5) The mean salary of 5 employees is $40300. The median is $38500. The lowest paid employee's salary is $32000. If the lowest paid employee gets a $3100 raise, then ...

a) What is the new mean?
New Mean = $


b) What is the new median?
New Median = $

Answer:

After the 6th day, beginning with the 7th day, he is walking less than 0.5 km.

Step-by-step explanation:

To find a percent of a number, change the percent to a decimal and multiply by the number. To find 90% of a number, change 90% to a decimal and multiply by the number.

90% = 0.9

He first walks 1 km. The next day, he walks 90% of 1 km. To find 90% of 1 km, multiply 0.9 by 1 km. It is 0.9 km. For the next day, he walks 90% of 0.9 km, which is 0.9 * 0.9 km = 0.81 km. To find how much he walks each day, multiply what he walked on the previous day by 0.9.

Now you can find out how much he walks each day until you see he walks less than 0.5 km.

Day 1: 1 km

Day 2: 1 km * 0.9 = 0.9 km

Day 3: 0.9 km * 0.9 = 0.81 km

Day 4: 0.81 km * 0.9 = 0.729 km

Day 4: 0.729 km * 0.9 = 0.6561

Day 5: 0.6561 km * 0.9 = 0.59049 km

Day 6: 0.59049 km * 0.9 = 0.531441 km

Day 7: 0.531441 km * 0.9 = 0.4782969 km

On the 6th day, he is still walking more than 0.5 km, but by the 7th day, he is walking less than 0.5 km.

Answer: After the 6th day, beginning with the 7th day, he is walking less than 0.5 km.

After the 6th day, beginning with the 7th day, he exists walking less than 0.5 km.

To find a percent of a number, change the percent to a decimal and multiply by the number. To discover 90% of a number, change 90% to a decimal and multiply by the number.

90% = 0.9

He first walks 1 km. The next day, he walks 90% of 1 km.

To discover 90% of 1 km, multiply 0.9 by 1 km. It is 0.9 km.

For the subsequent day, he walks 90% of 0.9 km, which exists

0.9 [tex]*[/tex] 0.9 km = 0.81 km.

To find how much he walks each day, multiply what he walked on the last day by 0.9.

Now find out how much he walks each day until sees he walks less than 0.5 km.

Day 1: 1 km

Day 2: 1 km [tex]*[/tex] 0.9 = 0.9 km

Day 3: 0.9 km [tex]*[/tex] 0.9 = 0.81 km

Day 4: 0.81 km [tex]*[/tex] 0.9 = 0.729 km

Day 4: 0.729 km [tex]*[/tex] 0.9 = 0.6561

Day 5: 0.6561 km [tex]*[/tex] 0.9 = 0.59049 km

Day 6: 0.59049 km [tex]*[/tex] 0.9 = 0.531441 km

Day 7: 0.531441 km [tex]*[/tex] 0.9 = 0.4782969 km

On the 6th day, he stands still walking more than 0.5 km, but by the 7th day, he exists walking less than 0.5 km.

Answer: After the 6th day, beginning with the 7th day, he exists walking less than 0.5 km.

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Answer:

160/1001, 175/1001

Step-by-step explanation:

i) There are:

₈C₁ ways to choose 1 new camera from 8 new cameras

₆C₃ ways to choose 3 refurbished cameras from 8 refurbished cameras

₁₄C₄ ways to choose 4 cameras from 14 cameras

The probability is:

P = ₈C₁ ₆C₃ / ₁₄C₄

P = 8×20 / 1001

P = 160 / 1001

P ≈ 0.160

ii) At most one new camera means either one new camera or no new cameras.  We already found the probability of one new camera.  The probability of no new cameras is the same as the probability of choosing 4 refurbished cameras:

P = ₆C₄ / ₁₄C₄

P = 15 / 1001

So the total probability is:

P = 160/1001 + 15/1001

P = 175/1001

P ≈ 0.175

To find the probability that one new camera will be selected, use the binomial coefficient and calculate the probability of selecting one new camera and three cameras that are not new. To find the probability of at most one new camera, calculate the probabilities of selecting zero new cameras and one new camera and add them together.

To find the probability that one new camera will be selected, we need to calculate the probability of selecting one new camera and three cameras that are not new. The total number of ways to select four cameras from 14 without replacement is given by the binomial coefficient 14 choose 4, which is equal to 14!/(4!(14-4)!). The probability of selecting one new camera is given by the product of the probability of selecting one new camera (8/14) and the probability of selecting three cameras that are not new (6/13 * 5/12 * 4/11). To find the probability that at most one new camera will be selected, we need to calculate the probabilities of selecting zero new cameras and one new camera and add them together.

