Answer:
1.5cm
Step-by-step explanation:
I TOOK THE TEST
The product of a number and 2.5 is 10 find the number
The number whose product with 2.5 result 10 is 4
What is an equation?An equation is the equality between two algebraic expressions, which have at least one unknown or variable.
In mathematics, when we express "product" we mean multiplication. So the expression gave "the product of a number and 2.5 is 10" is represented by:
x*2.5 = 10
Clearing the variable we have:
x = 10/2.5
x = 4
Substituting the value in the equation we have:
x*2.5 = 10
4*2.5 = 10
10 = 10
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equals 20+20=40 Expression value is
Answer:
expression is 20. +20...
There are many cones with a height of 12 inches. Let r represent the radius and v represent the volume of these cones
The relationship between the radius (r) and the volume (v) of these cones is v = 4πr^2.
We have multiple cones with a height of 12 inches, and we are representing the radius of the cones as r and the volume as v. We need to determine the relationship between the radius and the volume of these cones.
To find the relationship between the radius and the volume of the cones, we can follow these steps:
The formula for the volume of a cone is given by V = (1/3) * π * r^2 * h, where V is the volume, π is a constant (approximately 3.14), r is the radius, and h is the height.
Since we have multiple cones with the same height of 12 inches, we can substitute h = 12 into the volume formula: V = (1/3) * π * r^2 * 12.
Simplify the equation: V = 4πr^2.
Therefore, the relationship between the radius (r) and the volume (v) of these cones is v = 4πr^2.
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what is the probability that washing dishes tonight will take me between 15 and 16 minutes? give your answer accurate to two decimal places.
The probability of washing dishes by me tonight will take between 15 and 16 minutes is 14.29%.
The term "uniform distribution" refers to a type of probability distribution in which the likelihood of each potential result is equal.
Let's consider, the lower limit for this distribution as a and the upper limit for this distribution as b.
The following formula gives the likelihood that we will discover a value for X between c and d,
\(P(c\leq X\leq d)=\frac{d-c}{b-a}\)
Given the time it takes me to wash the dishes is 11 minutes and 18 minutes. From this, a = 11 and b = 18.
Then,
\(P(15\leq X\leq 16)=\frac{16-15}{18-11}=0.1429=14.29\%\)
The answer is 14.29%.
The complete question is -
The time it takes me to wash the dishes is uniformly distributed between 11 minutes and 18 minutes. What is the probability that washing dishes tonight will take me between 15 and 16 minutes?
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Which set of line segments could create a right triangle?
As we know in a right angled triangle the sum of two sides will always be greater than the third side.
We will use the Pythagoras Theorem here:
c² = a² + b²; where c is the longest side.
The only segments satisfying this are
15, 36, 39
Answer:
4) 5, 12, 13
Step-by-step explanation:
Which of the following pairs are NOT equivalent statements?
In x = y and x = e^y
log⬇️p N = b and b^p = N
x = sqrt y and x = y^1/2
log⬇️b N = p and B^p = N
Answer:
the answer is b
Step-by-step explanation:
bc they are not both equivalent.
The pair (B) is not equivalent after using the property of logarithm option (B) is correct.
What is a logarithm?It is another way to represent the power of numbers, and we say that 'b' is the logarithm of 'c' with base 'a' if and only if 'a' to the power 'b' equals 'c'.
\(\rm a^b = c\\log_ac =b\)
It is given that:
The expression is given in the options.
As we know, the expression can be defined as the combination of constants and variables with mathematical operators.
It is required to find the equivalent pairs:
A) In(x) = y and \(\rm x = e^y\)
As we know from the log property:
logₙm = a
\(\rm \rm n^a = m\)
The pairs are equivalent.
B) \(\rm log_p N = b \ and \ b^p = N\)
Again using the log property:
logₙm = a
\(\rm \rm n^a = m\)
The pairs are not equivalent.
C) \(\rm x = \sqrt y \ and \ x = y^{\dfrac{1}{2}}\)
Using the property of exponent we can write the above expression:
The pairs are equivalent.
D) \(\rm log_b N = p \ and \ b^p = N\)
logₙm = a
\(\rm \rm n^a = m\)
The pairs are equivalent.
Thus. the pair (B) is not equivalent after using the property of logarithm option (B) is correct.
