B. Calculate the distance Kinsey jogs if
she jogs'mile each day for 9 days.

Answer: 0.06 ft long

Step-by-step explanation:

Porcelli's board = 2% feet long; to convert 2% into fraction, divide by 100= 2/100 = 0.02 ft

Smith's board, from the question is 3 times porcelli's board = 3 x 0.02= 0.06 ft

I hope this helps.

Answer:

graph a

Step-by-step explanation:

The required shortest length to go from A to G is given as 8. Option C is correct.

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

Here,
The shortest distance from the figure can be given as,
= AE + EG
= 1 + 7
= 8

Thus, the required shortest length to go from A to G is given as 8. Option C is correct.

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Answer:

Her z-score is 2.03.

Step-by-step explanation:

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Michelle’s height is 1.758 meters. What is her z-score?

Michelle's is a woman.

The average height of women is [tex]\mu = 1.618[/tex] and the standard deviation is [tex]\sigma = 0.069[/tex]

This is Z when X = 1.758. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{1.758 - 1.618}{0.069}[/tex]

[tex]Z = 2.03[/tex]

Her z-score is 2.03.

Michelle's z-score is approximately 2.03, which means her height is 2.03 standard deviations above the mean height for American women.

To calculate Michelle’s z-score for her height, we use the formula for the z-score:

Z = (X - μ) / σ

Where Z is the z-score, X is the value (Michelle's height), μ is the mean, and σ is the standard deviation. Since Michelle is a woman, we will use the mean and standard deviation for women. The mean height for American women, μ, is 1.618 meters, and the standard deviation, σ, is 0.069 meters.

Plugging in the values we get:

Z = (1.758 - 1.618) / 0.069

Z = 0.14 / 0.069

Z ≈ 2.029

Michelle's height is approximately 2.03 standard deviations above the mean height for American women.

Correct Answer: 98.78

A. 98. 78

A) Part 2:  ​No, because only a small percentage of women are not allowed to join this branch of the military because of their height.

B. 57.8 ,    68.3

Approximately 98.8% of women meet the current military height requirement of 58-80 inches. If the military change to exclude the shortest 1% and tallest 2%, the height requirements change to approximately 58.11 to 68.38 inches.

To answer part (a) of your question, we must utilize the notion of a Z-score. A Z-score standardizes or normalizes a data point in terms of how many standard deviations away it is from the mean. A standard deviation of 2.4 and a mean of 63.4 inches would be the reference points. Using the formula ((X - Mean) / Std Deviation) we calculate the Z-score which gives us the percentage.

For 58 inches, we get a Z-score of -2.25, and using a Z-score table, this corresponds approximately to 1.2%. For 80 inches, the Z-score is 6.92, which is practically at the extreme end of the curve, so we can take this as 100%. Hence, almost 98.8% of women fall within the height requirements of 58 inches to 80 inches.

Coming to part (b) we need to find the height that represents the shortest and tallest 1% and 2% respectively. Using the Z-score table, the Z-score for 1% is -2.33 and for 2%, it's 2.05. Hence, the height cut-off for the shortest 1% will be Mean - 2.33*Std Deviation = 58.11 inches approx, and the tallest 2% will be Mean + 2.05*Std Deviation = 68.38 inches approx. This means, if the military changes their requirements to exclude the shortest 1% and tallest 2%, their new height requirements would approximately be from 58.11 inches to 68.38 inches.

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Answer:

5/10 & 1/2: anything that simplifies to 1/2 or .5

Final answer:

The expression 7/10-2/10 simplifies to 5/10, which can be further reduced to 1/2 by dividing both the numerator and denominator by their greatest common divisor, 5.

Explanation:

The expression 7/10-2/10 involves the subtraction of two fractions with the same denominator. When subtracting fractions with the same denominator, you simply subtract the numerators and keep the denominator the same. Thus, the calculation would be:

Numerator: 7 - 2 = 5

Denominator: 10 (remains the same)

Therefore, the expression 7/10-2/10 is equivalent to 5/10. This fraction can be further simplified by dividing the numerator and the denominator by their greatest common divisor, which in this case is 5. This gives us:

(5 ÷ 5) / (10 ÷ 5) = 1/2

So, 7/10-2/10 simplifies to 1/2, which is the equivalent fraction.

