Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:
yy'' + (y')^ 2 = 0, t > 0. (3)
Then show that for any nonzero constants c1 and c2, c1 + c2t^1/2 is not a solution of this equation.

Ansoogawer:

Step-by-step explanation:

Answer:

y = -3000x + 25,635

Step-by-step explanation:

well if the inital value of the car is $25,635 this means that when

x = 0   y = 25,635

(0 , 25635)

this will be our first point

Now if you tell us that in 1 year depreciate in value $3,000

this means thaw when

x = 1  y = 25,635 - 3,000

x = 1 y = 22,635

(1, 22635)

Now that we have 2 points we can have the equation

First we take the slope as follows

m = (y2 - y1) / (x2 - x1)

m = (22,635 - 25,635) / (1 - 0)

m = -3000 / 1

m = -3000

after calculating the slope we have to replace it in the following formula

(y - y1) = m (x - x1)

y - 25,635 = -3000 ( x - 0)

y - 25,635 = -3000x

y = -3000x + 25,635

Finally we replace the value of y by 10000

10,000 = -3000x + 25,635

10,000 - 25,635 = -3000x

-15,635 = -3000x

-15,635/-3000 = x

5.21167 = x       years

These are the years it would take for the value to be 10,000

to know the days we simply multiply by 365

5.21167 * 365 = 1902.26      days

Final answer:

Andy's expected profit per game is $1.73 when playing the described gambling game.

Explanation:

To find Andy's expected profit per game, we need to calculate the probability and corresponding profit for each possible outcome. Let's create a probability model:

- Andy draws a number card (2-10): This has a probability of 36/52 since there are 36 number cards in a standard deck of 52 cards. The profit for this outcome is $0.

- Andy draws a face card: This has a probability of 12/52 since there are 12 face cards in a standard deck. The profit for this outcome is $3 if the coin lands on heads, $2 if it lands on tails.

- Andy draws an ace: This has a probability of 4/52 since there are 4 aces in a standard deck. The profit for this outcome is $5, except if he draws the ace of clubs, in which case the profit is $25.

To find the expected profit, we multiply each probability by the corresponding profit and sum them up:

(36/52) * $0 + (12/52) * (($3 * 1/2) + ($2 * 1/2)) + (4/52) * (($5 * 3/4) + ($25 * 1/4)) = $0 + $0.69 + $1.04 = $1.73

Therefore, the expected profit per game for Andy is $1.73.

Answer:

15.9% of babies are born with birth weight under 6.3 pounds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6.8 pounds

Standard Deviation, σ = 0.5

We are given that the distribution of  birth weights is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P(birth weight under 6.3 pounds)

P(x < 6.3)

[tex]P( x < 6.3) = P( z < \displaystyle\frac{6.3 - 6.8}{0.5}) = P(z < -1)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x < -1) = 0.159 = 15.9\%[/tex]

15.9% of babies are born with birth weight under 6.3 pounds.

To find the proportion of babies born with a weight less than 6.3 pounds, we calculate the Z-score which results in -1. This Z-score corresponds to about 16% of the population in a standard normal distribution. Therefore, roughly 16% of babies are born weighing less than 6.3 pounds.

To solve this problem, you can use the properties of a normal distribution. In a normal distribution, scores fall within one standard deviation of the mean approximately 68% of the time, within two standard deviations about 95% of the time, and within three standard deviations about 99.7% of the time.

In this scenario, we would find the Z-score, a measure that describes a value's relationship to the mean of a group of values. The formula for the Z-score is (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.

With a mean (μ) = 6.8 pounds, standard deviation (σ) = 0.5 pounds, and X = 6.3 pounds, the calculation for the Z-score would be (6.3-6.8)/0.5 = -1. This means that 6.3 pounds is one standard deviation below the mean. Referring to the standard normal distribution table, a Z-score of -1 corresponds to approximately 16% or 0.16 of the population. Therefore, approximately 16% of babies are born with a weight of less than 6.3 pounds.

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Answer:

8

Step-by-step explanation:

Sin 30 = k/16

k = 16 x sin 30

k = 16 x (0.5) = 8

Answer:

P(A&B) = 0.4

Explanation:

Because it is a random process and there are no special constraints the probability for everybody is the same, the probability of choosing a particular site is 1/7, the person originally seated in chair number seven has 5/7 chance of not seating in chair number six and seven, the same goes for the person originally seated in chair number six; Because we want the probability of the two events happening, we want the probability of the intersection of the two events, and because the selection of a chair change the probability for the others (Dependents events) the probability P(A&B) = P(A) * P(B/A) where P(A) is 5/7 and the probability of choosing the right chair after the event A is 4/7, therefore, P(A&B) = 4/7*5/7 = 0.4.

