Jayanta is raising money for the​ homeless, and discovers each church group requires 2 hr of letter writing and 1 hr of​ follow-up calls, while each labor union needs 2 hr of letter writing and 3 hr of ​follow-up. She can raise ​$125 from each church group and ​$175 from each union. She has a maximum of 20 hours of letter writing and 14 hours of ​follow-up available each month. Determine the most profitable mixture of groups she should contact and the most money she can raise in a month.

Step-by-step explanation:

mass of the object=m=4 kg

the net force exerted on it=F=24N=kgm/s²

the acceleration of the object=a=?

formula for calculating the acceleration=

F=ma

plug in the values

24kgm/s²=4kga

divide each side by 4

24kgm/s²/4 kg=4kga/4kg

a=6m/s²

Final answer:

The acceleration of the 4.0-kg object with a net force of 24 N is 6 m/s².

Explanation:

Acceleration: To find the acceleration, we use Newton's Second Law: F = ma. Given a net force of 24 N on a 4.0-kg object, the acceleration can be calculated as follows:

a = F/m = 24 N / 4.0 kg = 6 m/s²

Therefore, the acceleration of the object is 6 m/s².

Answer:

The correct option is 4.

Step-by-step explanation:

It is given that 10 men and 12 women will be seated in a row of 22 chairs.

Total possible ways to arrange n terms is n!.

Similarly,

Total possible ways to place 22 people on 22 chairs = 22!

[tex]\text{Total outcomes}=22![/tex]

It is given that all men will be seated side by side in 10 consecutive positions.

Total possible ways to place 10 people on 10 chairs = 10!

Let 10 men = 1 unit because all men will be seated side by side in 10 consecutive positions. 12 women = 12 units because women can any where.

Total number of units = 12 + 1 = 13.

Total possible ways to place 13 units = 13!

Total possible ways to place 10 men and 12 women, when all men will be seated side by side in 10 consecutive positions is

[tex]\text{Favorable outcomes}=10!\cdot 13![/tex]

The probability that all men will be seated side by side in 10 consecutive positions

[tex]P=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}=\frac{10!\cdot 13!}{22!}[/tex]

Therefore the correct option is 4.

Answer:

100000

Step-by-step explanation:

I did the operations in the picture, you only have to know that 0+1=01, 1+1=10

and 1+1+1=11.

Now, I don't know if you need to calculate the total add, I calculated it.

In this case, you need to know that 1+1+1+1=100.

The sum of the two Base 2 numbers is: [tex]\[ {10001001_2} \][/tex].

The sum of the given Base 2 (binary) numbers is calculated as follows:

 11,110,110

+ 101,101,111

Starting from the rightmost digit (least significant bit) and moving left, we add the digits:

- In the rightmost column, 0 + 1 = 1.

- In the next column, 1 + 1 = 10 (which is 0 in the current column and carry over 1 to the next column).

- Continuing this process, we add the digits along with any carry from the previous column.

 Let's continue the addition:

 11,110,110

+ 101,101,111

--------------

 1,000,1001

Here's the step-by-step process:

- 0 + 1 = 1 (no carry).

- 1 + 1 = 10 (0 in this column, carry 1).

- 1 + 0 + 1 (carry) = 10 (0 in this column, carry 1).

- 1 + 1 + 1 (carry) = 11 (1 in this column, carry 1).

- 1 + 0 + 1 (carry) = 10 (0 in this column, carry 1).

- 1 + 1 + 1 (carry) = 11 (1 in this column, carry 1).

- 1 + 0 + 1 (carry) = 10 (0 in this column, carry 1).

- 1 + 1 + 1 (carry) = 11 (1 in this column, carry 1).

- 1 + 0 + 1 (carry) = 10 (0 in this column, carry 1).

- 1 + 1 + 1 (carry) = 11 (1 in this column, carry 1).

- Finally, we have a carry of 1 that we add to the leftmost digit, giving us 1 + 1 = 10 (0 in this column, carry 1).

Since there are no more digits to add, we write down the 1 at the beginning:

 1,000,1001

Therefore, the sum of the two Base 2 numbers is:

[tex]\[ {10001001_2} \][/tex].

Answer:

P=0.3698  or 36.98%

Step-by-step explanation:

Complete the table by adding the totals to each column and row.

                                Pedestrian     Pedestrian

                                Intoxicated    Not intoxicated Totals

Driver Intoxicated           56                     71               127

Driver Not intoxicated   292                   522             814

Totals                            348                    593             941

The probability that the pedestrian was intoxicated or the driver was intoxicated is the opposite event of neither of them was intoxicated.  The total of cases when neither of them was intoxicated is 593. So the probability is:

P1=593/941=0.6302

The probability of the opposite event is one minus the probability calculated:

P=1-0.6302=0.3698

And this is the probability that the pedestrian was intoxicated or the driver was intoxicated.