The probability of guessing exactly 18 out of 27 questions correctly is calculated using the binomial probability formula. Use the formula: P(x=k) = (n choose k) * (p^k) * (1-p)^(n-k), where n is the number of trials (27), p is the probability of success (1/5), and k is the number of successes (18). The answer will be a very small number.

The problem involves the use of binomial probability because we have a fixed number of trials (27 questions), each of which is independent and has only two outcomes (correct or incorrect) with constant probabilities. In this case, the probability of a success (choosing the correct answer) is 1/5 and the probability of a failure (choosing an incorrect answer) is 4/5.

We wish to find P(x=18), the probability of having exactly 18 successes. Using the formula for the probability of a binomial distribution:

P(x=k) = (n choose k) * [(p)^(k)] * [(1-p)^(n-k)]

Where:
- (n choose k) = n! / (k!(n-k)!)
- p is the probability of success
- k is the number of successes
- n is the number of trials.

Substituting the values:

P(x=18) = (27 choose 18) * ((1/5)^18) * ((4/5)^(27-18))

You can use a calculator to solve for the answer. Please note that the probability would be very small, as expected, because guessing 18 questions correctly out of 27 is not a very likely occurrence when each question has 5 choices.

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Answer:

Step-by-step explanation:

Derive the validity of universal form of part(a) of the elimination rule from the validity of universal instantiation and the valid argument called elimination in Section 2.3.

P(x)∨Q(x)

~Q(x)

∵ P(x)

Universal Instantiation has the following argument form

∀ x ∈ D, P (x)

P(a) for a particular a∈D

Universal Elimination Rule:

∀x, P(x)

∵~ P(a)

Here is a particular value.

P(a) For a particular  a∈D

Since the universal elimination is same as universal instantiation.

Therefore, Universal elimination is valid when universal instantiation and elimination rule are  valid

Answer:

yes :- P ( Ei ∩ Ej ) = 0

Step-by-step explanation:

These are mutually exclusive events

that is, a customer can only go to a gas station to fill his car with gas par time

P(E1) = 1/4

P(E2) = 1/4

P(E3) = 1/4

P(E4) = 1/4

summation of the four probabilities give 1

Final answer:

The outcomes E1, E2, E3, and E4 are mutually exclusive due to the physical limitation of the car's gas tank being already full, which prevents multiple fill-ups in immediate succession.

Explanation:

The concept of mutually exclusive outcomes in the context of choosing a gas station to fill up a car is referring to the idea that a customer cannot fill their car with gas at one station and then immediately go and fill it at another station. This is because once the car's tank is full, there is no capacity to add more fuel until it has been used.

Therefore, the outcomes E1, E2, E3, and E4, if defined as filling up at station 1, station 2, station 3, and station 4 respectively, are indeed mutually exclusive. When a customer chooses one station, the other stations are no longer an option for fueling during that particular instance, as the car's tank can only be filled once until some gas is used.

In economics, understanding market dynamics like price elasticity and strategies employed by gas stations in an oligopoly to maintain profits can influence consumer behavior and competition. These concepts help in analyzing why a customer might not be able to take advantage of another gas station's prices immediately after filling their tank.

Answer:

D.) 0.7471

Step-by-step explanation:

Mean=μ=100

Standard deviation=σ=15

We know that IQ score are normally distributed with mean 100 and standard deviation 15.

n=50

According to central limit theorem, if the population is normally distributed with mean μ and standard deviation σ then the distribution of sample taken from this population will be normally distributed with mean μxbar and standard deviation σxbar=σ/√n.

Mean of sampling distribution=μxbar=μ=100.

Standard deviation of sampling distribution=σxbar=σ/√n=15/√50=2.1213.

We are interested in finding the probability of sample mean between 98 and 103.

P(98<xbar<103)=?

