Arrivals at a fast-food restaurant follow a Poisson distribution with a mean arrival rate of 16 customers per hour. What is the probability that in the next hour there will be exactly 9 arrivals?

a. 0.7500

b. 0.1322

c. 0.0000

d. 0.0213

e. none of the above

Answer:

There are 400 micrograms of riboflavin in a multiple vitamin capsule containing 1/5 of the prophylactic dose

Step-by-step explanation:

First step: How many mg of riboflavin are there in the capsule?

This can be solved as a rule of three problem.

In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.

When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.

When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.

In this step, the measures are:

-The percentage of the dose.

-The number of mg of the dose.

As one measure increases, so do the other. It means that we have a direct rule of three.

One dose has 2mg. How many mg are in [tex]\frac{1}{5} = 0.2[/tex] of the dose.

1dose - 2mg

0.2dose - x

x = 2*0.2

x = 0.4mg

Final step: Converting 0.4mg to micrograms.

Unit conversion is an example of a direct rule of three. Each mg has 1000 micrograms, so.

1mg - 1000 micrograms

0.4mg - x micrograms

x = 1000*0.4

x = 400 micrograms

Final answer:

To calculate the amount of riboflavin in micrograms for a capsule containing 1/5 of the prophylactic dose of 2 mg, we convert 2 mg to 2,000 mcg and then calculate 1/5 of that amount, resulting in 400 mcg of riboflavin per capsule.

Explanation:

If the prophylactic dose of riboflavin is 2 mg, we need to calculate the amount in micrograms for a capsule containing 1/5 of this dose. First, let's convert the dose in milligrams to micrograms.

1 milligram (mg) = 1,000 micrograms (mcg)

Therefore, 2 mg = 2 × 1,000 mcg = 2,000 mcg.

Now, to find 1/5 of the prophylactic dose:

(1/5) × 2,000 mcg = 400 mcg

So, a multiple vitamin capsule containing 1/5 of the prophylactic dose would have 400 micrograms of riboflavin.

Answer:

D does not implies that the sequence is divergent. All others statements do.

Step-by-step explanation:

Statement D: "[tex](a_n)[/tex] has both an increasing subsequence and a decreasing subsequence" does not necessarily implies that the sequence is divergent. For example, let [tex](a_n)[/tex] the sequence given by:

[tex]a_n=\frac{1}{n}[/tex] if n is odd

[tex]a_n=-\frac{1}{n}[/tex] if n is even

We can see that the subsequence [tex]a_{2n-1}[/tex] is a decreasing sequence   (the subsequence given by odd indexes). And the subsequence [tex]a_{2n}[/tex] is an increasing sequence   (the subsequence given by even indexes).

However, [tex](a_n)[/tex] is a convergent sequence with limit zero.

Answer:

[tex]97.0\frac{kg}{mol*dL}[/tex]

Step-by-step explanation:

we know that 1 kg is equal to 1000 g and also we know that 1 L is equal to 10 dL then using a conversion factor we have:

[tex]( 9.70\times 10^{5} \dfrac{1}{10000} \frac{kg}{mol*dL} )[/tex]

Now we can eliminate the units of g and L and we have

][tex]( 9.70\times 10^{5} \dfrac{1}{10000} \frac{kg}{mol*dL} )=( 9.70\times 10^{5} \dfrac{1}{1\times 10^{4}} \frac{kg}{mol*dL} )=(\dfrac{9.70\times 10^{5}}{1\times 10^{4}} \frac{kg}{mol*dL} )[/tex]

Using the properties of exponents we have

[tex](\dfrac{9.70\times 10^{5}}{1\times 10^{4}} \frac{kg}{mol*dL} )=9.70\times 10^{1}\frac{kg}{mol*dL}=97.0\frac{kg}{mol*dL}[/tex]

Answer with Step-by-step explanation:

The given differential equation is variable separable in nature and hence will be solved accordingly as follows:

