Answer:
fafafsghdjdbdbdgsysvdve. wwmsk
Graphs that represent functions have one input corresponding to one output. Examples include straight lines, parabolas, and sine waves.
Graphs that represent functions are those in which every input has exactly one output. In other words, there can only be one value of y for each value of x. For example, a straight line, a parabola, or a sine wave are graphs that represent functions.
On the other hand, graphs that do not represent functions may have one input value mapping to multiple output values or no output values at all. Examples of such graphs include circles, ellipses, or a graph with one vertical line intersecting it at multiple points.
It's important to note that in a function, the vertical line test can be used to determine if a graph represents a function. If any vertical line intersects the graph at more than one point, then the graph does not represent a function.
#SPJ2
Answer:
We cannot calculate the probability that a randomly selected passenger weighs more than 200 pounds
Step-by-step explanation:
We cannot calculate the probability that a randomly selected passenger weighs more than 200 pounds because we do not know the number of possible outcomes, the events , sample space or the sample size. Probability is calculated with frequency or occurrences or how much certainty there is.It is a number between 0 and 1. 1 indicates certainty and 0 indicates impossibility. Without a range or frequency how can we depict the possibility or impossibility of an occurrence of 200 pounds.
You cannot calculate the probability that a randomly selected passenger weighs more than 200 pounds without sufficient data on the weight distribution of the population. Weight can widely vary due to individual factors, making it hard to have a definitive measurement. Accurate data and appropriate statistical methods are necessary.
The process of calculating the probability that a randomly selected passenger weighs more than 200 pounds would be seemingly simple deductive reasoning. However, it's impossible without access to sufficient data that provides information about the population's weight distribution. Since people's weights are variable and oftentimes private information, it would not be straightforward to obtain accurate and representative data.
For instance, while we can calculate the probability of drawing a certain card from a deck because we know the total number of cards and the number of each type of card, determining the likelihood of a randomly chosen passenger weighs over 200 pounds requires knowledge of the weight distribution of all potential passengers.
Moreover, weight can vary significantly among individuals due to factors like age, gender, health status, and so on. This makes it a continuous variable, meaning it's also affected by dimensions like decimal form and scientific notation when measuring. We'd need accurate data and appropriate statistical methodologies to consider all possible weight ranges and their frequencies for a reliable calculation of such probability.
#SPJ3
Answer:
yes they are equal
Step-by-step explanation:
5(p+4) when it is distributed it equals 5p+20 which equals to 5p+20
The expressions 5(p + 4) and 5p + 20 are equivalent at all times due to the distributive property of multiplication over addition in mathematics. These two expressions are the same, and their equivalence doesn't depend on the value of 'p'.
The expressions 5(p + 4) and 5p + 20 are equivalent at all times. This is due to the distributive property of multiplication over addition in mathematics.
This means that when you have an expression like 5(p + 4), you multiply each quantity inside the parentheses by the number outside, which gives you 5p + 20.
So these two expressions are actually the same and are equivalent no matter what the value of p is.
#SPJ2
please help me
Here, we are required to find the value of the other angles in the diagram attached to this answer sheet.
The value of the angles are;
The understanding of the triangle theorem and line theorems is very important to resolve this.
m∠1 + m∠2 = 180°(Theorem:sum of angles on a straight line =180⁰)
m∠1 + 98 = 180
m∠1 = 180 - 98
m∠2 + m∠3 + m∠7 = 180° (Theorem=sum ofangles in a triangle = 180⁰)
98 + 23 + m∠7 = 180
m∠7 + 121 = 180
m∠7 = 180 - 121
m∠4 = m∠7 (Theorem = alternate angles are equal)
m∠6 + m∠7 + m∠8 = 180° (Theorem = sum of angles on a straight line = 180)
m∠6 + 59 + 70 = 180
m∠6 + 129 = 180
m∠6 = 180 - 129
m∠4 + m∠8 + m∠9 = 180° (Theorem = sum of angles in a triangle = 180)
59 + 70 + m∠9 = 180
m∠9 + 129 = 180
m∠9 = 180 - 129
m∠4 + m∠5 = 180° (Theorem = sum of angles on a straight line = 180)
m∠5 + 59 = 180
m∠5 = 180 - 59
m∠10 + m∠9 = 180° (Theorem = sum of angles on a straight line = 180⁰)
m∠10 + 51 = 180
m∠10 = 180 - 51
Read more:
Answer:
The answer is below
Step-by-step explanation:
The complete question is given in the image attached below
m∠1 + m∠2 = 180° (sum of angles on a straight line)
m∠1 + 98 = 180
m∠1 = 180 - 98
m∠1 = 82°
m∠2 + m∠3 + m∠7 = 180° (sum of angles in a triangle)
98 + 23 + m∠7 = 180
m∠7 + 121 = 180
m∠7 = 180 - 121
m∠7 = 59°
m∠4 = m∠7 (alternate angles)
m∠4 = 59°
m∠6 + m∠7 + m∠8 = 180° (sum of angles on a straight line)
m∠6 + 59 + 70 = 180
m∠6 + 129 = 180
m∠6 = 180 - 129
m∠6 = 51°
m∠4 + m∠8 + m∠9 = 180° (sum of angles in a triangle)
59 + 70 + m∠9 = 180
m∠9 + 129 = 180
m∠9 = 180 - 129
m∠9 = 51°
m∠4 + m∠5 = 180° (sum of angles on a straight line)
m∠5 + 59 = 180
m∠5 = 180 - 59
m∠5 = 121°
m∠10 + m∠9 = 180° (sum of angles on a straight line)
m∠10 + 51 = 180
m∠10 = 180 - 51
m∠10 = 129°
Step-by-step explanation:
Confidence interval = 90% = 0.90.
1 - 0.90 = 0.10
Mean of 15 batch:
724+718+776+760+745+759+795+756+742+740+761+749+739+747+742 = 11253/15
= 750.2
Mean of second batch:
735+775+729+755+783+760+738+780
= 6055/8
= 756.875
Sd = 20
Z-alpha/2 = 1.64
= (750.2-756.875)-1.64*√20²/15 + 20²/8
= -21.0348
(750.2-756.875)+1.64*√20²/15 + 20²/8
= 7.68
2. The upper limit of the interval is less than 10 so in conclusion the difference in mean batch viscosity is about 10 or less than 10