If the events were independent the probability would be 0.51.

Answer:

[tex] z = \frac{205-233.3396}{37.498}= -0.76[/tex]

So then the value of 205 it's 0.76 deviations below the population mean on this case

Step-by-step explanation:

For this case we have the following data given:

177, 205, 210, 210, 232, 205, 185, 185, 178, 210, 206, 212, 184, 174, 185, 242, 188, 212, 215, 247, 241, 223, 220, 260, 245, 259, 278, 270, 280, 295, 275, 285, 290, 272, 273, 280, 285, 286, 200, 215, 185, 230, 250, 241, 190, 260, 250, 302, 265, 290, 276, 228, 265

And these values represent the weigths of all the team members of the Arizona Cardinals

Now if we organize the data values from the smallest to the largest we have:

174 177 178 184 185 185 185 185 188 190 200 205 205 206 210 210 210 212 212 215 215 220 223 228 230  232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290  290 295 302

For this case we can calculate the mean with the following formula:

[tex] \mu = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And if we replace we got:

[tex] \mu = 233.3396[/tex]

And for the deviation we can use the following formula:

[tex] \sigma =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

And if we replace we got:

[tex] \sigma = 37.498[/tex]

And in order to calculate How many standard deviations above or below the mean was he we can use the z score formula given by:

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

And we assume that x=205 and if we replace we have:

[tex] z = \frac{205-233.3396}{37.498}= -0.76[/tex]

So then the value of 205 it's 0.76 deviations below the population mean on this case

Answer:

1) m=[1,4]

2) m=[-3,0,1]

Step-by-step explanation:

for y= e^(m*x) , then

y′=m*e^(m*x)

y′′=m²*e^(m*x)

y′′′=m³*e^(m*x)

thus

1) y′′+3y′−4y=0

m²*e^(m*x) + 3*m*e^(m*x) - 4*e^(m*x) =0

e^(m*x) *(m²+3*m-4) = 0 → m²+3*m-4 =0

m= [-3±√(9-4*1*(-4)] /2 → m₁=-4 , m₂=1

thus m=[1,4]

2) y′′′+2y′′−3y′=0

m³*e^(m*x) + 2*m²*e^(m*x) - 3*m*e^(m*x) =0

e^(m*x) *(m³+2*m²-3m) = 0 → m³+2*m²-3m=0

m³+2*m²-3m= m*(m²+2*m-3)=0

m=0

or

m= [-2±√(4-4*1*(-3)] /2  → m₁=-3 , m₂=1

thus m=[-3,0,1]

Answer:395.75lb

Step-by-step explanation:see attachment

Answer:

A) True

B) True

C) False

Step-by-step explanation:

Knowing that the collinear points are all those that pass through a line, we have:

A) given two points they form a line, by themselves they are collinear  (graph 1)

B) Can be or not can be (graph 2)

C) Can be not must be (graph 3)

We want to see if the given statements are true or false.

We will see that:

Two or more points are collinear if we can draw a line that connects them.

A) Whit that in mind, the first statement is true, 2 points is all we need to draw a line, thus two different points are always collinear, so the first statement is true.

B) For the second statement suppose you have a line already drawn, then you can draw 4 points along the line, if you do that, you will have 4 collinear points, so yes, 4 points can be collinear.

C) For the final statement, again assume you have a line, you used 2 points to draw that line (because two points are always collinear). Now you could have more points outside the line, thus, the set of all the points is not collinear (not all the points are on the same line).

So sets of 3 or more points can be collinear, but not "must" be collinear, so the last statement is false.

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Answer:

99.7% of the distribution will be between 600 hours and 900 hours.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 750

Standard deviation = 50

What percent of the distribution will be between 600 hours and 900 hours?

600 = 750 - 3*50

600 is 3 standard deviations below the mean

900 = 750 + 3*50

900 is 3 standard deviations above the mean

By the Empirical Rule, 99.7% of the distribution will be between 600 hours and 900 hours.

Answer:

(a-1) 22 rookies receiving a rating of 4 or better.

(a-2) 14 rookies received a rating of 2 or worse.

(b-1) Constructed below in explanation.

(b-2) 8% of total rookies received a rating of 5.

Step-by-step explanation:

We are provided the rating of Fifty pro-football rookies on a scale of 1 to 5 based on performance at a training camp as well as on past performance.

