the existence of at least one point c in (-128, 128) such that f'(c) = (f(128) - f(-128))/(128 - (-128)).
Let's calculate the values:
f(128) = [tex]4(128)^(1/7)[/tex]≈ 4.5534
f(-128) = [tex]4(-128)^(1/7)[/tex]≈ -4.5534
f'(c) = (4.5534 - (-4.5534))/(128 - (-128)) = 9.1068/256 ≈ 0.0356
Therefore, by the Mean Value Theorem, there exists at least one point c in the interval (-128, 128) such that f'(c) ≈ 0.0356.
Please note that the specific value of c cannot be determined without further analysis or calculations. The Mean Value Theorem guarantees its existence but does not provide an exact value.
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AX and EX are secant segments that intersect at point X.
What is the length of DE?
1 unit
3 units
4.5 units
4 2/3 units
Answer:
DE=4
Step-by-step explanation:
got it right on edg
The length of DE is given as: 3 Units. (Option B)
A line that joins two points on a curve is called a Secant Line.
What is the Secant Theorem?
Two secant theorem states that if two secant lines are drawn from a point outside a circle a relationship is formed between the line segments.
How do we arrive at the length of DE?Based on the Secant Theorem, from the figure, we know that:
AX * BX = EX * DX
Assuming that the length of DE equals X,:
AX * BX = EX * DX
Equals
(7+2) X (2) = (x + 3) (3)
To solve for x we expand the brackets to state:
9 * 2= 3x + 9
3x = 18-9
3x= 9
x = 9/3
X which is same as DE = 3 Units.
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In a large population of college-educated adults, the mean IQ is 112 with standard deviation 50.62. Suppose 30 adults from this population are randomly selected for a market research campaign. The distribution of the sample mean IQ is: a. approximately Normal, with mean 112 and standard deviation 1.443. b. approximately Normal, with mean 112 and standard deviation 4.564. c. approximately Normal, with mean equal to the observed value of the sample mean and standard deviation 25. d. approximately Normal, with mean 112 and standard deviation 9.241.
Given: Population mean IQ = 112Population standard deviation IQ = 50.62Sample size (n) = 30To find: Distribution of the sample mean IQ
The Central Limit Theorem states that for a large sample size, the distribution of sample means will be approximately Normal with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size . Let's calculate the standard deviation of the sample mean IQ:
Standard deviation of sample mean IQ = (Population standard deviation IQ) / √n= 50.62 / √30= 9.241 (approx.)Therefore, the distribution of the sample mean IQ is approximately Normal, with mean 112 and standard deviation 9.241. The correct option is (d).
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What is the distance between A(7, 4) and B(2, −8)?
Answer:
The distance would 13 square units
Step-by-step explanation:
Well follow the formula
=√(2−1)^2+(2−1)^2
Plug in everything and solve, remeber to follow order of operations
what is a shape that has no sides the same length that is a quadrilateral
Answer:
Step-by-st
Bitly: URL Shortener - Short URLs & Custom Free Link ...bitlep explanation:
Which of the expressions below is equal to 10x+30? Select all that apply.
A metal bar weighs 24 ounces. 15% of the bar is gold. How many ounces of gold are in the bar? *
Answer:
7.6 ounces of silver
Step-by-step explanation:
Hope this helps :)
There ae 3.6 ounces of Gold in the metal bar.
What is mean by Percentage?A number or ratio that can be expressed as a fraction of 100 or a relative value indicating hundredth part of any quantity is called percentage.
To Calculate the percent of a number , divide the number by whole number and multiply by 100.
Given that;
A metal bar weighs 24 ounces.
And, 15% of the bar is gold.
Now, We can formulate as;
Amount of gold in the bar is,
⇒ 15% of 24
⇒ 15/100 × 24
⇒ 3.6 ounces
Hence, There ae 3.6 ounces of Gold in the bar.
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ayudaaaaaaaaaaa por favor
Answer:
ayuda te solo no eres tonto o si?
Step-by-step explanation:
divide 150 toys amongst two groups of children in the ratio 6:9
Answer:
60 and 90
Step-by-step explanation:
150/15 = 10
6 x 10 = 60
9 x 10 = 90
The required number of toys that can be divided among two groups is 60 and 90.
Given that,
To divide 150 toys amongst two groups of children in a ratio of 6:9.
The ratio can be defined as the proportion of the fraction of one quantity towards others. e.g.- water in milk.
Here,
Ratio = 6 / 9
Total of ratio = 15
Now,
For 1 group toys = 150 [6 / 15] = 60
For group 2 toys = 150 [9 / 15] = 90
Thus, the required number of toys that can be divided among two groups is 60 and 90.