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What is the equation of the line tangent to the curve y + e^x = 2e^xy at the point (0, 1)? Select one: a. y = x b. y = -x + 1 c. y = x - 1 d. y = x + 1
The equation of the tangent line to the curve y + e^x = 2e^xy at the point (0, 1) is y = -x + 1. The correct answer is (b).
To find the equation of the tangent line to the curve y + e^x = 2e^xy at the point (0, 1), we need to find the slope of the tangent line at that point.
First, we can take the derivative of both sides of the equation with respect to x using the product rule:
y' + e^x = 2e^xy' + 2e^x
Next, we can solve for y' by moving all the terms with y' to one side:
y' - 2e^xy' = 2e^x - e^x
Factor out y' on the left side:
y'(1 - 2e^x) = e^x(2 - 1)
Simplify:
y' = e^x / (1 - 2e^x)
Now we can find the slope of the tangent line at (0, 1) by plugging in x = 0:
y'(0) = 1 / (1 - 2) = -1
So the slope of the tangent line at (0, 1) is -1.
To find the equation of the tangent line, we can use the point-slope form of a line:
y - 1 = m(x - 0)
Substituting m = -1:
y - 1 = -x
Solving for y:
y = -x + 1
Therefore, the equation of the tangent line to the curve y + e^x = 2e^xy at the point (0, 1) is y = -x + 1. The correct answer is (b).
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difference between linear and projectile motion. Which component will usually remain at a constant velocity? Why?
The difference between the linear and projectile motion is that "Linear-motion" refers to motion of object in a "straight-line", while "projectile-motion" refers to motion of an object that is thrown into air, in a curved path.
In linear motion, the object moves along a straight line, with its velocity and acceleration aligned in the same direction. The object's speed and direction may change, but its motion remains linear.
In projectile motion, the object moves along a curved path under the influence of gravity. The object is launched into the air with an initial velocity, and then gravity causes it to follow a parabolic path until it lands back on the ground. The motion of the object is influenced by both its initial velocity and the force of gravity.
In both linear and projectile motion, the "horizontal-component" of velocity will usually remain constant because there is no external force acting on the object in horizontal direction, and thus no acceleration.
Therefore, the object will continue to move at a constant velocity in the horizontal direction, as long as there is no external force acting on it.
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Find the points on the given curve where the tangent line is horizontal or vertical. (Order your answers from smallest to largest r, then from smallest to largest theta.)
r = 1 + cos(theta) 0 ≤ theta < 2
horizontal tangent
(r, theta)=
(r, theta)=
(r, theta)=
vertical tangent
(r, theta)=
(r, theta)=
(r, theta)=
The points on the curve where the tangent line is horizontal or vertical are (0, π/2) and (2, 3π/2).
To find the points where the tangent line is horizontal or vertical, we need to determine the values of r and θ that satisfy these conditions. First, let's consider the horizontal tangent lines.
A tangent line is horizontal when the derivative of r with respect to θ is equal to zero. Taking the derivative of r = 1 + cos(θ) with respect to θ, we have
dr/dθ = -sin(θ). Setting this equal to zero, we get -sin(θ) = 0, which implies that sin(θ) = 0. The values of θ that satisfy this condition are θ = 0, π, 2π, etc. However, we are given that 0 ≤ θ < 2, so the only valid solution is θ = π. Substituting this back into the equation r = 1 + cos(θ), we find r = 2.
Next, let's consider the vertical tangent lines. A tangent line is vertical when the derivative of θ with respect to r is equal to zero. Taking the derivative of r = 1 + cos(θ) with respect to r, we have
dθ/dr = -sin(θ)/(1 + cos(θ)). Setting this equal to zero, we have -sin(θ) = 0. The values of θ that satisfy this condition are θ = π/2, 3π/2, 5π/2, etc. Again, considering the given range for θ, the valid solution is θ = π/2. Substituting this back into the equation r = 1 + cos(θ), we find r = 0.
Therefore, the points on the curve where the tangent line is horizontal or vertical are (0, π/2) and (2, 3π/2).
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Let f(x) = 2x² - 3x and g(x) = 5x - 1.
Find g[f(2)].
g[f(2)] =
Answer:
Step-by-step explanation:
To find g[f(2)], we need to evaluate the composite function g[f(2)] by first finding f(2) and then substituting the result into g(x).