The frequency distribution constructed is given below:

 Rating          Frequency

     1                       4

     2                     10       where ranking of 1 indicate a poor prospect whereas

     3                     14         ranking of 5 indicate an excellent prospect.

     4                     18

     5                      4

(a-1) Rookies receiving a rating of 4 or better = Rating of 4 + Rating of 5

       So, by seeing the frequency distribution 18 rookies received a rating of

         4 and 4 rookies received a rating of 5.

Hence, Rookies receiving a rating of 4 or better = 18 + 4 = 22 rookies.

(a-2) Number of rookies received a rating of 2 or worse = Rating of 2 +

                                                                                                 Rating of 1

     So, by seeing the frequency distribution 10 rookies received a rating of

      2 and 4 rookies received a rating of 1.

Hence, Rookies receiving a rating of 2 or worse = 10 + 4 = 14 rookies.

(b-1) Relative Frequency is calculated as = Each frequency value /

                                                                         Total Frequency

   Rating        Frequency(f)        Relative Frequency

     1                       4                          4 / 50 = 0.08

     2                     10                         10 / 50 = 0.2

     3                     14                          14 / 50 = 0.28

     4                     18                          18 / 50 = 0.36

     5                     4                           4 / 50  = 0.08

                     [tex]\sum f[/tex]= 50                

Hence, this is the required relative frequency distribution.

(b-2) To calculate what percent received a rating of 5 is given by equation :

             x% of 50 = 4 {Here 4 because 4 rookies received rating of 5}

         x = [tex]\frac{4*100}{50}[/tex] = 8% .

Therefore, 8% of total rookies received a rating of 5.

22 rookies received a performance rating of 4 or better, and there are 14 rookies that received a rating of 2 or worse. The relative frequency distribution rounded to 2 decimal places for ratings 1, 2, 3, 4, and 5 is 0.08, 0.2, 0.28, 0.36, and 0.08 respectively. Finally, 8% of rookies received a rating of 5.

To answer question a-1, we add the frequencies of the ratings 4 and 5 together. So 18 (the number of rookies who received a 4) plus 4 (the number of rookies who received a 5) equals 22. Therefore, 22 rookies received a rating of 4 or better.

For question a-2, we add the frequencies of the ratings 1 and 2 together. So 4 (the number of rookies who received a 1) plus 10 (the number of rookies who received a 2) equals 14. Thus, 14 rookies received a rating of 2 or worse.

Next, for question b-1, we find the relative frequency distribution by dividing each frequency by the total number of players (50). So, the relative frequency for 1 would be 4/50 ≈ 0.08, for 2 it's 10/50=0.2, for 3 it's 14/50 ≈ 0.28, for 4 it's 18/50 = 0.36, and for 5 it's 4/50 ≈ 0.08. These are all rounded to 2 decimal places.

Finally, for question b-2, to find the percent of players who received a rating of 5, we take the frequency of 5 which is 4, and divide it by the total number of rookies, which is 50, then multiply by 100 to convert it to a percentage (4/50 * 100). The answer is 8%.

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Answer:

Car A. would travel the farthest

Answer:

Step-by-step explanation:

i need help to show the work

The Possible outcomes and associated values in the sample space along with X are SSS (X = 3), SSF, SFS, FSS, SFF (X = 1), FSF, FFS,

FFF (X = 0).

We have,

Let's list all the possible outcomes in the sample space when three concrete beams are selected and their failure types (shear or flexure) are determined.

For each outcome, we'll also determine the value of X, which represents the number of beams that failed by shear.

Let S represent shear failure and F represent flexure failure.

Possible outcomes and associated values of X:

SSS (All three beams failed by shear)

X = 3

SSF (Two beams failed by shear, one by flexure)

X = 2

SFS (Two beams failed by shear, one by flexure)

X = 2

FSS (Two beams failed by shear, one by flexure)

X = 2

SFF (One beam failed by shear, two by flexure)

X = 1

FSF (One beam failed by shear, two by flexure)

X = 1

FFS (One beam failed by shear, two by flexure)

X = 1

FFF (All three beams failed by flexure)

X = 0

These are the eight possible outcomes in the sample space along with the associated values of X, representing the number of beams that failed by shear in each outcome.

Thus,

Possible outcomes and associated values of X: SSS (X = 3), SSF, SFS, FSS, SFF (X = 1), FSF, FFS, FFF (X = 0).