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[5 points) X is places in an account which carries a nominal annual interest rate of 2.5% compounded monthly. After five years, the accumulated value is places in an account which earns a nominal annual interest rate of 3.2% compounded quarterly. The value of this account in eight years is $10,000. Find X.
The initial value of X can be determined using compound interest calculations. X is invested in an account with a nominal annual interest rate of 2.5%, compounded on a monthly basis, for a period of five years. After the initial period, X is then transferred to another account with a nominal annual interest rate of 3.2%, compounded on a quarterly basis, for a total duration of eight years. The approximate value of X at the end of this investment period is $6,573.83.
To solve for X, we will substitute the first equation into the second equation and solve for X. Let's proceed with the calculations:
The first equation is: FV = X(1 + 0.025/12)^(12*5)
The second equation is: $10,000 = X(1 + 0.032/4)^(48)(1 + 0.025/12)^(125)
We can substitute the first equation into the second equation:
$10,000 = [X(1 + 0.025/12)^(125)] * (1 + 0.032/4)^(48)
$10,000 = X * (1 + 0.025/12)^(125) * (1 + 0.032/4)^(48)
Now we can simplify the equation:
$10,000 = X * (1.002083)^60 * (1.008)^32
Divide both sides of the equation by [(1.002083)^60 * (1.008)^32] to solve for X:
X = $10,000 / [(1.002083)^60 * (1.008)^32]
Using a calculator, we can find the value of X:
X ≈ $10,000 / (1.138877 * 1.335893)
X ≈ $10,000 / 1.521364
X ≈ $6,573.83
Therefore, the value of X is approximately $6,573.83.
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Find the measure of the exterior angle.
What is the ratio of yellow butterflies to total butterflies? Choose the correct option
Answer:2/3
Step-by-step explanation: there is 2 yellow out of 3 butterfiles
Let f: X+R be a linear function, where X is a topological vector space. (a) Suppose that f is bounded above on a neighborhood V of the origin. That means there exists 7>0 such that f(x) < y for all x EV. Prove that there exists a neighborhood W of the origin such that f(x)
Given a linear function f: X -> R that is bounded above on a neighborhood V of the origin, we need to prove that there exists a neighborhood W of the origin such that f(x) < y for all x in W.
To prove this statement, we can use the properties of linear functions and the topology of X. Since f is linear, we know that f(0) = 0.
Let's consider the open ball B(0, y) centered at the origin with radius y. Since f is bounded above on V, there exists a constant M > 0 such that f(x) < M for all x in V.
Now, let's define a neighborhood W = V ∩ f^(-1)(B(0, y/M)). W is the intersection of V and the inverse image of the open ball B(0, y/M) under the function f.
By the properties of inverse images, we know that f(x) < y/M for all x in W.
Therefore, we have found a neighborhood W of the origin such that f(x) < y for all x in W, satisfying the condition we needed to prove.
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Which property of equality would be used to solve 3x=81
Answer:
Division
Step-by-step explanation:
Consider the function represented in the table.
Which point of the given function corresponds with the
minimum value of its inverse function?
X
-10-20
3 8
0
-2.
5 8
4.5-6
A (-20, 8)
B (-10,3)
C (0, -2)
D (8,-6)
HELPPP
Answer:
The real answer is (-20, 8).
Step-by-step explanation:
I just did the unit test practice or whatever and used the answer above and got it wrong. This is the correct answer.
What is the value of b for b' - 36/64
i need help please help me
-6x - 14 > 10 what is the answer to this problem, please helt
Answer:
Inequality Form:
x < - 4
Interval Notation:
( − ∞ , − 4 )
Step-by-step explanation:
n the problem of estimating total hospitalization costs for kidney stone patients, suppose Muscat and Dhofar regions were selected as strata because they have very different incident rates for the disease, and the estimates for each region was needed separately. Also, this stratification into geographic regions simplified the sampling procedures. The sample data are summarized as follows: Muscat Dhofar ni=260 12=150 Mean cost ý,=170 RO Mean cost y =125 RO S =3050 $ -2525 G = 745 C-10 RO n 02-680 C2=12 RO A previous study showed the number of kidney stone incidents in the Muscat to be 325 out of 100,000 population and the number in the Dhofar to be 320 out of 100,000. The population of the Muscat was 775,878, and the population of the Dhofar was 249,729, according to the 2010 census. a) Obtain the estimates of N, and N2, the numbers of kidney stone patients expected to be found in the Muscat and Dhofar regions. b) Obtain the average annual cost of hospitalization of the kidney patients of the two region combined with 95% confidence interval and interpret the results. c) Find the appropriate sample size n and stratum sample size ni and n2 for estimating the population mean with a bound on the error of estimation equal to 50 RO, considering proportional allocation.
