A football is kicked upward from a height of 6 feet with an initial speed of 70 feet per second. Use the formula to find the balls height 3 seconds after it was kicked

Answer:

3 meters.

Step-by-step explanation:

400 rev / minute = 400 × 60 rev / 60 minutes

= 24,000 rev / hour

24,000 × C = 72,000 m : C is the circumference

C = 3 meters

Answer:

3 meters

Step-by-step explanation:

72 km / hour * 1 hour/ 60 min  * 1000m/ 1 km

72000 meters /60 minute

1200 meters / minute

velocity = radius * w

Where w is 2*pi * the revolutions per minute

1200 = r * 2 * pi *400

1200 / 800 pi = r

1.5 /pi = r meters

We want to find the circumference

C = 2 * pi *r

C = 2* pi ( 1.5 / pi)

C = 3 meters

Answer:

(1238.845 ;1285.376)

Step-by-step explanation:

Conditions for constructing a confidence interval :

Data must be random

Distribution should be normal and independent ;

Based on the conditions above ; data meets initial conditions ;

C. I = sample mean ± margin of error

Given the data :

1241 1210 1267 1314 1211 1299 1246 1280 1291

Mean, xbar = Σx / n = 11359 / 9 = 1262.11

The standard deviation, s = [√Σ(x - xbar)²/n - 1]

Using a calculator ; s = 37.525

The confidence interval :

C.I = xbar ± [Tcritical * s/√n]

Tcritical(0.10 ; df = n - 1 = 9 - 1 = 8)

Tcritical at 90% = 1.860

C. I = 1262.11 ± [1.860 * 37.525/√9]

C.I = 1262.11 ± 23.266

(1238.845 ;1285.376)

± 23.266

The margin of error :

[Tcritical * s/√n]

[1.860 * 37.525/√9]

C.I = ± 23.266

The length of wire used for the equilateral triangle is approximately 5.61 meters.

The remaining length of wire used for the circle will be 9 - 5.61 ≈ 3.39 meters.

Here,

To minimize the total area of both figures, we need to find the optimal cut point for the wire.

Let's assume the length of the wire used for the equilateral triangle is x meters, and the remaining length of the wire used for the circle is (9 - x) meters.

For the equilateral triangle:

An equilateral triangle has all three sides equal in length.

Let's call each side of the triangle s meters. Since the total length of the wire is x meters, each side will be x/3 meters.

The formula to find the area of an equilateral triangle with side length s is:

Area = (√(3)/4) * s²

Substitute s = x/3 into the area formula:

Area = (√(3)/4) * (x/3)²

Area = (√(3)/4) * (x²/9)

Now, for the circle:

The circumference (perimeter) of a circle is given by the formula:

Circumference = 2 * π * r

Since the remaining length of wire is (9 - x) meters, the circumference of the circle will be 2π(9 - x) meters.

The formula to find the area of a circle with radius r is:

Area = π * r²

To find the area of the circle, we need to find the radius.

Since the circumference is equal to 2πr, we can set up the equation:

2πr = 2π(9 - x)

Now, solve for r:

r = (9 - x)

Now, substitute r = (9 - x) into the area formula for the circle:

Area = π * (9 - x)²

Now, we want to minimize the total area, which is the sum of the areas of the triangle and the circle:

Total Area = (√(3)/4) * (x²/9) + π * (9 - x)²

To find the optimal value of x that minimizes the total area, we can take the derivative of the total area with respect to x, set it to zero, and solve for x.

d(Total Area)/dx = 0

Now, find the critical points and determine which one yields the minimum area.

Taking the derivative and setting it to zero:

d(Total Area)/dx = (√(3)/4) * (2x/9) - 2π * (9 - x)

Setting it to zero:

(√(3)/4) * (2x/9) - 2π * (9 - x) = 0

Now, solve for x:

(√(3)/4) * (2x/9) = 2π * (9 - x)

x/9 = (8π - 2πx) / (√(3))

Now, isolate x:

x = 9 * (8π - 2πx) / (√(3))

x(√(3)) = 9 * (8π - 2πx)

x(√(3) + 2π) = 9 * 8π

x = (9 * 8π) / (√(3) + 2π)

Now, we can calculate the value of x:

x ≈ 5.61 meters

So, the length of wire used for the equilateral triangle is approximately 5.61 meters.