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The outcomes in the sample space for three concrete beams failing are: (S,S,S) with X = 3, (S,S,F),(S,F,S),(F,S,S) with X = 2, (S,F,F),(F,F,S),(F,S,F) with X = 1, and (F,F,F) with X = 0. S represents a shear failure, F a flexure failure, and X the number of shear failures.

The sample space for this problem includes all possible outcomes for the three concrete beams that can fail. The possible outcomes are:

Here, S represents a shear failure and F represents a flexure failure. The number specified by X in each scenario represents how many beams among the three randomly selected ones failed by shear.

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Answer:

The augmented matrix has been given in the attachment

Step-by-step explanation:

The steps for the determination of INCONSISTENCY  are as shown in the attachment.

Answer:

a) [tex] E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu[/tex]

So then we conclude that [tex] \hat \theta_1[/tex] is an unbiased estimator of [tex]\mu[/tex]

[tex] E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu[/tex]

So then we conclude that [tex] \hat \theta_2[/tex] is an unbiased estimator of [tex]\mu[/tex]

b) [tex] Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2} [/tex]

[tex] Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8} [/tex]

Step-by-step explanation:

For this case we know that we have two random variables:

[tex] X_1 , X_2[/tex] both with mean [tex]\mu = \mu[/tex] and variance [tex] \sigma^2[/tex]

And we define the following estimators:

[tex] \hat \theta_1 = \frac{X_1 + X_2}{2}[/tex]

[tex] \hat \theta_2 = \frac{X_1 + 3X_2}{4}[/tex]

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

[tex] E(\hat \theta_i) = \mu , i = 1,2 [/tex]

So let's find the expected values for each estimator:

[tex] E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})[/tex]

Using properties of expected value we have this:

[tex] E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu[/tex]

So then we conclude that [tex] \hat \theta_1[/tex] is an unbiased estimator of [tex]\mu[/tex]

For the second estimator we have:

[tex]E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})[/tex]

Using properties of expected value we have this:

[tex] E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu[/tex]

So then we conclude that [tex] \hat \theta_2[/tex] is an unbiased estimator of [tex]\mu[/tex]

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

[tex] Var(aX) = a^2 Var(X)[/tex]

For the first estimator we have:

[tex] Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})[/tex]

[tex] Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)] [/tex]

Since both random variables are independent we know that [tex] Cov(X_1, X_2 ) = 0[/tex] so then we have:

[tex] Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2} [/tex]

For the second estimator we have:

[tex] Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})[/tex]

[tex] Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)] [/tex]

Since both random variables are independent we know that [tex] Cov(X_1, X_2 ) = 0[/tex] so then we have:

[tex] Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8} [/tex]

Both θ₁^ and θ₂^ are unbiased estimators of μ. The variance of θ₁^ is σ² / 2, while the variance of θ₂^ is 5σ² / 8.

Let's analyze the provided estimators of the mean (">").

a) Unbiased Estimators

An estimator  heta is unbiased if E[θ] = μ. We have:

θ₁^ = (X₁ + X₂) / 2

The expected value of θ₁^ is:

E[θ₁^] = E[(X₁ + X₂) / 2] = (E[X₁] + E[X₂]) / 2 = (μ + μ) / 2 = μ

Thus, θ₁^ is an unbiased estimator of μ.

θ₂^ = (X₁ + 3X₂) / 4

The expected value of θ₂^ is:

E[θ₂^] = E[(X₁ + 3X₂) / 4] = (E[X₁] + 3E[X₂]) / 4 = (μ + 3μ) / 4 = 4μ / 4 = μ

Thus, θ₂^ is also an unbiased estimator of μ.

b) Variance of Each Estimator

The variance of an estimator θ is given by Var(θ). Considering the given variances of X₁ and X₂ (both σ²):

Var(θ₁^) = Var[(X₁ + X₂) / 2] = (1/2)²Var(X₁) + (1/2)²Var(X₂) = (1/4)σ² + (1/4)σ² = σ² / 2

Var(θ₂^) = Var[(X₁ + 3X₂) / 4] = (1/4)²Var(X₁) + (3/4)²Var(X₂) = (1/16)σ² + (9/16)σ² = (1/16 + 9/16)σ² = (10/16)σ² = 5σ² / 8

Thus, the variance of θ₁^ is σ² / 2 and the variance of θ₂^ is 5σ² / 8.

The given statement is true, and the further discussion can be defined as follows:

That's why the given statement is "true".

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Final answer:

True. Parental care is considered a moderating variable that can influence the relationship between eating a hot breakfast and a child's school performance.