[tex]\frac{dy}{dx}=\frac{y}{x}\\\\=\frac{dy}{y}=\frac{dx}{x}\\\\\int \frac{dy}{y}=\int \frac{dx}{x}\\\\ln(y)=ln(x)+ln(c)\\\\ln(y)=ln(cx)\\\\(\because ln(ab)=ln(a)+ln(b))\\\\\therefore y=cx[/tex]

where 'c' is constant of integration whose value shall be obtained using the given condition [tex]y(1)=-2[/tex]

Thus we have

[tex]-2=c\times 1\\\\\therefore c=-2\\[/tex]

Thus solution becomes

[tex]y=-2x[/tex]

Answer:

The ball will take 6.3 seconds to reach the maximum height and hit the ground.

Step-by-step explanation:

a). When a ball was thrown upwards with an initial velocity u then maximum height achieved h will be represented by the equation

v² = u² - 2gh

where v = final velocity at the maximum height h

and g = gravitational force

Now we plug in the values in the equation

At maximum height final velocity v = 0

0 = (28)² - 2×(9.8)h

19.6h = (28)²

h = [tex]\frac{(28)^{2}}{19.6}[/tex]

  = [tex]\frac{784}{19.6}[/tex]

  = 40 meter

B). If the ball misses the building and hits the ground then we have to find the time after which the ball hits the ground that will be

= Time to reach the maximum height + time to hit the ground from the maximum height

Time taken by the ball to reach the maximum height.

Equation to find the time will be v = u - gt

Now we plug in the values in the equation

0 = 28 - 9.8t

t = [tex]\frac{28}{9.8}[/tex]

 = 2.86 seconds

Now time taken by the ball to hit the ground from its maximum height.

H = ut + [tex]\frac{1}{2}\times g\times (t)^{2}[/tex]

(17 + 40) = 0 + [tex]\frac{1}{2}\times g\times (t)^{2}[/tex]

57 = 4.9(t)²

t² = [tex]\frac{57}{4.9}[/tex]

t² = 11.63

t = √(11.63)

 = 3.41 seconds

Now total time taken by the ball = 2.86 + 3.41

                                                      = 6.27 seconds

                                                      ≈ 6.3 seconds

Therefore, the ball will take 6.3 seconds to reach the maximum height and hit the ground.

Answer:

6.25 hours

Step-by-step explanation:

As we know that,

1 hour = 60 minutes

⇒ [tex]1 \ minute = \frac{1}{60} \ hour[/tex]

⇒ [tex]375 \ minutes =375\times \frac{1}{60} \ hours[/tex]

⇒ 375 mintes = 6.25 hours

Thus, 375 mintes = 6.25 hours

For making mathematical induction, we need:

An [tex]n_0[/tex] for which the relation holds true

if its true for [tex]n_i[/tex], then, is true for [tex]n_{i+1}[/tex]

the relationship is not true for 1 or 2

[tex]1^3-1 = 0 < 4*1[/tex]

[tex]2^3-1 = 8 -1 = 7 < 4*2 = 8[/tex]

but, is true for 3

[tex]3^3-1 = 27 -1 = 26 > 4*3 = 12[/tex]

lets say that the relationship is true for n, this is

[tex]n^3 -1 \ge 4 n[/tex]

lets add 4 on each side, this is

[tex]n^3 -1 + 4 \ge 4 n + 4[/tex]

[tex]n^3 + 3 \ge 4 (n + 1)[/tex]

now

[tex](n+1)^3 = n^3 +3 n^2 + 3 n + 1[/tex]

[tex](n+1)^3 \ge n^3 + 3 n [/tex]

if [tex]n \ge 1[/tex] then [tex]3 n \ge 3[/tex] , so

[tex](n+1)^3 \ge n^3 + 3 n \ge n^3 + 3 [/tex]

[tex](n+1)^3 \ge  n^3 + 3 \ge 4 (n + 1)  [/tex]

[tex](n+1)^3  \ge 4 (n + 1)  [/tex]

and this is what we were looking for!