The remaining length of wire used for the circle will be 9 - 5.61 ≈ 3.39 meters.

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[tex] \sf \: d \: = 3.7m \\ \sf \: r \: = \frac{3.7}{2} = 1.85 \: m\\ \\ \sf \: c \: = \pi {r}^{2} \\ \\ \sf \: c \: = \pi ({1.85})^{2} \\ \sf c = 1.85 \times 1.85 \times \pi \\ \sf \: c = \boxed {\underline{ \bf a. \: 3.4225\pi \: m ^{2} }}[/tex]

Answer:

1+i

Step-by-step explanation:

I do believe i to be the imaginary unit.

Let's write out some partial sums from power=0 to power=7 or whatever we need to see a pattern.

i^0=1

i^0+i^1=1+i

i^0+i^1+i^2=1+i+-1=i

i^0+i^1+i^2+i^3=i+i^3=i+-i=0

i^0+i^1+i^2+i^3+i^4=0+i^4=0+1=1

Hmmm.... we might see 1+i, then i, then 0 again.... let's see.

i^0+i^1+i^2+i^3+i^4+i^5=1+i

Coolness so we should see a pattern

Sum from power=0 to power=multiples of 4 will give us 1.

Sum from power=0 to power=remainder of 1 when final power is divided by 4 gives us 1+i.

Sum from power=0 to power=remainder of 2 when final power is divided by 4 gives us i.

Sum from power=0 to power=remainder of 3 when final power is divided by 4 gives us 1

0.

So 2021 divided by 4....

Since 2020 is a multiple of 4, then 2021 has a remainder of 1 when divided by 4.

So the answer is 1+i.

Answer:

I think around $36

Hope it helps!

Answer:

It depends...

Step-by-step explanation:

It depends how much weeks are in the month if there are three weeks and no extra days then you would have an answer of about 1093 (exact: 1093.33333333). just divide the number of weeks by the number of money.

9514 1404 393

Answer:

  (d)  -1/32

Step-by-step explanation:

It may be easier to rearrange the expression so it has positive exponents.

  [tex]\dfrac{1}{2^{-2}x^{-3}y^5}=\dfrac{2^2x^3}{y^5}=\dfrac{4(2)^3}{(-4)^5}=-\dfrac{4\cdot8}{1024}=\boxed{-\dfrac{1}{32}}[/tex]

Answer:

30 clocks

Step-by-step explanation:

Set up an equation:

Variable x = number of clocks

1200 + 20x = 60x

Isolate variable x:

1200 = 60x - 20x

1200 = 40x

Divide both sides by 40:

30 = x

Check your work:

1200 + 20(30) = 60(30)

1200 + 600 = 1800

1800 = 1800

Correct!

Answer:

106 people.

Step-by-step explanation:

Logistic equation:

The logistic equation is given by:

[tex]P(t) = \frac{K}{1+Ae^{-kt}}[/tex]

In which

[tex]A = \frac{K - P_0}{P_0}[/tex]

K is the carrying capacity, k is the growth/decay rate, t is the time and P_0 is the initial value.

Suppose a rumor is going around a group of 191 people. Initially, only 38 members of the group have heard the rumor.

This means that [tex]K = 191, P_0 = 38[/tex], so:

[tex]A = \frac{191 - 38}{38} = 4.03[/tex]

Then

[tex]P(t) = \frac{191}{1+4.03e^{-kt}}[/tex]

3 days later 68 people have heard it.

This means that [tex]P(3) = 68[/tex]. We use this to find k.

[tex]P(t) = \frac{191}{1+4.03e^{-kt}}[/tex]

[tex]68 = \frac{191}{1+4.03e^{-3k}}[/tex]

[tex]68 + 274.04e^{-3k} = 191[/tex]

[tex]e^{-3k} = \frac{191-68}{274.04}[/tex]

[tex]e^{-3k} = 0.4484[/tex]

[tex]\ln{e^{-3k}} = \ln{0.4484}[/tex]

[tex]-3k = \ln{0.4484}[/tex]

[tex]k = -\frac{\ln{0.4484}}{3}[/tex]

[tex]k = 0.2674[/tex]

Then

[tex]P(t) = \frac{191}{1+4.03e^{-0.2674t}}[/tex]

How many people are expected to have heard the rumor after 6 days total have passed since it was initially spread?