Answer:

The probability is 0.0008.

Step-by-step explanation:

Let X represents the event of defective bulb,

Given, the probability of defective bulb, p = 20 % = 0.2,

So, the probability that bulb is not defective, q = 1 - p = 0.8,

The number of bulbs drawn, n = 10,

Since, binomial distribution formula,

[tex]P(x=r) = ^nC_r p^r q^{n-r}[/tex]

Where, [tex]^nC_r = \frac{n!}{r!(n-r)!}[/tex]

Hence, the probability that exactly 7 bulbs from the sample are defective is,

[tex]P(X=7)=^{10}C_7 (0.2)^7 (0.8)^{10-7}[/tex]

[tex]=120 (0.2)^7 (0.8)^3[/tex]

[tex]=0.000786432[/tex]

[tex]\approx 0.0008[/tex]

Calculate the probability of exactly 7 defective bulbs in a sample of 10 using the binomial distribution formula.

Binomial distribution:

P(X = 7) = C(10, 7) * (0.2)^7 * (0.8)^3 ≈ 0.2013

Answer:

5/2

Step-by-step explanation:

We are looking for p+q+r.

We are given:

p+q    =4          Equation 1.

    q+r=-2         Equation 2.                  

p     +r=3.          Equation 3.

I'm going to solve the first equation for p giving me p=4-q.

I'm going to solve the second equation for r giving me r=-2-q.

I'm going to plug this into the third equation.

p      +r=3

(4-q)+(-2-q)=3

4-q+-2-q=3

Combine like terms:

2-2q=3

Subtract 2 on both sides:

 -2q=1

Divide both sides by -2:

    q=-1/2.

Plug into the first equation to find p now.

First equation is p+q=4 with q=-1/2.

p+q=4

p+(-1/2)=4

Add 1/2 on both sides

p=4+1/2

p=9/2.

Let's find r now.

Using the second equation with p=9/2 and/or q=-1/2 we have:

q+r=-2  

-1/2+r=-2

Add 1/2 on both sides

      r=-2+1/2

      r=-3/2

So what is p+q+r?

Let's plug in our p,q, and r and find out!

p+q+r

9/2+-1/2+-3/2

9/2-4/2

5/2

a.

[tex]x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}[/tex]

By Fermat's little theorem, we have

[tex]x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5[/tex]

[tex]x^{11}\equiv x^7x^4\equiv x^5\equiv6\pmod 7[/tex]

5 and 7 are both prime, so [tex]\varphi(5)=4[/tex] and [tex]\varphi(7)=6[/tex]. By Euler's theorem, we get

[tex]x^4\equiv1\pmod5\implies x\equiv3^{-1}\equiv2\pmod5[/tex]

[tex]x^6\equiv1\pmod7\impleis x\equiv6^{-1}\equiv6\pmod7[/tex]

Now we can use the Chinese remainder theorem to solve for [tex]x[/tex]. Start with

[tex]x=2\cdot7+5\cdot6[/tex]

[tex]x=2\cdot7\cdot4\cdot2+5\cdot6[/tex]

[tex]x=2\cdot7\cdot4\cdot2+5\cdot6\cdot4\cdot6[/tex]

[tex]\implies x\equiv832\pmod{5\cdot7}\implies\boxed{x\equiv27\pmod{35}}[/tex]

b.

[tex]x^5\equiv3\pmod{64}[/tex]

We have [tex]\varphi(64)=32[/tex], so by Euler's theorem,

[tex]x^{32}\equiv1\pmod{64}[/tex]

Now, raising both sides of the original congruence to the power of 6 gives

[tex]x^{30}\equiv3^6\equiv729\equiv25\pmod{64}[/tex]

Then multiplying both sides by [tex]x^2[/tex] gives

[tex]x^{32}\equiv25x^2\equiv1\pmod{64}[/tex]

so that [tex]x^2[/tex] is the inverse of 25 mod 64. To find this inverse, solve for [tex]y[/tex] in [tex]25y\equiv1\pmod{64}[/tex]. Using the Euclidean algorithm, we have

64 = 2*25 + 14

25 = 1*14 + 11

14 = 1*11 + 3

11 = 3*3 + 2

3 = 1*2 + 1

=> 1 = 9*64 - 23*25

so that [tex](-23)\cdot25\equiv1\pmod{64}\implies y=25^{-1}\equiv-23\equiv41\pmod{64}[/tex].

So we know

[tex]25x^2\equiv1\pmod{64}\implies x^2\equiv41\pmod{64}[/tex]

Squaring both sides of this gives

[tex]x^4\equiv1681\equiv17\pmod{64}[/tex]

and multiplying both sides by [tex]x[/tex] tells us

[tex]x^5\equiv17x\equiv3\pmod{64}[/tex]

Use the Euclidean algorithm to solve for [tex]x[/tex].