So, for any natural equal or greater than 3, the relationship is true.

Answer:

90 mi/h

Step-by-step explanation:

Given,

For first 30 miles, her speed is 30 miles per hour,

Let x be her speed in miles per hour for another 30 miles,

Since, here the distance are equal in each interval,

So, the average speed of the entire journey

[tex]=\frac{\text{Average speed for first 30 miles + Average speed for another 30 miles}}{2}[/tex]

[tex]=\frac{30+x}{2}[/tex]

According to the question,

[tex]\frac{30+x}{2}=60[/tex]

[tex]30+x=120[/tex]

[tex]\implies x = 90[/tex]

Hence, she needs to go 90 miles per hour for remaining 30 miles.

Answer:

Eli walked 12 feet down the hall of his house to get to the door. He continued in a straight line out the door and across the yard to the mailbox, a distance of 32 feet.

He came straight back across the yard 14 feet and stopped to pet his dog.

Part A : find the image attached.

Part B : [tex]12+32+14=58[/tex] feet

Part C : [tex]32-14=18[/tex] feet from his house.

Answer:

Step-by-step explanation:

Let M be the heights of men in the United States and W be the heights of women in the United States

Given that M is N(69.1, 2.9) and W (63.7, 2.7)

For a husband and wife we have average height as

[tex]\frac{x+y}{2}[/tex]=Z (say)

[tex]Mean =E(z) = \frac{1}{2} [E(M)+E(W)] = 66.4 inches[/tex]

Var (Z) =[tex]Var (\frac{x+y}{2} )=\frac{1}{4}[Var(x)+ Var(y)+2 cov (x,y)]\\[/tex]

=[tex]0.25[2.9^2+2.7^2+2*r*sx*sy]\\= 0.25(20.398)\\=5.0995[/tex]

Mean = 66.4" and std dev = 2.258

Answer:

  (√138)/24

Step-by-step explanation:

[tex]\displaystyle\sqrt{\frac{46}{192}}=\sqrt{\frac{46\cdot 3}{192\cdot 3}}=\sqrt{\frac{138}{24^2}}=\frac{\sqrt{138}}{24}[/tex]

The child ingested approximately 20.18 milligrams of ferrous sulfate, which is determined by unit conversion.

Given data:

First, let's convert the 1.6 fluid ounces to millilitres:

[tex]1.6 \text{ fluid ounces} \times 29.5735 \text{ ml/oz} \approx 47.3176 \text{ ml}[/tex]

Now, calculate the amount of ferrous sulfate in milligrams:

[tex]\text{Concentration of ferrous sulfate} = \frac{2}{3} \text{ gr per 5 ml} \\= \frac{2}{3} \times \frac{1}{5} \text{ gr/ml}[/tex]

Since 1 grain (gr) is approximately equal to 64.79891 milligrams (mg), we'll convert the concentration to milligrams per millilitre (mg/ml):

[tex]\text{Concentration of ferrous sulfate} = \frac{2}{3} \times \frac{1}{5} \times 64.79891 \text{ mg/ml}[/tex]

To calculate the total amount of ferrous sulfate ingested:

[tex]\text{Amount of ferrous sulfate} = 47.3176 \text{ ml} \times \frac{2}{3} \times \frac{1}{5} \times 64.79891 \text{ mg/ml}[/tex]

Calculating this gives:

[tex]\text{Amount of ferrous sulfate} \approx 20.1817 \text{ mg}[/tex]

Therefore, the child ingested approximately 20.18 milligrams of ferrous sulfate.

Learn more about the unit conversion here:

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Answer:

$102,000

Step-by-step explanation:

As provided we have,

Variable cost components

Direct material = $6.10

Direct labor = $3.60

Variable manufacturing overhead = $1.40

Sales Commission = $1.10

It is assumed that sales commission is based on number of units and therefore, will be charged as variable cost and not the fixed cost.