Let's start by finding f(2):
f(x) = 2x² - 3x
f(2) = 2(2)² - 3(2)
= 2(4) - 6
= 8 - 6
= 2
Now that we have the value of f(2) as 2, we can substitute it into g(x):
g(x) = 5x - 1
g[f(2)] = g(2)
= 5(2) - 1
= 10 - 1
= 9
Therefore, g[f(2)] is equal to 9.
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Hey, my sis can't register rn so I'm asking this question for her. She's learning PEMDAS and she was given 4 numbers, 18, 2, 3, and 4, and 3 symbols, ×, +, and ÷. She has to use PEMDAS and get to the number 12 at the end of it.
Answer:
(18 + 2 * 3) / 3 + 4
Step-by-step explanation:
If you can’t use parentheses,
here is another solution:
18 + 3*4 / 3 + 2
10. A line has equation y=3kx−2k and a curve has equation y=x 2
−kx+2, where k is a constant. a) Find the set of values of k for which the line and curve meet at two distinet points. b) For cach of two particular values of k, the line is a tangent to the curve. Show that these two tangents meet on the x-axis. 11. The equation x 2
+px+q=0, where p and q are constants, has roots −3 and 5 . a) Find the values of p and q. b) Using these values of p and q, find the value of the constant r for which the equation x 2
+px+q+r=0 has equal roots. 12. A curve has equation y=x 2
−4x+4 and a line has the equation y=mx, where m is a constant. a) For the case where m=1, the curve and the line intersect at the point A and B. b) Find the coordinates of the mid-point of AB. c) Find the non-zero value of m for which the line is the tangent to the curve, and find the coordinates of the point where the tangent touches the curve. Answer: 1. ( 2
1
,0) 9. a) 25−(x−5) 2
2. a) (3x− 2
5
) 2
− 4
25
b) (5,25) b) − 3
1
3
10. a) k>1,k<− 2
1
a) The set of values of k for which the line and curve meet at two distinct points is k < -2/5 or k > 2.
To find the set of values of k for which the line and curve meet at two distinct points, we need to solve the equation:
x^2 - kx + 2 = 3kx - 2k
Rearranging, we get:
x^2 - (3k + k)x + 2k + 2 = 0
For the line and curve to meet at two distinct points, this equation must have two distinct real roots. This means that the discriminant of the quadratic equation must be greater than zero:
(3k + k)^2 - 4(2k + 2) > 0
Simplifying, we get:
5k^2 - 8k - 8 > 0
Using the quadratic formula, we can find the roots of this inequality:
\(k < (-(-8) - \sqrt{((-8)^2 - 4(5)(-8)))} / (2(5)) = -2/5\\ or\\ k > (-(-8)) + \sqrt{((-8)^2 - 4(5)(-8)))} / (2(5)) = 2\)
Therefore, the set of values of k for which the line and curve meet at two distinct points is k < -2/5 or k > 2.
b) To find the two values of k for which the line is a tangent to the curve, we need to find the values of k for which the line is parallel to the tangent to the curve at the point of intersection. For m to be the slope of the tangent at the point of intersection, we need to have:
2x - 4 = m
3k = m
Substituting the first equation into the second, we get:
3k = 2x - 4
Solving for x, we get:
x = (3/2)k + (2/3)
Substituting this value of x into the equation of the curve, we get:
y = ((3/2)k + (2/3))^2 - k((3/2)k + (2/3)) + 2
Simplifying, we get:
y = (9/4)k^2 + (8/9) - (5/3)k
For this equation to have a double root, the discriminant must be zero:
(-5/3)^2 - 4(9/4)(8/9) = 0
Simplifying, we get:
25/9 - 8/3 = 0
Therefore, the constant term is 8/3. Solving for k, we get:
(9/4)k^2 - (5/3)k + 8/3 = 0
Using the quadratic formula, we get:
\(k = (-(-5/3) ± \sqrt{((-5/3)^2 - 4(9/4)(8/3)))} / (2(9/4)) = -1/3 \\or \\k= 4/3\)
Therefore, the two values of k for which the line is a tangent to the curve are k = -1/3 and k = 4/3. To show that the two tangents meet on the x-axis, we can find the x-coordinate of the point of intersection:
For k = -1/3, the x-coordinate is x = (3/2)(-1/3) + (2/3) = 1
For k = 4/3, the x-coordinate is x = (3/2)(4/3) + (2/3) = 3
Therefore, the two tangents meet on the x-axis at x = 2.