Z-score associated with 98

Z-score=(xbar-μxbar)/σxbar

Z-score=(98-100)/2.1213

Z-score=-2/2.1213

Z-score=-0.94

Z-score associated with 103

Z-score=(xbar-μxbar)/σxbar

Z-score=(103-100)/2.1213

Z-score=3/2.1213

Z-score=1.41

P(98<xbar<103)=P(-0.94<Z<1.41)

P(98<xbar<103)=P(-0.94<Z<0)+P(0<Z<1.41)

P(98<xbar<103)=0.3264+0.4207

P(98<xbar<103)=0.7471

Thus, the probability that the sample mean is between 98 and 103 is 0.7471.

Answer:

For the velocity vector, we have

V(t) = dR

dt = (1, 2t, 3t

2

).

For the acceleration vector, we get

A =

dV

dt = (0, 2, 6t).

The velocity vector at t = 1 is

V(1) = (1, 2, 3).

The speed at t = 1 is

kV(1)k =

p

1

2 + 22 + 32 =

√

14.

Answer:

1/3

Step-by-step explanation:

60-90 is 30 numbers, right? So it is 30/90, or 1/3

Answer:

a) Attached

b) P=0.60

c) P=0.80

d) The expected flight time is E(t)=122.5

Step-by-step explanation:

The distribution is uniform between 1 hour and 50 minutes (110 min) and 135 min.

The height of the probability function will be:

[tex]h=\frac{1}{Max-Min}=\frac{1}{135-110} =\frac{1}{25}[/tex]

Then the probability distribution can be defined as:

[tex]f(t)=\frac{1}{25}=0.04 \,\,\,\,\\\\t\in[110,135][/tex]

b) No more than 5 minutes late means the flight time is 125 or less.

The probability of having a flight time of 125 or less is P=0.60:

[tex]F(T<t)=0.04(t-min)\\\\F(T<125)=0.04*(125-110)=0.04*15=0.60[/tex]

c) More than 10 minutes late means 130 minutes or more

The probability of having a flight time of 130 or more is P=0.80:

[tex]F(T>t)=1-0.04(t-110)\\\\F(T>130)=1-0.04*(130-110)=1-0.04*20=1-0.8=0.2[/tex]

d) The expected flight time is E(t)=122.5

[tex]E(t)=\frac{1}{2}(max+min)= \frac{1}{2}(135+110)=\frac{1}{2}*245=122.5[/tex]

Answer:

First part

[tex] P(X< 3.4-0.6*3.1) = P(X<1.54)[/tex]

And for this case we can use the z score formula given by:

[tex] z = \frac{x- \mu}{\sigma}[/tex]

And using this formula we got:

[tex] P(X<1.54) = P(Z<\frac{1.54 -3.4}{3.1})= P(Z<-0.6)[/tex]

And we can use the normal standard table or excel and we got:

[tex]P(Z<-0.6) = 0.274[/tex]

Second part

For the other part of the question we want to find the following probability:

[tex] P(-1.715 <X< 7.12)[/tex]

And using the score we got:

[tex] P(-1.715 <X< 7.12)=P(\frac{-1.715-3.4}{3.1} < Z< \frac{7.15-3.4}{3.1}) = P(-1.65< Z< 1.210)[/tex]

And we can find this probability with this difference:

[tex]P(-1.65< Z< 1.210)=P(Z<1.210)-P(z<-1.65) = 0.887-0.049=0.837[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the data of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(3.4,3.1)[/tex]  

Where [tex]\mu=3.4[/tex] and [tex]\sigma=3.1[/tex]

First part

And for this case we want this probability:

[tex] P(X< 3.4-0.6*3.1) = P(X<1.54)[/tex]

And for this case we can use the z score formula given by:

[tex] z = \frac{x- \mu}{\sigma}[/tex]

And using this formula we got:

[tex] P(X<1.54) = P(Z<\frac{1.54 -3.4}{3.1})= P(Z<-0.6)[/tex]

And we can use the normal standard table or excel and we got:

[tex]P(Z<-0.6) = 0.274[/tex]

Second part

For the other part of the question we want to find the following probability:

[tex] P(-1.715 <X< 7.12)[/tex]

And using the score we got:

[tex] P(-1.715 <X< 7.12)=P(\frac{-1.715-3.4}{3.1} < Z< \frac{7.15-3.4}{3.1}) = P(-1.65< Z< 1.210)[/tex]

And we can find this probability with this difference:

[tex]P(-1.65< Z< 1.210)=P(Z<1.210)-P(z<-1.65) = 0.887-0.049=0.837[/tex]

Answer:

P ( X > 5) = 0.0374

Step-by-step explanation:

Given:

n = 12

p = 0.23

Using Binomial distribution formula,

X ~ Binomial ( n = 12, p = 0.23)

[tex]=\frac{n!}{(n-x)! x!}. p^{x} q^{n-x}[/tex]

Substitute for n = 12, p = 0.23, q = 1-0.23 for  x = 6,7,8,9,10,11 and 12

P (X > 5)  = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) + P(X=11) + P(X=12)

P ( X > 5 ) = 0.0285 + 0.007299 + 0.00136 + 0.000181 + 0.0000162 + 1E-6 + 1E-6

P ( X > 5) = 0.0374

Answer:

[tex] E(X) =sum_{i=1}^n X_i P(X_i)[/tex]

And we know that each time that we win we recieve $1 and each time we loss we need to pay $1, so then the expected value would be given by:

[tex] E(X) = 1*\frac{4}{9} -1 \frac{5}{9} = -\frac{1}{9} =-0.11[/tex]

[tex] E(X^2) =sum_{i=1}^n X^2_i P(X_i)[/tex]

[tex] E(X^2) = 1^2*\frac{4}{9}+ (-1)^2 \frac{5}{9} = 1[/tex]

And the variance is defined as:

[tex] Var(X) = E(X^2) -[E(X)]^2 = 1 -[-\frac{1}{9}]^2 = \frac{80}{81}[/tex]

Step-by-step explanation:

For this case we know that we can win if we select 2 balls of the same color, so we can find the probability of win like this:

[tex] p = \frac{Possible}{ Total}= \frac{2C1 * (5C2)}{10C2}= \frac{2*10}{45}= \frac{4}{9}[/tex]

So then the probability of no win would be given by the complement:

[tex] q = 1-p = 1- \frac{4}{9}= \frac{5}{9}[/tex]

We can define the random variable X who represent the amount of money that we can win.

And we can use the definition of expected value given by:

[tex] E(X) =sum_{i=1}^n X_i P(X_i)[/tex]

And we know that each time that we win we recieve $1 and each time we loss we need to pay $1, so then the expected value would be given by:

[tex] E(X) = 1*\frac{4}{9} -1 \frac{5}{9} = -\frac{1}{9} =-0.11[/tex]

We can calculate the second monet like this:

[tex] E(X^2) =sum_{i=1}^n X^2_i P(X_i)[/tex]

[tex] E(X^2) = 1^2*\frac{4}{9}+ (-1)^2 \frac{5}{9} = 1[/tex]

And the variance is defined as:

[tex] Var(X) = E(X^2) -[E(X)]^2 = 1 -[-\frac{1}{9}]^2 = \frac{80}{81}[/tex]

A part of the question is missing and it says;

b) Calculate the variance of the amount you win.

Answer:

A) Expected value of Pay out;

(E(Y)) = - 0.11

B) Variance of the amount won; Var(Y) = 0.9879

Step-by-step explanation:

A) From the question, we have;

X ∈ {0,1,2} and by the nature of the question, X has a hypergeometric distribution as;

P(X =i) = [(5,i) (5, 2 - i)] / (10,2)

Furthermore, when we consider the random variable Y that marks the amount of money that we win, we'll get a function of X as;

Y = 1X [x∈{0,2}] - [1X(x=1)]

If we now use the linearity of expectation, we'll get;

E(Y) = E [X(x∈{0,2}) ] - [E(X(x=1))]

= P(X = 0) + P(X = 2) - P(X = 1)

For 2 balls with probability of a win, P = (possible outcome) /(total outcome) =

{(2C1) (5C2)}/(10C2) = (2 x 10)/45 = 20/45

While, for probability of no win;

P = 1 - (20/25) = 25/25

So, E(Y) =20/45 - 25/45 = - 5/45 =

- 0.11

B) Now let's calculate for the variance;

Var(Y) = E(Y(^2)) - E(Y)^(2)

Now, from question a, using the equation of Y, we can say;

(Y)^(2) = (1^2)X [x∈{0,2}] - [(1^2)X(x=1)]

And so;

E(Y^(2)) = P(X = 0) + P(X = 2) + P(X = 1) = (1^2)(20/45) + (-1^2)(25/45) = 45/45 = 1

So, Var(Y) = 1 - (-0.11)^2 = 0.9879

Answer:

The probability that the next customer will arrive in 3 minutes or less is 0.45.