Variable administrative expense = $0.55

As for 8,000 units this entire variable cost components will have same cost.

As the fixed cost is based on 10,000 units it will differ per unit on 8,000 units but the variable cost component will not change per unit.

Thus, total variable cost for 8,000 units = ($6.10 + $3.60 + $1.40 + $1.10 + $0.55) [tex]\times[/tex]  8,000 = $102,000

Answer:

2.063X10^-4

0.999

0.833

3.26X10^-3

9.85x10^-3

Step-by-step explanation:

B="At least one component needs service during the warranty period"

C="All three components need service during the warranty period"

D="Only the receiver needs service during the warranty period"

E="Exactly one of the three components needs service during the warranty period"

P(A1)=0.0595

P(A2)=0.0598

P(A3)=0.058

a) P(A1∩A2∩A3)=P(A1)P(A2)P(A3)=0.0595*0.0598*0.058=2.063X10^-4

b) P(B)=1-P(A1∩A2∩A3)=1-2.063X10^-4=0.999

c) P(C)=P(A1'∩A2'∩A3')=P(A1')P(A2')P(A3')=0.9405*0.9402*0.942=0.833

d) P(D)=P(A1'∩A2∩A3)=P(A1')P(A2)P(A3)=0.9405*0.0598*0.058=3.26X10^-3

e) P(E)=P(A1'∩A2∩A3)+P(A1∩A2'∩A3)+P(A1∩A2∩A3')=3.26X10^-3 + 0.0595*0.9402*0.058+0.0595*0.0598*0.942=9.85x10^-3

Answer:

a. 13 years

b. 27950 students

Step-by-step explanation:

Let x be the number of years since the initial year 2005. In 2005, [tex]x = 0[/tex], in 2006, [tex]x = 1[/tex], and so on.

For college "A" the number of students is given by the function:

[tex]NSA = 18200 + 750x[/tex]

For college "B" the number of students is given by the function:

[tex]NSB = 34450 - 500x[/tex]

Matching these expressions you have the value of x for which the two colleges have the same number of students.

[tex]NSA = 18200 + 750x = 34450 - 500x = NSB\\\\750x + 500x = 34450 - 18200\\\\1250x = 16250\\\\x = 16250/1250 = 13[/tex]

Then, the two colleges will have the same number of students 13 years later, that is, in 2018.

The number of students that each colleges will have in 2018 will be [tex]NSA = NSB = 18200 + 750 (13) = 27950[/tex] students.

Answer:

0=6

Step-by-step explanation:

-0.25yxyx0=6

Answer:

So you may know that that the time for downloading a file  of A (MB) is B (seconds), where A and B are numbers.

Then suppose if some file size is C, which is 10 times more than A, and you assume that the download speed sees this as downloading 10 times the A (MB) file. Thus the time for downloading this second file will be 10*B (seconds).

As, if                            [tex]C\ (MB) file=10*A\ (MB) file[/tex]

then                             [tex]C\ (MB)file =10*B\ (seconds)[/tex]

B(seconds) is the time required to download A (MB) file.

Answer:

75 is 10.34% of 725.50.

Step-by-step explanation:

Let 75 is x % of 725.50.

It can be written as mathematical equation.

x % of 725.50 = 15

[tex]\frac{x}{100}\times 725.50=75[/tex]

Multiply both sides by 100.

[tex]\frac{x}{100}\times 725.50\times 100=75\times 100[/tex]

[tex]x\times 725.50=7500[/tex]

Divide both sides by 725.50.

[tex]\frac{x\times 725.50}{725.50}=\frac{7500}{725.50}[/tex]

[tex]x=\frac{7500}{725.50}[/tex]

[tex]x=10.3376981392[/tex]

[tex]x\approx 10.34[/tex]

Therefore, 75 is 10.34% of 725.50.

Answer:

Ans. This company rate of return is 57.07% annual

Step-by-step explanation:

Hi, well, we need to set the following equation with the data of our problem. The equation as follows.