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The shop sells 216 "I LOVE NY" T-shirts each day. Therefore, over the course of one week, the shop sells 1512 T-shirts.

The shop in Times Square is open from 9 am to 9 pm, which means the shop operates for 12 hours. Since there are 60 minutes in an hour, this shop is open for a total of 720 minutes each day.

Three "I LOVE NY" T-shirts are sold every 10 minutes. So, in 720 minutes, the number of T-shirts sold would be 720 ÷ 10 = 72 sets of three T-shirts. Therefore, 72 sets x 3 shirts = 216 T-shirts are sold per day.

Finally, to calculate the weekly total, it is necessary to multiply the daily total by 7 (the number of days in a week). So, 216 T-shirts x 7 days = 1512 T-shirts sold in one week.

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Answer:

The right statement is sin(J) = cos(L) ⇒ answer D

Step-by-step explanation:

* Lets describe the figure

- LKJ is a right triangle, where K is a right angle

∵ m∠K = 90°

∵ LJ is opposite to angle K

∴ LJ is the hypotenuse

∵ LJ = 219

∵ KJ = 178

- By using Pythagoras Theorem

∵ (LJ)² = (LK)² + (KJ)²

∴ (219)² = (LK)² + (178)² ⇒ subtract (178)² from both sides

∴ (LK)² = (219)² - (178)²

∴ (LK)² = 16277

∴ LK = √16277 = 127.58

* Lets revise how to find the trigonometry function

# sin Ф = opposite/hypotenuse

# cos Ф = adjacent/hypotenuse

# tan Ф = opposite/adjacent

∵ LK is the opposite side to angle J

∵ LJ is the hypotenuse

∵ sin(J) = LK/LJ

∵ LK = 127.58 , LJ = 219

∴ sin(J) = 127.58/219 = 0.583

∵ LK is the adjacent side to angle L

∵ LJ is the hypotenuse

∵ cos(L) = LK/LJ

∵ LK = 127.58 , LJ = 219

∴ cos(L) = 127.58/219 = 0.583

∴ sin(J) = cos(L)

* The right statement is sin(J) = cos(L)

Answer:

B. 1.25

Step-by-step explanation:

Probability is as below

[tex]0 \leqslant p(a) \leqslant 1[/tex]

When P(A) = 0, it is an unlikely event

When P(A) = 1, it is a certain event

Answer:

  r = 11

Step-by-step explanation:

Substituting the given value for C, we have ...

  22€ = 2€·r

Dividing by the coefficient of r, we get

  22€/(2€) = r = 11

To find the value of 'r' using the equation C = 2€r; C = 22€, you simply need to substitute the given value of C into the equation and solve for r. In this case, r equals 11.

Given : C = 2€r; C = 22€

To find the value of r, substitute C = 22€ into the equation:

22€ = 2€r

Divide both sides by 2€ to isolate r:
r = 11

Answer:

Step-by-step explanation:

so u do 2+2 4=243==32===3=424=4=234=234=32=43=4=34

Answer:

20

Step-by-step explanation:

Answer:

a. The two events are dependent.

b. [tex]P(A\cap B)[/tex]= [tex]\frac{1}{220}[/tex].

Step-by-step explanation:

Given

Total coins =220

Number of Indian pennies= 6

A: When one of the 220 coins is randomly selected, it is one of the Indian pennies.

Therefore , the probability of getting an  Indian pennies=[tex]\frac{6}{220 }[/tex]

By using formula of probability=[tex]\frac{Number \; of\; favourable\; cases}{total\; number \; of \;cases}[/tex]

Probability of getting an  Indian pennies=[tex]\frac{3}{110}[/tex]

B: When another one of the 220 coins is randomly selected( with replacement) , It is also one of the Indian pennies.

Therefore, probability of getting an Indian pennies=[tex]\frac{6}{220}[/tex]

Probability of getting an Indian pennies =[tex]\frac{3}{110}[/tex]

[tex]A\cap B[/tex]: 1

[tex]P(A\cap B)=\frac{1}{220}[/tex]

If two events are independent. Then

[tex]P(A\cap B)= P(A)\times p(B)[/tex]

P(A).P(B)= [tex]\frac{3}{110} \times \frac{3}{110}[/tex]=[tex]\frac{9}{12100}[/tex]

Hence, [tex]P(A\cap B)\neq P(A).P(B)[/tex]

Therefore, the two events are dependent.

b. Probability that events A and B both occur

Number of favourable cases when both events A and B occur=1

Total coins=220

Probability=[tex]\frac{Number \; of\; favourable \; cases}{Total\; number\; of\; cases}[/tex]

[tex]P(A\cap B)=\frac{1}{220}[/tex]

Answer:

a) {{}, {a}}.

b) {{}, {a}, {b}, {a, b}}.

c) {{}, {1}, {2}, {1, 2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}, {4}, {1, 4}, {2, 4}, {1, 2, 4}, {3, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}.