This is P(6). So

[tex]P(6) = \frac{191}{1+4.03e^{-0.2674*6}} = 105.52[/tex]

Rounding to the nearest whole number, 106 people.

Answer:

a)  The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24,   P(2,0) = 0,   P(0,1) = 0,  P(1,1) = 0.24,  P(2,1)= 0.16

b)  P( W = 0 ) = 0.36,    P(W = 1 ) = 0.48,  P(W = 2 ) = 0.16

c) P ( z = 0 ) = 0.6

  P ( z = 1 ) = 0.4

Step-by-step explanation:

Number of head on first toss = Z

Total Number of heads on 2 tosses = W

% of head occurring = 40%

% of tail occurring = 60%

P ( head ) = 2/5 ,    P( tail ) = 3/5

a) Determine the joint probability distribution of W and Z

P( W =0 |Z = 0 ) = 0.6         P( W = 0 | Z = 1 ) = 0

P( W = 1 | Z = 0 ) = 0.4        P( W = 1 | Z = 1 ) = 0.6

P( W = 1 | Z = 0 ) = 0           P( W = 2 | Z = 1 ) = 0.4

The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24,   P(2,0) = 0,   P(0,1) = 0,  P(1,1) = 0.24,  P(2,1)= 0.16

B) Marginal distribution of W

P( W = 0 ) = 0.36,    P(W = 1 ) = 0.48,  P(W = 2 ) = 0.16

C) Marginal distribution of Z ( pmf of Z )

P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

Part(a): The required joint probability of W and Z is ,

[tex]P(0,0)=0.36,P(1,0)=0.24,P(2,0)=0,P(0,1)=0,P(1,1)=0.24,\\\\P(2,1)=0.16[/tex]

Part(b): The pmf (marginal distribution) of W is,

[tex]P(w=0)=0.36,P(w=1)=0.48,P(w=2)=0.16[/tex]

Part(c): The pmf (marginal distribution) of Z is,

[tex]P(z=0)=0.6,P(z=1)=0.4[/tex]

Part(a):

The joint distribution is,

[tex]P(w=0\z=0)=0.6,P(w=1|z=0)=0.4,P(w=2|z=0)=0[/tex]

Also,

[tex]P(w=0\z=1)=0,P(w=1|z=1)=0.6,P(w=2|z=1)=0.4[/tex]

Therefore,

[tex]P(0,0)=0.36,P(1,0)=0.24,P(2,0)=0,P(0,1)=0,P(1,1)=0.24,\\\\P(2,1)=0.16[/tex]

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Answer:

0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.

Step-by-step explanation:

We have the mean, which means that the Poisson distribution is used to solve this question.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

A mean of 3.5 customers arrive hourly at the drive-through window.

This means that [tex]\mu = 3.5[/tex]

What is the probability that, in any hour, more than 5 customers will arrive?

This is:

[tex]P(X > 5) = 1 - P(X \leq 5)[/tex]

In which

[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]

Then

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-3.5}*3.5^{0}}{(0)!} = 0.0302[/tex]

[tex]P(X = 1) = \frac{e^{-3.5}*3.5^{1}}{(1)!} = 0.1057[/tex]

[tex]P(X = 2) = \frac{e^{-3.5}*3.5^{2}}{(2)!} = 0.1850[/tex]

[tex]P(X = 3) = \frac{e^{-3.5}*3.5^{3}}{(3)!} = 0.2158[/tex]

[tex]P(X = 4) = \frac{e^{-3.5}*3.5^{4}}{(4)!} = 0.1888[/tex]

[tex]P(X = 5) = \frac{e^{-3.5}*3.5^{5}}{(5)!} = 0.1322[/tex]

Finally

[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0302 + 0.1057 + 0.1850 + 0.2158 + 0.1888 + 0.1322 = 0.8577[/tex]

[tex]P(X > 5) = 1 - P(X \leq 5) = 1 - 0.8577 = 0.1423[/tex]

0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.