[tex]FutureValue=\frac{A((1+r)^{n} -1}{r}[/tex]

Where:

A= $30,000

n= 5

r= the rate that we are looking for

Everything should look like this

[tex]450,000=\frac{30,000((1+r)^{5} -1)}{r}[/tex]

Now, you can see how hard it is to solve this equation for "r", therefore we go to MS excel so we can use the function "Seek Goal". Please see the attached excel spread sheet.

The instructions to the "seek goal" function are as follows.

Set Cell:                 C5

To Value:         450000

By changing cell: C4

The result is in yellow.

Ans. 57.07% annual

Best of luck.

Answer:

[tex]y(x)=Ce^{-x}[/tex]

Step-by-step explanation:

We are given that y'=-y

We have to find the solution of given differential equation

[tex]\frac{dy}{dx}=-y[/tex]

[tex]\frac{dy}{y}=-dx[/tex]

Integrating on both sides then we get

[tex]lny=-x+c[/tex]

[tex]if\;lna=b [/tex] then [tex] a=e^b[/tex]

[tex]y=e^{-x+c}=e^{-x}\cdot e^c[/tex]

[tex]y=Ce^{-x}[/tex] because [tex]e^c=Constant=C[/tex]

Hence,[tex]y(x)=Ce^{-x}[/tex] is the solution of given differential  equation  y'=-y

Answer:

a.Neither injective nor surjective

b.Injective but not surjective

c.Surjective but not injective

Step-by-step explanation:

Injective function:It is also called injective function.If f(x)=f(y)

Then, x=y

Surjective function:It is also called Surjective function.

If function is onto function then Range of function=Co-domain of function

a.We are given that

[tex]f:R\rightarrow R[/tex]

[tex]f(x)=x^2[/tex]

It is not injective because

f(1)=1 and f(-1)=1

Two elements have same image.

If function is one-to-one then every element have different image.

Function is not surjective because negative elements have not pre-image in R

Therefore, Co-domain not equal to range.

Given function neither injevtive nor surjective.

b.[tex]f:N\rightarrow N[/tex]

[tex]f(n)=n^2[/tex]

If [tex]f(n_1)=f(n_2)[/tex]

[tex]n^2_1=n^2_2[/tex]

[tex]n_1=n_2[/tex]

Because N={1,2,3,...}

Hence, function is Injective.

2,3,4,.. have no pre- image in N

Therefore, function is not surjective

Because Range not equal to co-domain.

Hence, given function is injective but not Surjective.

c.[tex]f:Z\times Z \rightarrow Z[/tex]

f(n,k)=n+k

It is not injective because

f(1,2)=1+2=3

f(2,1)=2+1=3

Hence, by definition of one-one function it is not injective.

For every element belongs to Z we can find pre- image in [tex]Z\times Z[/tex]

Hence, function is surjective.

For function (a) f(x) = x² and function (b) f(n) = n², neither is injective or surjective. Function (c) f(n, k) = n + k is both injective and surjective.

The functions injective and surjective are important concepts in mathematics that pertain to the type of mappings functions perform from one set to another. An injective, or one-to-one, function assigns distinct outputs to distinct inputs, while a surjective, or onto, function ensures that every element of the function's codomain is mapped to by at least one element of its domain.

Answer:

number of way is [tex]\frac{59!}{50)!9!}[/tex]

Step-by-step explanation:

given data:

total number of problems 100

total points for each problem 5

let ten problems are

[tex]x_1, x_2,........., x_{10}[/tex]

according to the given information

[tex]x_1 +x_2 +.......+x_{10} = 100[/tex]

[tex]x_i \geq 5[/tex]

where i =1 + 10

so, number of way integer point can assign are

   [tex]^{(n+r-1)}C_{(r-1)}[/tex]

where

r = 10

[tex]n = 100 - 10\times 5 = 50[/tex]

so, we have

[tex]^{(59)}C_{9}[/tex]

[tex]\frac{59!}{59-9)!9!}[/tex]

number of way is [tex]\frac{59!}{50)!9!}[/tex]

Answer:

  The product is approximately 3600.