Step-by-step explanation:

The power set of a set is the set of all subset of the set in question. The number of power sets (including the empty set) of a set with [tex]n[/tex] (where [tex]n \in \mathbb{Z}[/tex]) unique elements is [tex]2^{n}[/tex].

In other words, there shall be

This explanation shows how to find the power set using binary numbers (only 0 and 1.) (Credit: Mathsisfun.)

List all the binary numbers that are equivalent to decimals ranging from 0 to [tex]2 - 1 = 1[/tex].

[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 0 \\ 1 & 1\end{array}[/tex].

Reverse the original set. Each digit in the binary number corresponds to a member of the original set (i.e. a letter in a) and b) or a number in c).) 0 means that the element is absent in the subset and 1 means that the element is present.

[tex]\begin{array}{c|l}a & \text{Element of the Power Set}\\ 0 & \{\}\\ 1 & \{a\}\end{array}[/tex].

The power set of a) thus contains:

Similarly, list all the binary numbers that are equivalent to decimals ranging from 0 to [tex]4 - 1 = 3[/tex].

[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 00 \\ 1 & 01 \\ 2 & 10 \\ 3 & 11\end{array}[/tex].

[tex]\begin{array}{cc|l}b & a & \text{Element of the Power Set}\\ 0 & 0 & \{\}\\ 0 & 1 & \{a\}\\ 1 & 0 & \{b\} \\ 1 & 1 & \{a, b\}\end{array}[/tex].

The power set of b) thus contains:

Similarly, list all the binary numbers that are equivalent to decimals ranging from 0 to [tex]16 - 1 = 15[/tex].

[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 0000 \\ 1 & 0001 \\ 2 & 0010 \\ 3 & 0011\\4 & 0100 \\ 5 & 0101\\ 6 & 0110 \\ 7 & 0111\\ 8 & 1000\\ 9 & 1001\\ 10 & 1010\\ 11 & 1011\\ 12& 1100 \\13 & 1101 \\ 14 & 1110\\ 15 & 1111 \end{array}[/tex].

[tex]\begin{array}{cccc|l}4 & 3 & 2 &1& \text{Element of the Power Set}\\ 0 & 0 & 0 & 0 &\{\}\\ 0 & 0 & 0 & 1 & \{1\}\\ 0 & 0 & 1 & 0 & \{2\} \\ 0 & 0 &1 & 1 & \{1, 2\} \\ 0 & 1 & 0 & 0 & \{3\} \\ 0 & 1 & 0 & 1& \{1, 3\}\\ 0 & 1 & 1 & 0& \{2, 3\}\\ 0 & 1 & 1 & 1 & \{1, 2, 3\} \\ 1 & 0 & 0 & 0 & \{4\} \\ 1 & 0 & 0 & 1 & \{1, 4\}\\ 1& 0 & 1 &0&\{2, 4\}\\ 1 & 0 & 1 & 1 &\{1, 2, 4\}\\ 1 & 1 & 0 & 0 & \{3, 4\} \\ 1 & 1 & 0 & 1 & \{1, 3, 4\} \\ 1 & 1 & 1 & 0 & \{2, 3, 4\} \\ 1 & 1 & 1 & 1 & \{1, 2, 3, 4\}\end{array}[/tex].

The power set of c) thus contains:

Answer:

22 minutes after 7:00 P.M. they will be closest.

Step-by-step explanation:

A boat heading south travelling for t hours at the rate of 20 km/h, so the distance x = 20t

The another boat will reach the dock after travelling another 1-t hours at the rate of 15 km/h, so the distance =

y = 15 - 15t

D = d²  = x² + y²

D = (20t)² + (15 - 15t)²

dD/dt = -2(15² )( 1-t ) +2 × 20² × t

dD/dt = 2 (15² + 20²) × t -2 ( 15 )² = 0

t = [tex]\frac{2(15)^{2}}{(2\times15^{2}+2\times20^{2})}[/tex]

t = 0.36 hours = 0.36 × 60 = 21.6 minutes ≈ 22 minutes

Therefore, the distance is minimized 22 minutes after 7 pm.