Step-by-step explanation:

Let N (t) be a Poisson process with arrival rate λ. If X is the time of the next arrival then,

[tex]P(X>t)=e^{-\lambda t}[/tex]

Given:

[tex]\lambda=\frac{12}{60}[/tex]

t = 3 minutes

Compute the probability that the next customer will arrive in 3 minutes or less as follows:

P (X ≤ 3) = 1 - P (X > 3)

              [tex]=1-e^{-\frac{12}{60}\times3}\\=1-e^{-0.6}\\=1-0.55\\=0.45[/tex]

Thus, the probability that the next customer will arrive in 3 minutes or less is 0.45.

The probability that the next customer will arrive in 3 minutes or less is; 0.4512

This is a Poisson distribution problem with the formula;

P(X > t) = e^(-λt)

Where;

λ is arrival rate

t is arrival time

We are given;

λ = 12/60

t = 3 minutes

P (X ≤ 3) = 1 - (P(X > 3))

Thus;

P (X ≤ 3) = 1 - e^((12/60) × 3)

P (X ≤ 3) = 1 - 0.5488

P (X ≤ 3) = 0.4512

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Answer:

The probability of the number of starfish dropping below 3 is 0.0463.

Step-by-step explanation:

Let X represent the number of starfish in the tide pool.

X follows a Poisson distribution with mean 6.4.

The formula for Poisson distribution is as follows.

[tex]P(X=x) = e^{-6.4} * [6.4^{x} / x!] when x = 0, 1, ...[/tex]

[tex]P(X = x) = 0[/tex] otherwise

We need to find the probability that the number of starfish in the tide pool drops below 3.

Therefore, the required probability is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X=2)

[tex]P(X < 3) = e^{-6.4} + (e^{-6.4}*6.4) + (e^{-6.4}*6.4^{2}/2)[/tex]

P(X < 3) = 0.00166 + 0.01063 + 0.03403

P (X < 3) = 0.0463

Answer:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]

And for this case [tex] r =0.45[/tex]

The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]0.45^2 =0.2025[/tex], so then the % of variation explained is 20.25%.

The proportion of the variability seen in final grade performance that can be predicted by math ability scores is 20.25%.

Step-by-step explanation:

For this case we asume that we fit a linear model:

[tex] y = mx+b[/tex]

Where y represent the final grade and x the math ability scores

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]  

Where:  

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]  

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]  

[tex]\bar x= \frac{\sum x_i}{n}[/tex]  

[tex]\bar y= \frac{\sum y_i}{n}[/tex]  

And we can find the intercept using this:  

[tex]b=\bar y -m \bar x[/tex]  

The correlation coeffcient is given by:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]

And for this case [tex] r =0.45[/tex]

The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]0.45^2 =0.2025[/tex], so then the % of variation explained is 20.25%.

The proportion of the variability seen in final grade performance that can be predicted by math ability scores is 20.25%.

Answer:

x = 6 units.

Step-by-step explanation:

By Geometric mean property:

[tex]x = \sqrt{3 \times 12} = \sqrt{36} = 6 \\ \hspace{20 pt} \huge \orange{ \boxed{ \therefore \: x = 6}}[/tex]

Hence, x = 6 units.

Answer:

[tex] P(X >2) [/tex]

And we can calculate this with the complement rule like this:

[tex] P(X>2) = 1-P(X<2)[/tex]

And using the cdf we got:

[tex] P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480[/tex]

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

[tex]P(X=x)=\lambda e^{-\lambda x}, x>0[/tex]

And 0 for other case. Let X the random variable of interest:

[tex]X \sim Exp(\lambda=\frac{1}{2.725})[/tex]

Solution to the problem

We want to calculate this probability:

[tex] P(X >2) [/tex]

And we can calculate this with the complement rule like this:

[tex] P(X>2) = 1-P(X<2)[/tex]

And using the cdf we got:

[tex] P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480[/tex]

Final answer:

The student's question pertains to calculating the probability of selecting two individuals from a sorority consisting of full members and pledges. The calculation involves using combinatorial formulas to determine the likelihood of different types of selections.