Answer:  

a) h(x; 6, 9, 17).

b) P[X=2] = 0.2036

P[X ≤ 2] = 0.2466

P[X ≥ 2] = 0.9570.

c) Mean  = 3.176.

Variance = 1.028.

Standard deviation = 1.014.

Step-by-step explanation:

From the given details K=6, n=9, N=-17.

We conclude that it is the hypergeometric distribution:  

a) h(x; 6, 9, 17).

b)

[tex]P[X=2]=\frac{(^{g}C_{2})^{17-9}C_{6-2}}{^{17}C_{6}\textrm{}}[/tex]

P[X=2] = 0.2036

P[X ≤ 2] = P(x=0)+ P(x=1) + P(x=2)

P[X ≤ 2] = 0.2466

P[X ≥ 2] = 1-[P(x=0)+P(x=1)]

P[X ≥ 2] = 0.9570.

c)

Mean= [tex]n\frac{K}{N}[/tex]

            = 3.176.

Variance = [tex]n\frac{K}{N}( \frac{N-K}{N})(\frac{N-n}{n-1} )[/tex]

               = 2.824 x 0.6471 x 0.5625

               = 1.028.

Standard deviation = [tex]\sqrt{1.028}[/tex] = 1.014.

Step-by-step explanation:

-27 okay 3^-3 its same as 3^3

Answer: A)

[tex]3^{-3}[/tex]

[tex]3^{-3}=\frac{1}{3^3}[/tex]

[tex]=\frac{1}{3^3}[/tex]

[tex]3^3=27[/tex]

[tex]=\frac{1}{27}[/tex]

OAmalOHopeO

Let x, y, and z be the amounts (in liters, L) of the 25%, 35%, and 55% solutions that the chemist used.

She ended up with 120 L of solution, so

x + y + z = 120 … … … [1]

x L of 25% acid solution contains 0.25x L of acid. Similarly, y L of 35% solution contains 0.35y L of acid, and z L of 55% solution contains 0.55z L of acid. The concentration of the new solution is 40%, so that it contains 0.40 (120 L) = 48 L of acid, which means

0.25x + 0.35y + 0.55z = 48 … … … [2]

Lastly,

z = 3y … … … [3]

since the chemist used 3 times as much of the 55% solution as she did the 35% solution.

Substitute equation [3] into equations [1] and [2] to eliminate z :

x + y + 3y = 120

x + 4y = 120 … … … [4]

0.25x + 0.35y + 0.55 (3y) = 48

0.25x + 2y = 48 … … … [5]

Multiply through equation [5] by -2 and add that to [4] to eliminate y and solve for x :

(x + 4y) - 2 (0.25x + 2y) = 120 - 2 (48)

0.5x = 24

x = 48

Solve for y :

x + 4y = 120

4y = 72

y = 18

Solve for z :

z = 3y

z = 54

Answer:

2 pages per minute

Step-by-step explanation:

Take the number of pages and divide by the number of minutes

46 pages / 23 minutes

2 pages per minute

2 Pages per Minute

46 ÷ 23 = 2

Anthony can read 2 pages per minute.

Answer:

1.  c. Poisson

2. 0.9592 = 95.92% probability that in any one minute at least one purchase is​ made.

3. 0.0017 = 0.17% probability that no one makes a purchase in the next 2​ minutes.

Step-by-step explanation:

We have only the mean, which means that the Poisson distribution is used to solve this question, and thus the answer to question 1 is given by option c.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

Mean of 3.2 minutes:

This means that [tex]\mu = 3.2n[/tex], in which n is the number of minutes.

2. What is the probability that in any one minute at least one purchase is​ made?

[tex]n = 1[/tex], so [tex]\mu = 3.2[/tex].

This probability is:

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-3.2}*3.2^{0}}{(0)!} = 0.0408[/tex]

So

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0408 = 0.9592[/tex]

0.9592 = 95.92% probability that in any one minute at least one purchase is​ made.

3. What is the probability that no one makes a purchase in the next 2​ minutes?

2 minutes, so [tex]n = 2, \mu = 3.2(2) = 6.4[/tex]

This probability is P(X = 0). So

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-6.4}*6.4^{0}}{(0)!} = 0.0017[/tex]

0.0017 = 0.17% probability that no one makes a purchase in the next 2​ minutes.