The two boats were closest together 12 minutes after 7:00 PM.

To find the time when the two boats were closest together, we can first determine the position of each boat at 8:00 PM. The boat traveling south will have traveled for 1 hour at a speed of 20 km/h, so it would be 20 km south of the dock. The boat traveling east will have traveled for 1 hour at a speed of 15 km/h, so it would be 15 km east of the dock. We can then calculate the distance between the two boats by using the Pythagorean theorem. The distance is the square root of the sum of the squares of the distances traveled south and east, which is approximately 25 km. Since both boats started at the dock at 7:00 PM, to find the time when they were closest together, we can subtract the time traveled by the boat heading south until it reaches the closest point to the other boat from 60 minutes. The boat heading south will have traveled (20/25) * 60 minutes, which is 48 minutes. So, the two boats were closest together 12 minutes after 7:00 PM.

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Answer:IF each vendor has his own price or (ppower) so far every single vendor will have his own price.

Step-by-step explanation:

The graph show\ing the demand (D) and supply (S = MC) curves in the market for hot dogs indicate: Competitive market.

In a market were their is competition, when demand and supply curves intersect this indicate market equilibrium.

Based on the graph the market equilibrium price will be $1.50 per hot dog while on the other hand the market equilibrium quantity will be 250 hot dogs which  is the point were demand and supply intersect.

Inconclusion the market for hot dogs indicate: Competitive market.

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First, lets convert them into multipliers:

The multiplier for a:

                             25% increase = 1.25

                             40% decrease = 0.6

                             40% increase  = 1.4

Now to work out the overall percentage change, we just times all of the multipliers together, and convert it back to a percentage:

1.25 x 0.6 x 1.4 = 1.05    

So the overall multiplier is 1.05

And a multiplier of 1.05 = a 5% increase.

That means that the percentage change is + 5%

_________________________________________

Answer:

The percentage change in shares from the start of 2013 to the end of 2015 is:

+ 5%

_______________________________________

Note: if you haven't been taught multipliers - then ask and I'll try my best to explain!

Take up to the third-order derivative:

[tex]f(x)=(-3x+15)^{3/2}[/tex]

[tex]f'(x)=\dfrac32(-3x+15)^{1/2}(-3)=-\dfrac92(-3x+15)^{1/2}[/tex]

[tex]f''(x)=-\dfrac94(-3x+15)^{-1/2}(-3)=\dfrac{27}4(-3x+15)^{-1/2}[/tex]

[tex]f'''(x)=-\dfrac{27}8(-3x+15)^{-3/2}(-3)=\dfrac{81}8(-3x+15)^{-3/2}[/tex]

Evaluate each derivative at [tex]x=a=2[/tex]:

[tex]f(2)=9^{3/2}=27[/tex]

[tex]f'(2)=-\dfrac929^{1/2}=-\dfrac{27}2[/tex]

[tex]f''(2)=\dfrac{27}4\dfrac1{9^{1/2}}=\dfrac94[/tex]

[tex]f'''(2)=\dfrac{81}8\dfrac1{9^{3/2}}=\dfrac38[/tex]

Then the Taylor polynomial is

[tex]P_3(x)=f(2)+f'(2)(x-2)+\dfrac{f''(2)}2(x-2)^2+\dfrac{f'''(2)}6(x-2)^3[/tex]

[tex]P_3(x)=27-\dfrac{27}2(x-2)+\dfrac98(x-2)^2+\dfrac1{16}(x-2)^3[/tex]

Answer:

There are two times for the ball to reach a height of 64 feet:

1 second after thrown ⇒ the ball moves upward

2 seconds after thrown ⇒ the ball moves downward

Step-by-step explanation:

* Lets explain the function to solve the problem

- h(t) models the height of the ball above the ground as a function

 of the time t

- h(t) = -16t² + 48t + 32

- Where h(t) is the height of the ball from the ground after t seconds

- The ball is thrown upward with initial velocity 48 feet/second

- The ball is thrown from height 32 feet above the ground

- The acceleration of the gravity is -32 feet/sec²

- To find the time when the height of the ball is above the ground

  by 64 feet substitute h by 64

∵ h(t) = -16t² + 48t + 32

∵ h = 64

∴ 64 = -16t² + 48t + 32 ⇒ subtract 64 from both sides

∴ 0 = -16t² + 48t - 32 ⇒ multiply the both sides by -1

∴ 16t² - 48t + 32 = 0 ⇒ divide both sides by 16 because all terms have

  16 as a common factor

∴ t² - 3t + 2 = 0 ⇒ factorize it

∴ (t - 2)(t - 1) = 0

- Equate each bracket by zero to find t

∴ t - 2 = 0 ⇒ add 2 to both sides

∴ t = 2

- OR

∴ t - 1 = 0 ⇒ add 1 to both sides

∴ t = 1

- That means the ball will be at height 64 feet after 1 second when it

 moves up and again at height 64 feet after 2 seconds when it

 moves down

* There are two times for the ball to reach a height of 64 feet

  1 second after thrown ⇒ the ball moves upward

  2 seconds after thrown ⇒ the ball moves downward

Answer:

20m/s

Step-by-step explanation:

This can be solved using the acceleration / velocity equations.