Explanation:

The student is asking about finding the probability of selecting two persons at random from a group consisting of full members and pledges in a sorority. With 38 members in total, 28 full members and 10 pledges, we need to calculate the probability of different pairs of members being selected. This is a combinatorial probability question.

In such problems, if there is no specific requirement about who needs to be picked (like how many full members or pledges), the total number of ways to select two members from the group without regard to order is calculated using the combination formula C(n, k) = n! / [k!(n-k)!], where 'n' is the total number of items, and 'k' is the number of items to choose.

For instance, the probability of selecting two full members can be calculated by finding the number of ways to choose two full members from the 28 available, divided by the number of ways to choose any two members from the entire group of 38.

Answer:

Step-by-step explanation:

The given polygon is a square. To determine the apotherm which is the perpendicular line from the midpoint of the square, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

Let a represent the apotherm

Apotherm = length of each side of the square.

Therefore

8² = a² + a² = 2a²

64 = 2a²

a² = 64/2 = 32

a = √32

The formula for determining the area of a polygon is

Area of polygon is

area = a^2 × n ×tan 180/n

Where n is the number of sides

(n = 4)

Area = √32² × 4 × tan(180/4)

Area = 128 × 1

Area = 128

The formula for determining the perimeter of a regular polygon is

P = 2 × area/apotherm

Perimeter = 2 × 128/√32

Perimeter = 45.3

Using the normal distribution, it is found that:

a) P(X < 230) = 0.734.

b) P(180 < X < 245) = 0.6885.

c) P( X >190) = 0.734.

d) X = 151.76.

e) X = 188.56.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem:

Item a:

This probability is the p-value of Z when X = 230, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{230 - 210}{32}[/tex]

[tex]Z = 0.625[/tex]

[tex]Z = 0.625[/tex] has a p-value of 0.734.

Hence:

P(X < 230) = 0.734.

Item b:

This probability is the p-value of Z when X = 245 subtracted by the p-value of Z when X = 180, hence:

X = 245

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{245 - 210}{32}[/tex]

[tex]Z = 1.09[/tex]

[tex]Z = 1.09[/tex] has a p-value of 0.8621.

X = 180

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{180 - 210}{32}[/tex]

[tex]Z = -0.94[/tex]

[tex]Z = -0.94[/tex] has a p-value of 0.1736.

0.8621 - 0.1736 = 0.6885.

Then:

P(180 < X < 245) = 0.6885.

Item c:

This probability is 1 subtracted by the p-value of Z when X = 190, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{190 - 210}{32}[/tex]

[tex]Z = -0.625[/tex]

[tex]Z = -0.625[/tex] has a p-value of 0.266.

1 - 0.266 = 0.734.

Hence:

P( X >190) = 0.734.

Item d:

This is X = c when Z has a p-value of 0.0344, hence X when Z = -1.82.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.82 = \frac{X - 210}{32}[/tex]

[tex]X - 210 = -1.82(32)[/tex]

[tex]X = 151.76[/tex]

Item e:

This is X when Z has a p-value of 1 - 0.7486 = 0.2514, hence X when Z = -0.67.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.67 = \frac{X - 210}{32}[/tex]

[tex]X - 210 = -0.67(32)[/tex]

[tex]X = 188.56[/tex]

More can be learned about the normal distribution at https://brainly.com/question/24663213

The question involves calculating probabilities and specific values (c) for a given normal distribution with mean 210 and standard deviation 32. The probabilities for certain ranges and tail ends are computed using the normal cumulative distribution function and its inverse.

The question pertains to finding probabilities and specific values associated with a normal distribution X which is denoted as N(210, 32), meaning it has a mean (μ) of 210 and a standard deviation (σ) of 32.

Answer:

A) from the line of best fit, the approximately y-intercept is (0,1.8). This means without any practice, 1h.8 games are won.