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Find the value of x.
A 15.0
B. 13.5
C. 9.0
D. 7.5
Answer:
D
Step-by-step explanation:
The segment inside the triangle is an angle bisector and divides the opposite side into segments that are proportional to the other 2 sides, that is
\(\frac{x}{7}\) = \(\frac{15}{14}\) ( cross- multiply )
14x = 105 ( divide both sides by 14 )
x = 7.5 → D
Which property would allow you to use mental computation to simplify the problem 27 + 15 + 3 + 5?
commutative property of addition
distributive property
additive identity
additive inverse
Answer:
commutative property of addition
Step-by-step explanation:
Answer:
commutative property of addition
Step-by-step explanation:
The picturegram shows information about CDs sold in a shop.
1 . How manny CDs were sold on Wednesday | Key = 3 |
2. How manny more CDs were sold on Thursday than Wednesday?
**If you know the answer let me know!**
Answer:
i believe number 1 is 18 and number 2 is 9.
Step-by-step explanation:
if one full circle represents 6 CDs then on wednesday 18 Cds were sold because 6+6+6=18 and on thursday they sold 9 more Cds than on wednesday because they sold 6+6+6+6+3 which equals 9.
please help me!!
20 points
Answer:
182 \(yd^{2}\)
Step-by-step explanation:
Area of a parallelogram: bh
bh
= 13(14)
= 182 sq. yds
A
B
What is the distance between Points C and
D?
-4 units
4 units
6 units
2 units
Math homework PLS HELP What is the distance between points C and D?
If A+B= 90°, then
\( \frac{\tan A \tan B + \tan A \cot B}{ \sin A \sec B } - \frac{{ \sin}^{2} B}{ \cos^{2} A} \: \text{is equal to }\)
Answer:
A + B = 90° => A = 90 - B
So Tan A = Cot (90 - A) = Cot B
So Tan B = Cot (90 - B) = Cot A
SecB = Cosec (90 -B) = Cosec A
CosA = Sin (90 -A) = Sin B
The value of the angle given regarding the tangent will be tan²A.
How to explain the angle?From the information, TanA TanB + TanACot B/SinA SecB - Sin²B/Cos²A
= TanA CotA + TanATanA/AinA CosecA - Sin²B/Sin²B
= 1 + Tan²A - 1
= Tan²A
In conclusion, the correct option is Tan²A.
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which set represents a Pythagorean triple
A. 1,2,3
B.9,12,16
C. 30,40,50
D.38,44,49
The set that represents a Pythagorean Triple is given as follows:
C. 30,40,50
What is the Pythagorean Theorem?The Pythagorean Theorem states that for a right triangle, the length of the hypotenuse squared is equals to the sum of the squared lengths of the sides of the triangle.
For item c, we have that:
30² + 40² = 50²
900 + 1600 = 2500
2500 = 2500.
Which is a true statement, hence item c represents the Pythagorean Triple in the context of this problem.
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-5x + 3y = 11 and x - 2y = 2 How would I solve by substitution?
Answer:
x=-4,y=-3
Step-by-step explanation:
A triangle has side lengths of (3a – 7),
(a + 14), and (2a + 1). What simplified
expression represents the perimeter of the
triangle?
Answer:
6a+8
Step-by-step explanation:
To find the perimeter of a shape add all the sides up
To do this just add everything together
this means we have
3a-7+a+14+2a+1
Add like terms and get
6a+8
what is the scale factor of 3x + 2, 3y -1?
(will give brainliest)
21 Line is mapped onto line m by a dilation centered at the origin with a scale factor of 2. The equation of line is 3x − y = 4. Determine and state an.
Two trains are traveling at constant speeds on different tracks. Trains Which train is traveling faster? Explain your reasoning. Train is traveling faster, because it will go 100 meters in only seconds, compared to Train
Answer:
Train A is traveling faster
Step-by-step explanation:
Train A has more distance than train B.
Train A is traveling faster, because it will go 100 meters in only seconds, compared to Train B.
The angle of elevation from a viewer to the top of a flagpole is 50°. The viewer is 80 ft away and the viewer's eyes are 5.5 ft from the ground. How high
is the pole to the nearest tenth of a foot?