Specifically,

v² = u² + 2as

Where

v = final velocity = what we need to find

u = initial velocity = given as 10.0m/s

a = acceleration = given as 0.5m/s²

s = distance = 300m

Hence,

v² = 10² + (2)(0.5) (300)

    = 100 + 300

    =400

v = √400 = 20 m/s

Jessica has a total of 22,275 different possible schedules to choose from for her next semester given the number of sections for each class and assuming there are no time conflicts.

Jessica is creating her semester schedule and there are 15 sections of English 101, 9 sections of Spanish 102, 11 sections of Biology 102, and 15 sections of College Algebra. To figure out how many different possible schedules are available, we need to multiply the number of sections for each class.

Therefore, the total number of different possible schedules Jessica can choose is calculated as follows:

15 (English 101) * 9 (Spanish 102) * 11 (Biology 102) * 15 (College Algebra) = 22,275 possible schedules.

This is under the assumption that there are no time conflicts between the different classes.

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[tex]f(x_1,\ldots,x_n)=x_1+\cdots+x_n=\displaystyle\sum_{i=1}^nx_i[/tex]

[tex]{x_1}^2+\cdots+{x_n}^2=\displaystyle\sum_{i=1}^n{x_i}^2=4[/tex]

The Lagrangian is

[tex]L(x_1,\ldots,x_n,\lambda)=\displaystyle\sum_{i=1}^nx_i+\lambda\left(\sum_{i=1}^n{x_i}^2-4\right)[/tex]

with partial derivatives (all set equal to 0)

[tex]L_{x_i}=1+2\lambda x_i=0\implies x_i=-\dfrac1{2\lambda}[/tex]

for [tex]1\le i\le n[/tex], and

[tex]L_\lambda=\displaystyle\sum_{i=1}^n{x_i}^2-4=0[/tex]

Substituting each [tex]x_i[/tex] into the second sum gives

[tex]\displaystyle\sum_{i=1}^n\left(-\frac1{2\lambda}\right)^2=4\implies\dfrac n{4\lambda^2}=4\implies\lambda=\pm\frac{\sqrt n}4[/tex]

Then we get two critical points,

[tex]x_i=-\dfrac1{2\frac{\sqrt n}4}=-\dfrac2{\sqrt n}[/tex]

or

[tex]x_i=-\dfrac1{2\left(-\frac{\sqrt n}4\right)}=\dfrac2{\sqrt n}[/tex]

At these points we get a value of [tex]f(x_1,\cdots,x_n)=\pm2\sqrt n[/tex], i.e. a maximum value of [tex]2\sqrt n[/tex] and a minimum value of [tex]-2\sqrt n[/tex].

Answer:

450 km

Step-by-step explanation:

Let's say Va is the speed of the car from city A, Ta is the time it spent traveling, and Da is the distance it traveled.

Similarly, Vb is the speed of the car from city B, Tb is the time it spent traveling, and Db is the distance it traveled.

Given:

Va = Vb - 10

Ta₁ = Tb₁ = 5

Ta₂ = Tb₂ + 4.5

Db₂ = 150

Find:

D = Da₁ + Db₁ = Da₂ + Db₂

Distance = rate × time

In the first scenario:

Da₁ = Va Ta₁

Da₁ = (Vb - 10) (5)

Da₁ = 5Vb - 50

Db₁ = Vb Tb₁

Db₁ = Vb (5)

Db₁ = 5Vb

So:

D = Da₁ + Db₁

D = 10Vb - 50

In the second scenario:

Da₂ = Va Ta₂

Da₂ = (Vb - 10) (Tb₂ + 4.5)

Da₂ = Vb Tb₂ + 4.5Vb - 10Tb₂ - 45

Db₂ = Vb Tb₂

150 = Vb Tb₂

Substituting:

Da₂ = 150 + 4.5Vb - 10Tb₂ - 45

Da₂ = 105 + 4.5Vb - 10Tb₂

Da₂ = 105 + 4.5Vb - 10 (150 / Vb)

Da₂ = 105 + 4.5Vb - (1500 / Vb)

So:

D = Da₂ + Db₂

D = 105 + 4.5Vb - (1500 / Vb) + 150

D = 255 + 4.5Vb - (1500 / Vb)

Setting this equal to the equation we found for D from the first scenario:

10Vb - 50 = 255 + 4.5Vb - (1500 / Vb)

5.5Vb - 305 = -1500 / Vb

5.5Vb² - 305Vb = -1500

5.5Vb² - 305Vb + 1500 = 0

11Vb² - 610Vb + 3000 = 0

(Vb - 50) (11Vb - 60) = 0

Vb = 50, 5.45

Since Vb > 10, Vb = 50 km/hr.

So the distance between the cities is:

D = 10Vb - 50

D = 10(50) - 50

D = 450 km

Answer:

The solution is [tex]x\leq 1[/tex]

All real numbers less than or equal to 1

The graph in the attached figure

Step-by-step explanation:

we have

[tex]8\geq 3x+5[/tex]

Subtract 5 both sides

[tex]8-5\geq 3x[/tex]

[tex]3\geq 3x[/tex]

Divide by 3 both sides

[tex]1\geq x[/tex]

Rewrite

[tex]x\leq 1[/tex]

The solution is the interval ------> (-∞,1]

All real numbers less than or equal to 1

In a number line the solution is the shaded area at left of x=1 (close circle)

see the attached figure

Answer:

The equation of tangent line is [tex]y=5x-33  [/tex]

Step-by-step explanation:

We need to find out the equation of tangent line.

Given :- g(6)=−3  and  g'(6)=  5

If  g(6)=−3

then the point on the line for the required tangent is  (6,−3)

If  g'(6)=  5

then the slope of the tangent at that point is  45

The tangent line can be specified by the slope-point form of the equation:

[tex](y-y_1)=m(x-x_1)[/tex]

which in this case is

[tex](y-(-3))=5(x-6)[/tex]

[tex](y+3)=(5x-30)[/tex]

subtract both the sides by 3,

[tex]y+3-3=5x-30-3[/tex]

[tex]y=5x-33[/tex]

Therefore, the equation of tangent line is [tex]y=5x-33[/tex]

Final answer:

The equation of the tangent line to the graph of y = g(x) at the point where x = 6 is y = 5x - 33, using the point-slope form and the given point (6, -3) with the slope of 5.

Explanation:

To find the equation of the tangent line to the graph at a particular point, we use the point-slope form of a line, given by y - y1 = m(x - x1), where (x1, y1) is the point on the graph and m is the slope at that point. Given that g(6) = -3 and g'(6) = 5, we can substitute these values into the point-slope form to get the equation of the tangent line. The equation is then y + 3 = 5(x - 6), which simplifies to y = 5x - 33.

Answer: 0.008

Step-by-step explanation:

Given: Mean : [tex]\mu=1200[/tex]

Standard deviation : [tex]\sigma = 60[/tex]

Sample size : [tex]n=36[/tex]

The formula to calculate z-score is given by :_

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x= 1224, we have

[tex]z=\dfrac{1224-1200}{\dfrac{60}{\sqrt{36}}}=2.4[/tex]

The P-value = [tex]P(z>2.4)=1-P(z<2.4)=1-0.9918024=0.0081976\approx0.008[/tex]

Hence, the probability that the sample mean will be larger than 1224 =0.008

Answer:

In 20 years, the population will be about 80.3 thousand people

Step-by-step explanation:

If our first time is 0 and the population that goes along with that time is 70,000, we have a coordinate point where x is the time (0), and y is the population at that time (70).  Our next time is 10 years later, when the population is 75,000.  The coordinate point for that set of data is (10, 75).  Now we will use those 2 points in the standard form of an exponential equation to write the model for this particular situation.  

Exponential equations are of the form

[tex]y=a(b)^x[/tex]

where x and y are the coordinates from our points, one at a time; a is the initial value, and b is the growth rate.  Filling in an equation with the first set of data:

[tex]70=a(b)^0[/tex]

Anything raised to the power of 0 = 1, so b to the power of 0 = 1 and we simply have that a = 70.