The estimated number of kidney stone patients in the Muscat region (N₂) is approximately 447,184, and in the Dhofar region (N₁) is approximately 128,694.
Given,
Estimation of total hospitalization costs for kidney stone patients .
Here,
To obtain the estimates of N and N, the numbers of kidney stone patients expected to be found in the Muscat and Dhofar regions, we can use the stratified sampling formula:
Nᵢ = (nᵢ / n) * N,
where:
- Nᵢ is the estimate of the population size in stratum i.
- nᵢ is the sample size in stratum i.
- n is the total sample size (sum of all stratum sample sizes).
- N is the total population size.
Given the information provided, we have the following data:
For Dhofar region:
- n₁ = 260 (sample size)
- N = 249,729 (population size according to the 2010 census)
For Muscat region:
- n₂ = 150 (sample size)
- N = 775,878 (population size according to the 2010 census)
Using the given formula, we can calculate the estimates for each region:
For Dhofar region:
N₁ = (n₁ / n) * N = (260 / (260 + 150)) * 249,729
For Muscat region:
N₂ = (n₂ / n) * N = (150 / (260 + 150)) * 775,878
To obtain the values, let's calculate them:
For Dhofar region:
N₁ = (260 / (260 + 150)) * 249,729 ≈ 128,694
For Muscat region:
N₂ = (150 / (260 + 150)) * 775,878 ≈ 447,184
Therefore, the estimated number of kidney stone patients in the Muscat region (N₂) is approximately 447,184, and in the Dhofar region (N₁) is approximately 128,694.
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(3/2)^4= in fraction form help pls
Answer:
81/16
Step-by-step explanation:
Sophia drove 63 miles. Sophia's car used 2 gallons of gas. How many miles per gallon did Sophia car get?
Answer:
31.5 mi/gal
Step-by-step explanation:
[tex]63[/tex]÷[tex]2[/tex][tex]=31.5[/tex]
So, Sophia got 31.5 miles per gallon
Answer: 31.5 I think.
Step-by-step explanation:
Give other person brainliest
I need help with this helpppp :(
Answer:
Domain: (-∞,+∞)
Range: (-∞,1)
y-intercept: (0,2)
Asymptote: I am not sure (sorry) I know they can be solved using the equation n(x)=0
Step-by-step explanation:
Domain: the set of all x-values
- this graph has arrows which means the domain is from -∞ to +∞ (-∞,+∞)
Range: the set of all y-values
- the graph extend continuously on the negative side so -∞ and it stops at 1 on the positive side (-∞,1)
Y-intercept: this point is where the graph crosses the y-axis, this is at (0,2)
solve the following equation on the interval [0°,360°). separate multiple answers with a comma. remember to include a degree symbol. 4cos2xtanx−2tanx=0
To solve the equation 4cos^2(x)tan(x) - 2tan(x) = 0 on the interval [0°, 360°), we can use algebraic manipulations and trigonometric identities. Let's simplify the equation step by step:
Start with the given equation: 4cos^2(x)tan(x) - 2tan(x) = 0.
Factor out the common term tan(x): tan(x)(4cos^2(x) - 2) = 0.
Set each factor equal to zero and solve separately:
a) tan(x) = 0:
Since tan(x) is zero at x = 0°, 180°, and 360°, we have x = 0°, 180°, 360° as solutions.
b) 4cos^2(x) - 2 = 0:
Add 2 to both sides: 4cos^2(x) = 2.
Divide by 4: cos^2(x) = 1/2.
Take the square root: cos(x) = ±√(1/2).
To find the values of x in the interval [0°, 360°), we need to consider both the positive and negative square root:
cos(x) = √(1/2):
x = 45°, 315° (since cos(45°) = cos(315°) = 1/√2)
cos(x) = -√(1/2):
x = 135°, 225° (since cos(135°) = cos(225°) = -1/√2)
Therefore, the solutions to the equation 4cos^2(x)tan(x) - 2tan(x) = 0 on the interval [0°, 360°) are: x = 0°, 45°, 135°, 180°, 225°, 315°, and 360°.
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4 out of the 80 students at a school assembly were first-grade students. What percentage of the students at the assembly were first-graders?
Answer:5
Step-by-step explanation:
Answer:
5 percent
Step-by-step explanation:
4/80 = 5 percent