Step-by-step explanation:

The whole number nearest the first factor is 6.

The hundred nearest the second factor is 600.

The product of these is 6×600 = 3600, your estimated product.

Proof with Step-by-step explanation:

Since it is given that f(x) is decreasing thus we infer that

[tex]\frac{d}{dx}f(x)=f'(x)<0[/tex]

Similarly it is given that g(x) is decreasing thus we infer that

[tex]\frac{d}{dx}g(x)=g'(x)<0[/tex]

Now let [tex]y=fog=f[g(x)]\\\\[/tex]

Differentiating both sides with respect to 'x' and using chain rule of differentiation we get

[tex]dy/dx=\frac{d}{dx}f[g(x)]\\\\y'=f'[g((x)]\times g'(x)\\\\\\[/tex]

Now since both f'(x) and g'(x) are both negative(<0) we can infer that their product is always positive (>0)

[tex]\therefore y'(x)>0[/tex]

Thus fog(x) is an increasing function.

In Mathematics, the composition of two decreasing functions, in this case f and g, results in an increasing function. This happens because if for any two numbers, the larger number results in a smaller function value in a decreasing function, then applying this principle twice (i.e., making a composition) reverses the order, therefore increasing the function.

For all real numbers, if f and g are decreasing functions, the composite function f o g is indeed increasing. The reason behind this is a principle in mathematics that states the composition of two decreasing functions is an increasing function.

To understand this, consider the fact that a function g is decreasing if for any two real numbers x1 and x2, if x1 < x2, then g(x1) > g(x2). Similarly, a function f is decreasing if for any two real numbers y1 and y2, if y1 < y2, then f(y1) > f(y2).

Now, consider the composite function f o g. For any two real numbers x1 and x2, if x1 < x2, then g(x1) > g(x2) (since g is decreasing). Since f is also a decreasing function, f(g(x1)) > f(g(x2)), meaning that the composite function increases as x increases.

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Answer:

a) The variable of the study is: milligrams of nitrogen per liter of water.

This is the amount that needs to be measured and analyzed to reach conclusions in the study.

b) The variable is quantitative. The quantitative variables are those that represent quantities. This variables can be measured on a continuous or discrete scale. Then, all the variables that you can measure or count are quantitative variables(height of trees, number of passengers per car, wind speed, milligrams of nitrogen per liter, etc). On the other hand, qualitative variables are those that can’ t be measured, and they represent attributes, like apple colors (red, green), size of trousers (small, medium, large) and so on.

c) The population under study is the milligrams of nitrogen per liter of water that are in the entire lake. You can estimate the parameters of the population by taking samples (In the example, 28 samples are taken).

Answer:

The frequency of the car is 0.0233743 laps/sec and 1.4024 laps were completed in 60 seconds

Step-by-step explanation:

We are given that he fastest ever qualifying lap was in 1987 by Bill Elliott, with a time of 42.782 seconds.

[tex]Frequency = \frac{1}{Time}[/tex]

[tex]Frequency = \frac{1}{42.782}[/tex]

[tex]Frequency = 0.0233743[/tex]

No. of laps completed in 42.782 seconds. = 1

No. of laps completed in 60 seconds. = [tex]\frac{1}{42.782} \times 60[/tex]

                                                                = [tex]1.4024[/tex]

Hence the frequency of the car is 0.0233743 laps/sec and 1.4024 laps were completed in 60 seconds

Answer:

  standard divisor: 525.00

Step-by-step explanation:

The standard divisor is found by dividing the total number of residents to be represented, by the number of representatives. Here, that is 7875 divided by 15. The resulting standard divisor is 525 (exactly).

By Hamilton's method, the number of residents in each district is divided by this number, and the integer and fraction of the quotient are recorded.