Now we use that value of a along with the x and y from the next coordinate pair to solve for b:

[tex]75=70(b)^{10}[/tex]

Begin by dividing both sides by 70 to get

[tex]1.071428571=b^{10}[/tex]

Undo the power of 10 on the right by taking the 10th root of both sides:

[tex](1.071428571)^{\frac{1}{10}}=(b^{10})^{\frac{1}{10}}[/tex]

On the right side we simply have b now, and on the left we have

1.006923142=b

Now we have a and b to write the model for this situation:

[tex]y=70(1.006923142)^x[/tex]

We need to find y, the population, in x = 20 years:

[tex]y=70(1.006923142)^{20}[/tex]

Raise the parenthesis to the 20th power giving you

y = 70(1.147959784) and

y = 80.3 thousand people

Final answer:

The exponential growth function that fits the given data is y(t) = a * (1 + r)^t. Using this function, we can find the population 20 years from now.

Explanation:

The exponential growth function that fits the given data is:

y(t) = a * (1 + r)^t

where:

To find the growth rate per year, we can use the formula: r = (P/P0)^(1/t) - 1

Given that the population is projected to be 75,000 in 10 years, we can substitute these values into the formula to find the growth rate:

r = (75,000/70,000)¹/¹⁰ - 1 ≈ 0.035

The exponential growth function becomes:

y(t) = 70,000 * (1 + 0.035)^t

To find the population 20 years from now, we can substitute t = 20 into the exponential growth function:

y(20) = 70,000 * (1 + 0.035)²⁰ ≈ 95,212

Answer:

Quotient will be 1.25

Step-by-step explanation:

First we convert decimal numbers to fractions. So write down the decimal divided by 1 and then multiply both top and bottom with 10 for every number after decimal point.

Here we found for  .4375  = [tex]\frac{4375}{10000}[/tex]

and .35 =   [tex]\frac{35}{100}[/tex]

Now we divide both the numbers as

= [tex]\frac{\frac{4375}{1000} }{\frac{35}{100} }[/tex]

= [tex]\frac{4375}{1000}[/tex] × [tex]\frac{100}{35}[/tex]

= [tex]\frac{125}{100}[/tex]

= 1.25

Quotient will be 1.25

Final answer:

The quotient of 0.4375 divided by 0.35 is 1.25, which rounded to the tenths place is 1.3.

Explanation:

The student is asking to find the quotient of two decimal numbers, which is a basic arithmetic operation involving division. The numbers are 0.4375 and 0.35. To find the quotient, simply divide 0.4375 by 0.35.

Using a calculator or performing the division manually, you would proceed as follows:

Adjust the decimals by multiplying both numbers by 100 to make them whole numbers, resulting in 43.75 divided by 35.

Perform the division to get the preliminary result: 43.75 / 35 = 1.25.

Since we need to round the final answer to the tenths place based on the least precise number given (35.5 g), round 1.25 to one decimal place, which is 1.3 (1.25 rounds up because the next digit, 5, is equal to or greater than 5).

Therefore, the quotient of 0.4375 divided by 0.35, rounded to the tenths place, is 1.3.

Answer: There is a probability of 40% of getting a school or district uses digital content and uses it as part of their curriculum.

Step-by-step explanation:

Since we have given that

Probability that schools or districts in a country use digital content = 80%

Probability that schools uses digital content as a part of their curriculum out of 80% = [tex]\dfrac{5}{10}[/tex]

So, the probability that a selected school or district uses digital content and uses it as  a part of their curriculum is given by

[tex]\dfrac{80}{100}\times \dfrac{5}{10}\\\\=0.8\times 0.5\\\\=0.4\\\\=40\%[/tex]

Hence, there is a probability of 40% of getting a school or district uses digital content and uses it as part of their curriculum.

The probability that a randomly selected school or district uses digital content and uses it as part of their curriculum is 40%.

To find the probability that a randomly selected school or district uses digital content and uses it as part of their curriculum, we need to multiply the probabilities of these events occurring.

Given that 80% of K-12 schools or districts use digital content and 5 out of 10 of these schools use it as part of their curriculum, we can calculate the probability as:

P(Uses digital content and uses it as part of curriculum) = P(Uses digital content) x P(Uses it as part of curriculum | Uses digital content)

Substituting the values, we have:

P(Uses digital content and uses it as part of curriculum) = 0.80 x 0.50 = 0.40 or 40%

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Answer:

[tex]\frac{dh}{dt}=\frac{160}{169\pi }  ft/min[/tex]

Step-by-step explanation:

This is a classic related rates problem.  Gotta love calculus!

Start out with the formula for the volume of a cone, which is

[tex]V=\frac{1}{3}\pi r^2h[/tex]

and with what we know, which is [tex]\frac{dV}{dt}=40[/tex]

and the fact that the diameter = height (we will come back to that in a bit).