Explanation:

Research indicates that children who eat a hot breakfast at home tend to perform better in school. However, this outcome is not solely based on the meal itself, but also on the parental care provided before the child goes to school. In this context, parental care acts as a moderating variable, potentially influencing the relationship between having a hot breakfast and a child's school performance. A moderating variable is one that affects the strength or direction of the relationship between an independent variable (hot breakfast) and a dependent variable (school performance). Therefore, the statement is True.

Answer:

P(A) = 400/924 = 100/231 or 0.4329

Step-by-step explanation:

The probability that both groups will have the same number of men P(A);

For the two groups to have the same number of men they must include 3 men and 3 women in each group.

P(A) = Number of possible selections of 3 men from 6 and 3 women from 6 into each of the two groups N(S) ÷ total number of possible selections of members into the two groups N(T).

P(A) = N(S)/N(T)

Since order is not important, we will use combination.

N(S) = 6C3 × 6C3 = 20 × 20 = 400

N(T) = 12C6 = 924

P(A) = 400/924 = 100/231 or 0.4329

The probability that both groups will have the same number of men is 0.43.

When dividing the group into two equal-sized groups of 6 each, the total number of ways to do this is represented by the binomial coefficient C(12, 6), which is equal to 924.

Now, to ensure that both groups have the same number of men, we need to consider the ways in which we can choose 3 men from the 6 available men, which is represented by C(6, 3), and similarly, we can choose 3 women from the 6 available women, which is also represented by C(6, 3).

The total number of ways to choose 3 men and 3 women for both groups is C(6, 3) * C(6, 3) = 20 * 20 = 400.

Since there are 2 equally likely outcomes (either both groups have the same number of men or they don't), the probability is 400/924, which simplifies to 0.43.

In summary, the probability that both groups will have the same number of men is 0.43 because there are 400 ways to select 3 men and 3 women for each group out of a total of 924 possible ways to divide the 12 people into two groups.

This results in a 0.43 probability for the desired outcome.

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Answer:

(a) P (X ≥ 4) = 0.972

(b) E (X) = 20

Step-by-step explanation:

Let X = number of people tested to detect the presence of gene in 2.

Then the random variable X follows a Negative binomial distribution with parameters r (number of success) and p probability of success.

The probability distribution function of X is:

[tex]f(x)={x-1\choose r-1}p^{r}(1-p)^{x-r}[/tex]

Given: r = 2 and p = 0.1

(a)

Compute the probability that four or more people will have to be tested before two with the gene are detected as follows:

P (X ≥ 4) = 1 - P (X = 3) - P (X = 2)

              [tex]=1-[{3-1\choose 2-1}(0.1)^{2}(1-0.1)^{3-2}]-[{2-1\choose 2-1}(0.1)^{2}(1-0.1)^{2-2}]\\=1-0.018-0.01\\=0.972[/tex]

Thus, the probability that four or more people will have to be tested before two with the gene are detected is 0.972.

(b)

The expected value of a negative binomial random variable X is:

[tex]E(X)=\frac{r}{p}[/tex]

The expected number of people to be tested before two with gene are detected is:

[tex]E(X)=\frac{r}{p}=\frac{2}{0.1}=20[/tex]

Thus, the expected number of people to be tested before two with gene are detected is 20.

To find out how many inches must fall in December so that the average monthly precipitation for October, November, and December exceeds 3.11 inches, subtract the precipitation from October and November from the total precipitation. Therefore, 3.3 inches must fall in December.

To find out how many inches must fall in December so that the average monthly precipitation for October, November, and December exceeds 3.11 inches, we can use the formula for calculating average:

Average = (Total precipitation) / (Number of months)

Let's solve for the total precipitation:

Total precipitation = Average precipitation * Number of months

Total precipitation = 3.11 inches * 3 months = 9.33 inches

To find out how many inches must fall in December, we subtract the precipitation from October and November from the total precipitation:

December precipitation = Total precipitation - (October precipitation + November precipitation)

December precipitation = 9.33 inches - (2.98 inches + 3.05 inches) = 3.3 inches

Therefore, in order for the average monthly precipitation for these months to exceed 3.11 inches,

3.3 inches

must fall in December.

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Answer: the cost of a DVD is $8 and the cost of a video game is $33

Step-by-step explanation:

Let x represent the cost of a video game.

Let y represent the cost of a DVD.