B) slope: (5.6-1.8)/(2-0) = 1.9

y = 1.9x + 1.8

(Line of best fit)

x = 13,

y = 1.9(13) + 1.8 = 26.5

Predicted no. of games won after 13 months of practice is 26.5

Final answer:

The y-intercept, representing initial games won and the equation for line of best fit predicting future games won, are determined from the graph data.

Explanation:

Part A:

Part B:

The equation for the line of best fit in slope-intercept form is y = 1.4x + 1.8.

To predict the number of games won after 13 months of practice, substitute x = 13 into the equation:

y = 1.4(13) + 1.8 = 19.5

So, the predicted number of games that could be won after 13 months of practice is 19.5.

Answer:

2522520

Step-by-step explanation:

Number of ways logs can be classified = 14C5* 9C4*5C3

= 2002*126*10

= 2522520

Number of ways to select 5 good = 14C5, out of remaining 9, number of ways to select acceptable log = 9C4, out of remaining 5, number of ways to select prime log = 5C3 and remaking two unacceptable in 2C2 ways

Final answer:

To find how many ways 14 logs can be classified into categories to match the usual counts, use the multinomial coefficient formula based on the given category counts, resulting in a single calculations. to be 54.6.

Explanation:

The question asks in how many ways 14 logs can be classified into four categories (Prime, Good, Acceptable, Not Acceptable) if we know the usual counts for three of these categories are 3 Prime, 5 Good, 4 Acceptable, and the rest are Not Acceptable. This is a combinatorial problem that can be solved using combinations.

First, we distribute the logs into the Prime, Good, and Acceptable categories as given. This leaves us with 14 - (3 + 5 + 4) = 2 logs to be classified as Not Acceptable. Hence, all of the logs are accounted for with the specified counts.

The number of ways to classify the logs can thus be calculated as the number of ways to choose 3 out of 14 for Prime, then 5 out of the remaining 11 for Good, then 4 out of the remaining 6 for Acceptable. The remaining 2 are automatically classified as Not Acceptable. However, since we're simply fulfilling given counts, and the categories are distinct without overlap, we actually approach this as a partition of 14 objects into parts of fixed sizes, which is a straightforward calculation given by the multinomial coefficient:

The formula for the calculation is: 14! / (3! × 5! × 4! × 2!)

Which simplifies to the total number of ways these logs can be classified according to the specified counts

= 54.6

Answer:

Check attachment for complete question

Step-by-step explanation:

Given that,

y=Coskt

We are looking for value of k, that satisfies 4y''=-25y

Let find y' and y''

y=Coskt

y'=-kSinkt

y''=-k²Coskt

Then, applying this 4y'"=-25y

4(-k²Coskt)=-25Coskt

-4k²Coskt=-25Coskt

Divide through by Coskt and we assume Coskt is not equal to zero

-4k²=-25

k²=-25/-4

k²=25/4

Then, k=√(25/4)

k= ± 5/2

b. Let assume we want to use this

y=ASinkt+BCoskt

Since k= ± 5/2

y=A•Sin(±5/2t)+ B •Cos(±5/2t)

y'=±5/2ACos(±5/2t)-±5/2BSin(±5/2t)

y''=-25/4ASin(±5/2t)-25/4BCos(±5/2t

Then, inserting this to our equation given to check if it a solution to y=ASinkt+BCoskt

4y''=-25y

For 4y''

4(-25/4ASin(±5/2t)-25/4BCos(±5/2t))

-25A•Sin(±5/2t)-25B•Cos(±5/2t).

Then,

-25y

-25(A•Sin(±5/2t)+ B •Cos(±5/2t))

-25A•Sin(±5/2t) - 25B •Cos(±5/2t)

Then, we notice that, 4y'' is equal to -25y, then we can say that y=Coskt is a solution to y=ASinkt+BCoskt

The family of functions y = A sin(kt) + B cos(kt) and their derivatives can be verified as solutions to the second-order differential equation associated with a simple harmonic oscillator equation, regardless of the frequency 'k'. This is due to the periodic nature of the sine and cosine functions.

The subject of this question concerns the family of functions y = A sin(kt) + B cos(kt) and their derivatives. It seems the student is asked to verify that for any value of 'k', representing the frequency of the sine and cosine functions, each function in this family is indeed a solution of some differential equations.