We need to find [tex]\frac{dh}{dt}[/tex] when h = 13

The thing we need to notice now is that there is no information given to us that involves the radius.  It does, however, give us a height.  We need to replace the r with something in terms of h.  Let's work on that first.

We know that d = h.  Because d = 2r, we can say that 2r = h, and solving for r gives us that [tex]r=\frac{h}{2}[/tex].

Now we can rewrite the formula with that replacement:

[tex]V=\frac{1}{3}\pi  (\frac{h}{2})^2h[/tex]

Simplify that all the way down to

[tex]V=\frac{1}{12}\pi  h^3[/tex]

The derivative of that function with respect to time is

[tex]\frac{dV}{dt}=\frac{1}{12}\pi(3h^2)\frac{dh}{dt}[/tex]

Filling in what we have gives us this:

[tex]40=\frac{1}{12}\pi (3)(13)^2\frac{dh}{dt}[/tex]

Solve that for the rate of change of the height:

[tex]\frac{dh}{dt}=\frac{160}{169\pi } \frac{ft}{min}[/tex]

or in decimal form:

[tex]\frac{dh}{dt}=.95\pi  \frac{ft}{min}[/tex]

This involves relationship between rates using Calculus.

dh/dt = 0.3 ft/min

Volumetric rate; dv/dt = 40 ft³/min

height of pile; h = 13 ft

We are not given the diameter here but as we are dealing with a right circular cone, we will assume that the diameter is equal to the height.

Thus; diameter; d = 13 ft

radius; r = h/2 = d/2 = 13/2

radius; r= 6.5 ft

V = ¹/₃πr²h

We want to find how fast the height is increasing and this is dh/dt.

Thus, we will need to express r in the volume formula in terms of h;

V = ¹/₃π(h/2)²h

V = ¹/₃π(h²/4)h

V = ¹/₁₂πh³

dV/dt = 3(¹/₁₂πh²)dh/dt

dV/dt = ¹/₄πh²(dh/dt)

Plugging in the relevant values, we have;

40 = ¹/₄π × 13² × (dh/dt)

dh/dt = (40 × 4)/(π × 13²)

dh/dt = 0.3 ft/min

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Check the pictures below.

if we knew the roots/solutions of the equation, we can set h(s) = 0 and solve for "s" to find out how many seconds is it when the height is 0.

if you notice in the first picture, when f(x) = 0, is when the parabola hits a root/solution or the ground, for David he'll be hitting the water surface, and the equation that has both of those roots/solutions conspicuous is

h(s) = -4.9(s - 2)(s + 1).

Answer:

D

Step-by-step explanation:

[tex]\[\frac{x}{x^{2} +3x+2} -\frac{1}{(x+2)(x+1)} =\frac{x}{x^2+3x+2} -\frac{1}{x(x+1)+2(x+1)} =\frac{x}{x^{2}+3x+2 } -\frac{1}{x^{2} +2x+x+2} =\frac{x}{x^{2} +3x+2} -\frac{1}{x^{2} +3x+2} =\frac{x-1}{x^{2} +3x+2} \][/tex]

Answer:

he is correct

Step-by-step explanation:

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Answer:

Nico invest [tex]\$2,100[/tex] at 7% and [tex]x=\$900[/tex] at 3%

Step-by-step explanation:

we know that

The simple interest formula is equal to

[tex]I=P(rt)[/tex]

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

At 7%

[tex]t=1\ years\\ P=\$x\\r=0.07[/tex]

substitute in the formula above

[tex]I1=x(0.07*1)[/tex]

[tex]I1=0.07x[/tex]

At 3%

[tex]t=1\ years\\ P=\$(x-1,200)\\r=0.03[/tex]

substitute in the formula above

[tex]I2=(x-1,200)(0.03*1)[/tex]

[tex]I2=0.03x-36[/tex]

The total interest is equal to

I=I1+I2

I=$174

substitute

[tex]174=0.07x+0.03x-36[/tex]

[tex]0.10x=174+36[/tex]

[tex]0.10x=210[/tex]

[tex]x=\$2,100[/tex]

[tex]x-1,200=2,100-1,200=\$900[/tex]

therefore

Nico invest [tex]\$2,100[/tex] at 7% and [tex]x=\$900[/tex] at 3%

Step-by-step explanation:

[tex]x-\text{the number}\\\\\text{Nine times the difference of a number and 3}:\\\\\boxed{9\times(x-3)=9(x-3)}[/tex]

[tex]9(x-3)\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\=9x+(9)(-3)\\\\=9x-27[/tex]