2 video games and 3 DVDs cost $90.00. This is expressed as

2x + 3y = 90 - - - - - - - - - - - 1

1 video game and 2 DVDs cost $49.00. This is expressed as

x + 2y = 49 - - - - - - - - - - - 2

Multiplying equation 1 by 1 and equation 2 by 2, it becomes

2x + 3y = 90 - - - - - - - - - - - -3

2x + 4y = 98 - - - - - - - - - - - - 4

Subtracting equation 4 from equation 3, it becomes

3y - 4y = 90 - 98

- y = - 8

y = 8

Substituting y = 8 into equation 2, it becomes

x + 2 × 8 = 49

x + 16 = 49

x = 49 - 16

x = 33

By solving a system of linear equations, it is determined that the cost of a video game is $33 and the cost of a DVD is $8.

The question involves solving a system of linear equations to determine the cost of a video game and a DVD.

Let's denote the cost of one video game as V and the cost of one DVD as D. Based on the given information, we can set up the following two equations:

To solve for D, we can multiply the second equation by 2 and subtract it from the first equation:

2V + 3D - (2V + 4D) = 90 - 98

This simplifies to -D = -8, which means D = $8.

Now that we know the cost of a DVD, we can substitute it back into the second equation:

V + 2(8) = 49

V + 16 = 49

V = 49 - 16

V = $33.

Therefore, the cost of a video game is $33, and the cost of a DVD is $8.

The correct statements regarding the secant function are as follows:

It is one divided by the cosine, that is:

[tex]\sec{x} = \frac{1}{\cos{x}}[/tex]

The vertical asymptotes are when the cosine is 0, that is, [tex]x = k\frac{\pi}{2}, k = 1, 2, ...[/tex]

As for the values of the function:

[tex]\sec{\left(\frac{\pi}{4}\right)} = \frac{1}{\cos{\left(\frac{\pi}{4}\right)}} = \sqrt{2}[/tex]

[tex]\sec{\left(\frac{\pi}{3}\right)} = \frac{1}{\cos{\left(\frac{\pi}{3}\right)}} =  2[/tex]

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Answer:

(a) We conclude after testing that mean diameter is 8.2500 cm.

(b) P-value of test = 2 x 0.0005% = 1 x [tex]10^{-5}[/tex] .

(c) 95% confidence interval on the mean diameter = [8.2525 , 8.2545]

Step-by-step explanation:

We are given with the population standard deviation, [tex]\sigma[/tex] = 0.0020 cm

Sample Mean, Xbar = 8.2535 cm   and Sample size, n = 15

(a) Let Null Hypothesis, [tex]H_0[/tex] : Mean Diameter, [tex]\mu[/tex] = 8.2500 cm

 Alternate Hypothesis, [tex]H_1[/tex] : Mean Diameter,[tex]\mu[/tex] [tex]\neq[/tex] 8.2500 cm{Given two sided}

So, Test Statistics for testing this hypothesis is given by;

                           [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] follows Standard Normal distribution

After putting each value, Test Statistics = [tex]\frac{8.2535-8.2500}{\frac{0.0020}{\sqrt{15} } }[/tex] = 6.778

Now we are given with the level of significance of 5% and at this level of significance our z score has a value of 1.96 as it is two sided alternative.

Since our test statistics does not lie in the rejection region{value smaller than 1.96} as 6.778>1.96 so we have sufficient evidence to accept null hypothesis and conclude that the mean diameter is 8.2500 cm.

(b) P-value is the exact % where test statistics lie.

For calculating P-value , our test statistics has a value of 6.778

So, P(Z > 6.778) = Since in the Z table the highest value for test statistics is given as 4.4172  and our test statistics has value higher than this so we conclude that P - value is smaller than 2 x 0.0005% { Here 2 is multiplied with the % value of 4.4172 because of two sided alternative hypothesis}

Hence P-value of test = 2 x 0.0005% = 1 x [tex]10^{-5}[/tex] .

(c) For constructing Two-sided confidence interval we know that:

    Probability(-1.96 < N(0,1) < 1.96) = 0.95 { This indicates that at 5% level of significance our Z score will lie between area of -1.96 to 1.96.