To prove this, we first observe the time derivative of this function family: y' = Ak cos(kt) − Bk sin(kt). A second derivative will yield y'' = −Ak² sin(kt) − Bk² cos(kt). By comparing y'' with the original function (y), we see that y'' is equivalent to -k²*y with y as the original function, which verifies the solution to a simple harmonic oscillator equation. Represented as y''+k²*y=0.

Oscillations are an inherent property of these functions - occurring because sin(kt) and cos(kt) are periodic functions, meaning they repeat their values in regular intervals or periods. The values of 'A' and 'B' simply shift the oscillation up or down (amplitude), and don't change the periodic nature. Therefore, it can be concluded that every member of the family of functions y = A sin(kt) + B cos(kt) is indeed a solution for any picked value of 'k'.

https://brainly.com/question/27235369

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Answer:

f(x) = 8 (x + ¼)² − ½

Step-by-step explanation:

f(x) = 8x² + 4x

Divide both sides by 8.

1/8 f(x) = x² + 1/2 x

Take half of the second coefficient, square it, then add to both sides.

(½ / 2)² = (1/4)² = 1/16

1/8 f(x) + 1/16 = x² + 1/2 x + 1/6

Factor the perfect square.

1/8 f(x) + 1/16 = (x + 1/4)²

Multiply both sides by 8.

f(x) + 1/2 = 8 (x + 1/4)²

Subtract 1/2 from both sides.

f(x) = 8 (x + 1/4)² − 1/2

Use simple unitary method:

∵ 1.5 cm on map = 25 miles in reality

∴ x cm on map = 190 miles in reality

[tex]\frac{1.5}{x} = \frac{25}{190}\\ x = 11.4 cm[/tex]

Thus, the two cities are 11.4 cm apart on the map

Final answer:

To determine the map distance between two cities that are 190 miles apart using the scale 1.5 cm = 25 miles, divide the actual distance by the scale ratio (16.67 miles/cm), resulting in 11.4 cm.

Explanation:

For the map with scale 1.5 cm = 25 miles, we first calculate how many miles one centimeter represents by dividing the miles by the centimeters in the scale:

25 miles / 1.5 cm = 16.67 miles/cm

Next, we find the map distance for 190 miles by dividing the actual distance by the distance one centimeter represents:

190 miles / 16.67 miles/cm = 11.4 cm

So, the two cities are 11.4 cm apart on the map.

Answer:

Percentage of scores that fall between 70 and 80 = 24.34%

Step-by-step explanation:

We are given a test with a population mean of 75 and standard deviation equal to 16.

Let X = Percentage of scores

Since, X ~ N([tex]\mu,\sigma^{2}[/tex])

The z probability is given by;

           Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)    where, [tex]\mu[/tex] = 75  and  [tex]\sigma[/tex] = 16

So, P(70 < X < 80) = P(X < 80) - P(X <= 70)

P(X < 80) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{80-75}{16}[/tex] ) = P(Z < 0.31) = 0.62172

P(X <= 70) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{70-75}{16}[/tex] ) = P(Z < -0.31) = 1 - P(Z <= 0.31)

                                              = 1 - 0.62172 = 0.37828

Therefore, P(70 < X < 80) = 0.62172 - 0.37828 = 0.24344 or 24.34%

Answer:

[tex]z=\frac{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24[/tex]  

[tex]p_v =P(z<-2.24)=0.0125[/tex]  

If we compare  the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6

Step-by-step explanation:

Assuming the following question: A presidential candidate's aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at least 60% . If 127 out of a random sample of 240 college students expressed an intent to vote, can we reject the aide's estimate at the 0.1 level of significance?

Data given and notation

n=240 represent the random sample taken

X=127 represent the college students expressed an intent to vote

[tex]\hat p=\frac{127}{240}=0.529[/tex] estimated proportion of college students expressed an intent to vote

[tex]p_o=0.6[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that at least 60% of students are intented to vote .:  

Null hypothesis:[tex]p \geq 0.6[/tex]  

Alternative hypothesis:[tex]p < 0.6[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

[tex]p_v =P(z<-2.24)=0.0125[/tex]  

If we compare  the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6