Answer:

Step-by-step explanation:

Answer: distributive property

Step-by-step explanation: the 4 is multiplied by everting in the parenthesis

Answer:

55

Step-by-step explanation:

55 appears 3 times, which is the most repetition in the data set

Answer:

55

Step-by-step explanation:

Mode = number that appears most often

The number 55 appears 3 times which is the most out of the other numbers

Hence mode = 55

Answer:

im not too sure but try using a cartesuan plane and measure it precisely using a protractor then key in the measurements. Im not entirely sure its the correct method tho

Answer:

8×1000 milligrams

8000 milligrams

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

For each subset it can either contain or not contain an element. For each element, there are 2 possibilities. Multiplying these together we get 27 or 128 subsets. For generalisation the total number of subsets of a set containing n elements is 2 to the power n.

total subsets

Answer:

Z =  -1.60

it is low ... it appears that for this problem 2 standard deviations below must be reached to be considered "unusual"

Step-by-step explanation:

Answer:

x=3

Step-by-step explanation:

7(x - 7) = -28

x - 7 = -4

x = 3

Answer:

x = 3

Step-by-step explanation:

Your goal is to isolate the x from the other numbers.

-28 = 7(x - 7)

Distribute the 7 to the (x - 7)

You will end up with:

-28 = 7x - 49

Add 49 to both sides of the equation to further isolate the x

21 = 7x

Finally, divide both sides by 7 so x is by itself

x = 3

Answer:

268.40

Step-by-step explanation:

We can write a ratio to solve

2.5 meters        22 meters

-----------------  = --------------

30.5 dollars       x dollars

Using cross products

2.5 * x = 30.5 * 22

2.5x =671

Divide each side by 2.5

2.5x / 2.5 = 671/2.5

x =268.4

Answer:

im pretty sure the degree is 4.

Step-by-step explanation:

Answer:

23) No

24) No

25) Yes

Step-by-step explanation:

Question 23)

We want to determine if a zero exists between 1 and 2 for the function:

[tex]f(x)=x^2-4x-5[/tex]

Find the zeros of the function. We can factor:

[tex]\displaystyle 0 = (x-5)(x+1)[/tex]

Zero Product Property:

[tex]x-5=0\text{ or } x+1=0[/tex]

Solve for each case. Hence:

[tex]\displaystyle x = 5\text{ or } x=-1[/tex]

Therefore, our zeros are at x = 5 and x = -1.

In conclusion, a zero does not exist between 1 and 2.

Question 24)

We have the function:

[tex]f(x)=2x^2-7x+3[/tex]

And we want to determine if a zero exists between 1 and 2.

Factor. We want to find two numbers that multiply to (2)(3) = 6 and that add to -7.

-6 and -1 suffice. Hence:

[tex]\displaystyle \begin{aligned} 0 & = 2x^2-7x + 3 \\ & = 2x^2 -6x -x + 3 \\ &= 2x(x-3) - (x-3) \\ &= (2x-1)(x-3) \end{aligned}[/tex]

By the Zero Product Property:

[tex]2x-1=0\text{ or } x-3=0[/tex]

Solve for each case:

[tex]\displaystyle x=\frac{1}{2} \text{ or } x=3[/tex]

Therefore, our zeros are at x = 1/2 and x = 3.

In conclusion, a zero does not exist between 1 and 2.

Question 25)

We have the function:

[tex]f(x)=3x^2-2x-5[/tex]

And we want to determine if a zero exists between -2 and 3.

Factor. Again, we want to find two numbers that multiply to 3(-5) = -15 and that add to -2.

-5 and 3 works perfectly. Hence:

[tex]\displaystyle \begin{aligned} 0&= 3x^2 -2x -5 \\ &= 3x^2 +3x - 5x -5 \\ &= 3x(x+1)-5(x+1) \\ &= (3x-5)(x+1)\end{aligned}[/tex]

By the Zero Product Property:

[tex]\displaystyle 3x-5=0\text{ or } x+1=0[/tex]

Solve for each case:

[tex]\displaystyle x = \frac{5}{3}\text{ or } x=-1[/tex]

In conclusion, there indeed exists a zero between -2 and 3.