P(-1.96 <  [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95

P([tex]-1.96\frac{\sigma}{\sqrt{n} }[/tex] < [tex]Xbar - \mu[/tex] < [tex]1.96\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.95

P([tex]-Xbar-1.96\frac{\sigma}{\sqrt{n} }[/tex] < [tex]-\mu[/tex] < [tex]1.96\frac{\sigma}{\sqrt{n} }-Xbar[/tex] ) = 0.95

P([tex]Xbar-1.96\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]Xbar+1.96\frac{\sigma}{\sqrt{n} }[/tex]) = 0.95

So, 95% confidence interval for [tex]\mu[/tex] = [[tex]Xbar-1.96\frac{\sigma}{\sqrt{n} }[/tex] , [tex]Xbar+1.96\frac{\sigma}{\sqrt{n} }[/tex]]

                                                        = [[tex]8.2535-1.96\frac{0.0020}{\sqrt{15} }[/tex] , [tex]8.2535+1.96\frac{0.0020}{\sqrt{15} }[/tex]]

                                                        = [8.2525 , 8.2545]

Here [tex]\mu[/tex] = mean diameter.

Therefore, 95% two-sided confidence interval on the mean diameter

           =  [8.2525 , 8.2545] .

Answer:

a

b

c

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Since the figure was not supplied, let's focus on that principle.

To calculate it simply subtract the area of the square minus the area of the circle. Given the side of the square 20 ft

1. Square Area

[tex]S= s^{2}\\S=20^{2} \Rightarrow S=400 ft^{2}[/tex]

2.Circle Area

Notice the radius is half the square side, i.e. 10 ft

[tex]S=\pi*R^{2}\\S=3.14*(10)^{2}\\S=314 \:ft^{2}\\[/tex]

Subtracting the area of the square and the area of the circle:

[tex]400-314=86 ft^{2}[/tex]

The area of the shaded is 86 feet².

Thus, option (B) is correct.

Let's assume the side length of the square is "s".

The area of the square is then given by s².

Substitute the side s = 20 into above formula as

Area of square = 20 x 20

                         = 400 square feet

Now, Area of circle = πr²

                                = 3.14 x (10)²

                                = 3.14 x 100

                                = 314 square feet.

Now, the area of the shaded region can be calculated as:

Area of shaded region = Area of square - Area of circle

                                      = 400 - 314

                                      = 86 feet²

Thus, option (B) is correct.

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Answer: the last option is the correct answer.

Step-by-step explanation:

When x = 3, height = 2 × 3 = 6

When x = 3, base = π × 3² = 9π

Volume = 1/3 × 6 × 9π = 18π

Therefore,

The formula for determining the volume of the cone is

Volume = 1/3 × height × area of base

Height is f(x) = 2x

Area of base is g(x) = πx²

Therefore, the equation that can be used to the volume of the cone, V(x) would be

1/3(f.g)(x)

Final answer:

The student needs to find an equation V(x) representing the volume of a cone. Using the volume formula V(x) = kx² and the given volume of 18 when x is 3, the constant k can be calculated. Thus, the equation for the volume is V(x) = 2x².

Explanation:

The student is asking for an equation that represents the volume of a cone, V(x), based on a given volume when x equals 3. The general formula for the volume of a cone is V = (1/3)πr²h, where r is the radius of the base and h is the height of the cone. Since the volume is given as a quadratic function of x, and we know that when x equals 3 the volume is 18, a possible representation for the volume as a function of x could be V(x) = kx², where k is a constant that we would solve for using the given volume at x = 3.

To find the constant k, we substitute x with 3 and V(x) with 18 in the equation V(x) = kx² and solve for k. Therefore, the equation becomes 18 = k(3²), which simplifies to 18 = 9k. Solving for k gives us k = 2, so the equation for the volume of the cone as a function of x is V(x) = 2x².

1 inch on the map = 50 miles on the Earth.

A certain trip on the map is 2-3/4 inches.

-- first inch = 50 miles

-- second inch = another 50 miles

-- 3/4 inch = 3/4 of 50 miles (37.5 miles)

Total:

Here's an equation;

1 map-inch = 50 real-miles

Multiply each side by 2-3/4 :

2-3/4 inches = (2-3/4) x (50 miles)

2-3/4 map-inches = 137.5 real-miles

The actual distance between Houston and Austin is 137.5 miles.

We are given that;

The distance between Houston and Austin = 2 3/4 inches

Now,

To find the actual distance between Houston and Austin, we need to multiply the map distance by the scale factor.

The map distance is 2 3/4 inches, which is equivalent to 11/4 inches. The scale factor is 1 inch = 50 miles, which means that every inch on the map corresponds to 50 miles in reality. So, we have:

11/4 x 50 = (11 x 50) / 4

= 550 / 4

= 137.5

Therefore, by unit conversion answer will be 137.5 miles.