If the total of integers is less than the total number of representatives, then the difference is apportioned to the districts according to the values of the fractions.

Here, the total of quotient integers is 13, so 2 additional representatives need to be apportioned. The two largest fractions are 0.77 associated with District 5, and 0.58 associated with District 1. These two districts, then, get the "extra" representatives. (1 is added to the "integer" for those districts.)

Final assignments of representatives to districts are seen in the rightmost column of the spreadsheet.

Answer:

D. H0​: p ≤0.36 H1​: p>0.36

[tex]$$(Right tail test) Z_{\alpha}=Z_{0.1}=1.28155[/tex]

[tex]Z_p=-0.25516[/tex]

D. Reject H0. There is sufficient evidence to support the hypothesis that the proportion of admissions officers who visit an applying​ students' social networking page has increased in the past year

Step-by-step explanation:

To solve this problem, we run a hypothesis test about the population proportion.

[tex]$$Sample proportion: $\bar P=0.47\\Sample size n=100$\\Significance level \alpha=0.10$\\(Left tail test) Z_{1-\alpha}=Z_{0.90}=-1.28155$\\(Right tail test) Z_{\alpha}=Z_{0.1}=1.28155$\\(Two-tailed test) $Z_{1-\alpha/2}=Z_{0.95}=-1.64485$ and $ Z_{\alpha/2}=Z_{0.05}=1.64485\\\\[/tex]

The appropriate hypothesis system for this situation is:

[tex]H_0:\pi_0=0.35\\H_a:\pi_0 > 0.35\\\\$Proportion in the null hypothesis is:\\\pi_0=0.35\\\\[/tex]

[tex]$$The test statistic is $Z=\frac{(\bar P-\pi_0)\sqrt{n}}{\sqrt{\pi_0(1-\pi_0)}}\\$The calculated statistic is Z_c=\frac{(0.47-0.35)\sqrt{100}}{\sqrt{0.35(1-0.35)}}=-0.25516\\p-value = P(Z \geq Z_c)=0.01620\\\\[/tex]

Since, the calculated statistic [tex]Z_c[/tex] is greater than critical [tex]Z_{\alpha}[/tex], the null hypothesis should be rejected. There is enough statistical evidence to state that the proportion of admissions officers who visit an applying​ students' social networking page has increased in the past year.

The vertex, y-intercept, and x-intercepts of the parabola f(x) = 4x² - 6x - 8 can be found using the formulas -b/2a for the vertex, f(0) for the y-intercept, and the quadratic formula for the x-intercepts.

The equation given is of the form f(x) = ax² + bx + c, which is a parabola's quadratic equation with a = 4, b = -6, and c = -8.

A) To find the vertex of the parabola, we use the formula -b/2a to calculate the x-coordinate of the vertex, and then substitute this value back into the function to get the y-coordinate. In this case, the vertex is at (-b/2a, f(-b/2a)).

B) The vertical intercept or the y-intercept of a parabola is simply the point at which it crosses the y-axis. This occurs at x = 0, so to find it, simply substitute x = 0 into the equation.

C) To find the x-intercept or 'a' intercepts, we solve the equation f(x) = 0, which takes the form of a quadratic equation ax² + bx + c = 0. We find the roots or solutions using the quadratic formula -b ± √b² - 4ac/2a.

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Answer:

205.4 miles

Step-by-step explanation:

As per the question,

We have been provided that an airplane started flies 165 miles from point A  in the direction of 130 degrees and after that, it changes its direction to 245 degrees and travel to 80 miles again, which is drawn in the figure as below:

From the figure,

Let XY be the plane.

Now,

By using the cosine rule in triangle APR, we get

AP² = AR² + PR² - 2 × AR × PR × cos 115°

Now put the value from figure:

AP² = 165² + 80² - 2 × 165 × 80 × (-0.325)

AP² = 42205

AP = 205.4 miles.

Hence, the airplane is 205.4 miles far from point A.