Answer:
Step-by-step explanation:
100.2 the ration
-9 + ( -16) =
( -4) – (-6) + (-5) – (8) =
( 3) – (-9) + (-3) – (-2) =
- ( 5) – (-7) + (-8) =
- ( -8) – (-9) + (-4) – (-1) =
( -5) – (-11) + (10) – (8) =
( -1) – (-2) + (-3) – (-6) =
( -4) – ( 4) – (6) + (-5) – (-8) =
(-6) + (-5) – (-8) =
Answer:
-9 + ( -16) = -25
( -4) – (-6) + (-5) – (8) =-11
( 3) – (-9) + (-3) – (-2) =11
- ( 5) – (-7) + (-8) =-6
- ( -8) – (-9) + (-4) – (-1) =14
( -5) – (-11) + (10) – (8) =8
( -1) – (-2) + (-3) – (-6) =4
( -4) – ( 4) – (6) + (-5) – (-8) =-11
(-6) + (-5) – (-8) =-3
Step-by-step explanation:
Hope it helps
Answers:
-9 + ( -16) = -25
( -4) – (-6) + (-5) – (8) = -11
( 3) – (-9) + (-3) – (-2) = 11
- ( 5) – (-7) + (-8) = -6
- ( -8) – (-9) + (-4) – (-1) = 14
( -5) – (-11) + (10) – (8) = 8
( -1) – (-2) + (-3) – (-6) = 4
( -4) – ( 4) – (6) + (-5) – (-8) = -11
(-6) + (-5) – (-8) = -3
how to find the turning point of a polynomial function?
the turning point(s) of a polynomial function by using below steps
To find the turning point of a polynomial function, follow these steps:
1. Determine the degree of the polynomial. The turning point will occur in a polynomial of odd degree (1, 3, 5, etc.) or at most one turning point in a polynomial of even degree (2, 4, 6, etc.).
2. Write the polynomial function in the form f(x) = axⁿ + bxⁿ⁻¹ + ... + cx + d, where n represents the degree of the polynomial and a, b, c, d, etc., are coefficients.
3. Find the derivative of the polynomial function, f'(x), by differentiating each term of the function with respect to x. This will give you a new function that represents the slope of the original polynomial function at any given point.
4. Set f'(x) equal to zero and solve for x to find the x-coordinate(s) of the turning point(s). These are the values where the slope of the polynomial function is zero, indicating a potential turning point.
5. Substitute the x-coordinate(s) obtained in step 4 into the original polynomial function, f(x), to find the corresponding y-coordinate(s) of the turning point(s).
6. The turning point(s) of the polynomial function is given by the coordinates (x, y), where x is the x-coordinate(s) found in step 4 and y is the y-coordinate(s) found in step 5.
By following these steps, you can find the turning point(s) of a polynomial function.
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The blood test for determining coagulation activity defects is called the Prothrombin Time (PT) test.
The blood test for determining coagulation activity defects is called the Prothrombin Time (PT) test. This test measures the time it takes for blood to clot and is used to assess the functioning of the clotting factors in the blood. It is commonly used to evaluate the extrinsic pathway of the coagulation cascade, which involves factors outside of the blood vessels.
The PT test is an important diagnostic tool in hematology and is used to diagnose or monitor conditions that affect blood clotting, such as bleeding disorders or the effectiveness of anticoagulant medications. By measuring the PT, healthcare professionals can determine if there are any abnormalities in the coagulation process and make appropriate treatment decisions.
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We have a one sample test for the population mean. The significance level is a fixed value. Suppose we increase the sample size. Assume the true mean equals the null mean.(a) The t critical value moves closer to zero.(b) The size of the rejection region decreases(c) The probability of a Type I error decreases(d) The probability of a Type I error increases
the question is that as we increase the sample size in a one sample test for the population mean, the size of the rejection region decreases. This means that the probability of rejecting the null hypothesis when it is actually true, also known as a Type I error, decreases.
The t critical value is a constant value determined by the sample size and the significance level. It is the point at which we decide to reject or fail to reject the null hypothesis. As we increase the sample size, the t critical value moves closer to zero, but this does not have a direct impact on the probability of a Type I error.
The size of the rejection region, on the other hand, is determined by the t critical value and the variability of the sample. As the sample size increases, the variability decreases, leading to a smaller rejection region. This means that it is less likely to reject the null hypothesis when it is actually true, resulting in a lower probability of a Type I error.