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Answer:

0.5 week

Step-by-step explanation:

The time spent by a pound of meat in this system is given by the average number of pounds of meat on inventory (2500 pounds) divided by the average weekly meat consumption (5000 pounds per week).

The time, in weeks, is:

[tex]t=\frac{2500\ pounds}{5000\ pounds/week} \\t= 0.5\ week[/tex]

The average time spent by a pound of meat in production is 0.5 week.

Answer:

[tex]240cm^2[/tex]

Step-by-step explanation:

The area of the rectangular face with dimension 8 by 8 is [tex]8*8=64cm^2[/tex]

The area of the rectangular face with dimension 10 by 8 is [tex]10*8=80cm^2[/tex]

The area of the rectangular face with dimension 6 by 8 is [tex]6*8=48cm^2[/tex]

The area of the two rectangular faces is [tex]2*\frac{1}{2}*8*6=48cm^2[/tex]

The total surface area is [tex]64+80+48+48=240cm^2[/tex]

Answer:

P ( disease | positive ) = 0.68

Step-by-step explanation:

This is a classic question of Bayes' Theorem, which calculates probability in a scenario where one event has already occured.

In this case, we have the following data:

P ( disease ) = 0.10

P (no disease) = 0.90

P ( positive | disease ) = 0.95

P ( positive | no disease) = 0.05

Formula to use:

[tex]P ( disease\ |\ positive ) = \frac{P(disease)\ P(positive\ |\ disease)}{P(disease)\ P(positive\ |\ disease) + P(no\ disease)\ P(positive\ |\ no\ disease)}[/tex]We substitute the values from our data in this formula:

[tex]\frac{(0.10) \times (0.95)}{(0.10)\times(0.95) + (0.90)\times(0.05)} \\\\\frac{0.095}{0.095+0.045}\\\\\frac{0.095}{0.14}\\ \\\\\=0.68[/tex] (Rounded off to two decimal places)

Hence, the probability of a person testing positive once they have the disease is 68%.

Answer:

The value of n is 5.

Step-by-step explanation:

It is provided that in there are n black and n white balls in an urn.

Three balls are selected at random without replacement.

And if all the three balls are white the probability is [tex]\frac{1}{12}[/tex]

The probability of selecting 3 white balls without replacement is:

[tex]P(3\ white\ balls)=\frac{n}{2n}\times \frac{n-1}{2n-1}\times \frac{n-2}{2n-2} \\\frac{1}{12}=\frac{n(n-1)(n-2)}{2n(2n-1)(2n-2)} \\\frac{1}{12}=\frac{(n-1)(n-2)}{2(2n-1)(2n-2)} \\2(4n^{2}-6n+2)=12(n^{2}-3n+2)\\8n^{2}-12n+4=12n^{2}-36n+24\\4n^{2}-24n+20=0\\n^{2}-6n+5=0\\[/tex]

Solve the resultant equation using factorization as follows:

[tex]n^{2}-6n+5=0\\n^{2}-5n-n+5=0\\n(n-5)-1(n+5)=0\\(n-1)(n-5)=0[/tex]

So the value of n is either 1 or 5.

Since 3 white balls are selected the value of n cannot be 1.

So the value of n is 5.

Check:

[tex]P(3\ white\ balls)=\frac{n}{2n}\times \frac{n-1}{2n-1}\times \frac{n-2}{2n-2} \\=\frac{5}{2\times5}\times \frac{5-1}{(2\times5)-1}\times \frac{5-2}{(2\times5)-2}\\=\frac{1}{2}\times\frac{4}{9}\times\frac{3}{8} \\=\frac{1}{12}[/tex]

Thus, the value of n is 5.

Answer:

a) a

b) ab

e) bbbbb

Step-by-step explanation:

We are given the following in the question:

[tex]a, b \in S[/tex]

[tex]x \in S \Rightarrow xb \in S[/tex]

a) a

It is given that [tex]a \in S[/tex]

b) ab

[tex]\text{If }a \in S\\\Rightarrow ab \in S[/tex]

Thus, ab belongs to S.

c) aba

This does not belong to S because we cannot find x for which xb belongs to S.

d) aaab

This does not belong to S because we cannot find x for which xb belongs to S.

e) bbbbb

[tex]\text{If }b \in S\Rightarrow bb \in S\\\text{If }bb \in S\Rightarrow bbb \in S\\\text{If }bbb \in S\Rightarrow bbbb \in S\\\text{If }bbbb \in S\Rightarrow bbbbb \in S[/tex]

Thus, bbbbb belongs to S.