Therefore, we can conclude that increasing the sample size in a one sample test for the population mean decreases the size of the rejection region and the probability of a Type I error.
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You are given \( f=10 \) meissurements: \( 3,5,4,6,10,5,6,9,2,13 \). (D) Calculate \( x_{2} \) \( \frac{2}{x}= \) (b) Firud m. \( m= \) (c) Find the mode. (If these is more than one mode, enter your answer
Two values have frequency 2, so both are modes. They are 5 and 6. Therefore, the mode is 5 and 6.
Given measure: 3, 5, 4, 6, 10, 5, 6, 9, 2, 13.(D) To calculate \(x_2\), first we need to sort the data in ascending order: 2, 3, 4, 5, 5, 6, 6, 9, 10, 13
Now, we need to find the median, which is the middle value of the data. Since the data has even number of values, we will calculate the mean of middle two values, that is:(5+6)/2 = 5.5 Therefore, \(x_2 = 5.5\).\( \frac{2}{x}= \) To find the value of x, we will first cross-multiply and then take the reciprocal of both sides:\[\frac{2}{x} = y \Rightarrow 2 = xy \Rightarrow x = \frac{2}{y}\] Therefore, \( \frac{2}{x}= \frac{2}{y}\).
(b) To calculate Fried m, we will use the formula: \[f_m = L + \frac{(n/2 - F)}{f} \times c\]where L is the lower limit of the modal class, F is the cumulative frequency of the class preceding the modal class, f is the frequency of the modal class, c is the class interval, and n is the total number of values.
First, we will calculate the class interval:c = (upper limit of class - lower limit of class) = (7-6) = 1 Next, we will construct a frequency table to find the modal class:| Class Interval | Frequency ||-------------------|------------|| 2-3 | 1 || 3-4 | 1 || 4-5 | 1 || 5-6 | 2 || 6-7 | 2 || 7-8 | 1 || 9-10 | 1 || 10-11 | 1 || 13-14 | 1 |The modal class is the class with highest frequency.
Here, two classes have frequency 2, so both are modes. They are 5-6 and 6-7.
Therefore, L = 5, F = 2, f = 2, n = 10, and c = 1. Substituting the values, we get:\[f_m = L + \frac{(n/2 - F)}{f} \times c = 5 + \frac{(10/2 - 2)}{2} \times 1 = 7\] Therefore, Fried m = 7.
(c) To find the mode, we look for the value(s) with highest frequency. Here, two values have frequency 2, so both are modes. They are 5 and 6. Therefore, the mode is 5 and 6.
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c. what is the probability of women who did not breast fed and had breast cancer? 1. .04 2. .20 3. .16 4. .80 d. what do you call this probability? 1. joint probability 2. conditional probability 3. additional rule 4. marginal probability
The probability of women who did not breastfeed and had breast cancer is not provided in the given options. However, the answer would depend on various factors and cannot be determined solely based on the information given.
It is important to note that breastfeeding is known to have a protective effect against breast cancer, but it does not guarantee that a woman will not develop breast cancer if she does not breastfeed.
Additionally, there are several other risk factors for breast cancer, such as family history, genetics, hormonal factors, and lifestyle choices, which can influence the probability. Therefore, without specific data or context, it is not possible to provide an accurate probability.
Conditional probability is the term used to describe the probability of an event occurring given that another event has already occurred. In this case, if we had information on the probability of breast cancer given that a woman did not breastfeed, we could calculate the conditional probability.
However, since the given information does not include such data, we cannot determine the probability or classify it as a specific type of probability (e.g., joint, conditional, additional rule, or marginal probability).
Learn more about Probability:
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at which angles are the x and y coordinate on the unit circle exactly the same? give your answer in radians and degrees
Answer:
In radian :
\(\frac{\pi }{4} \ \text{and}\ \frac{5\pi }{4}\)
In degree :
45° and 225°
Step-by-step explanation:
the x and y coordinate on the unit circle exactly the same at the angles :
In radian :
\(\frac{\pi }{4} \ \text{and}\ \frac{5\pi }{4}\)
In degree :
45° and 225°
At those angles :
\(\left( x,y\right) =\left( \frac{\sqrt{2} }{2} ,\frac{\sqrt{2} }{2} \right) \ or\ \left( x,y\right) =\left( -\frac{\sqrt{2} }{2} ,-\frac{\sqrt{2} }{2} \right)\)