64 = 3*17 + 13

17 = 1*13 + 4

13 = 3*4 + 1

=> 1 = 4*64 - 15*17

so that [tex](-15)\cdot17\equiv1\pmod{64}\implies17^{-1}\equiv-15\equiv49\pmod{64}[/tex], and so [tex]x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}[/tex]

Answer:

  (a) 120 square units (underestimate)

  (b) 248 square units

Step-by-step explanation:

(a) left sum

See the attachment for a diagram of the areas being summed (in orange). This is the sum of the first 4 table values for f(x), each multiplied by 2 (the width of the rectangle). Quite clearly, the curve is above the rectangle for the entire interval, so the rectangle area underestimates the area under the curve.

  left sum = 2(1 + 5 + 17 + 37) = 2(60) = 120 . . . . square units

(b) right sum

The right sum is the sum of the last 4 table values for f(x), each multiplied by 2 (the width of the rectangle). This sum is ...

  right sum = 2(5 +17 + 37 +65) = 2(124) = 248 . . . . square units

Answer:

5005

Step-by-step explanation:

In the question we have to form a committee of six from a club of consisting of 15 people.

This a simple case of selection of six people from a group of 15 people.

which can be done in

[tex]^{15}C_6= \frac{15!}{6!\times9!}=5005[/tex]

hence, the number of ways of forming committees of size six from a club of 15 members= 5005

Answer:

Giovanni was 0.5 miles from the finish line

Step-by-step explanation:

This is a problem of movement with constant velocity.

For this kind of problems, generally it is enough to remember the definition of average velocity v:

[tex]v=\frac{x}{t}[/tex]

Where x is the change in position that took place in an interval t.

First, find the time that Jean, who cycled at 24 miles per hour, spent on the race:

Isolating t from the last equation,

[tex]t=\frac{x}{v}[/tex], and replacing the data for Jean movement:

[tex]t=\frac{120}{24}=5h[/tex]

Second, find what was the distance that Giovanni had cycled when Jean crossed the line:

[tex]x=v*t\\ x=23.9*5=119.5[/tex]

When Jean crossed the line he had cycled 120 miles, and Giovanni 119.5; so Giovanni was 0.5 miles from the finish line.

Answer:

15 miles

Step-by-step explanation:

Let [tex]x[/tex] be the miles in the circular park path, [tex]t_{L}[/tex] the time Louisa takes to finish and [tex]t_{C}[/tex] the time Calvin takes to finish both in hours.

Then [tex]x[/tex], the longitude is equal to the velocity times the time used to finish. So

[tex]x=5t_{L}[/tex]

[tex]x=6t_{C}[/tex]

And the difference between Louisa's time and Calvin' time is 30 minutes, half an hour. So:

[tex]t_{C}=t_{L}-0.5[/tex]

Three equations, three unknowns, the system can be solved.

Equalizing the equation with x :

[tex]5t_{L}=6t_{C}[/tex]

In this last equation replace [tex]t_{C}[/tex]  with the other equation and solve:

[tex]5t_{L}=6(t_{L}-0.5)\\ 5t_{L}=6t_{L}-3\\ 3=6t_{L}-5t_{L}\\ 3=t_{L}\\ t_{L}=3[/tex]

With Louisa's time find x:

[tex]x=5t_{L}\\ x=5(3)\\ x=15[/tex]

The length of the circular path that Louisa ran is 15 miles when she completed one circular path.

Let's denote the length of the circular path as L miles. Louisa's speed is 5 miles per hour, and Calvin's speed is 6 miles per hour.

We'll use the equation for time, which is time = distance / speed.

For Louisa:

[tex]t_L = \frac{L}{5}[/tex]

Where:

For Calvin:

[tex]t_C = \frac{L}{6}[/tex]

Where:

According to the problem, it took Calvin 30 minutes (0.5 hours) less than it took Louisa to run the full path:

[tex]t_L = t_C + 0.5[/tex]

Substitute the expressions for [tex]t_L[/tex] and [tex]t_C[/tex] into the equation:

[tex]\frac{L}{5} = \frac{L}{6} + 0.5[/tex]

To solve for L, first find a common denominator for the fractions.

The common denominator for 5 and 6 is 30.

Therefore:

[tex]\frac{6L}{30} = \frac{5L}{30} + 0.5[/tex]

Multiply everything by 30 to eliminate the denominators:

[tex]6L = 5L + 15[/tex]

Subtract 5L from both sides:

[tex]L = 15[/tex]

The probability of guessing the correct seven-digit pin code on the first try is very low, approximately 0.000024%.

Given that,

A thief steals an ATM card.

The thief must guess the correct seven-digit pin code.

The pin code is entered using a 4-key keypad.

The probability of guessing the correct seven-digit pin code on the first try depends on a few factors.

To break it down,

if the thief has a 4-key keypad and repetition of digits is allowed, that means there are four options for each digit.

So, there are a total of 4⁷ (4 raised to the power of 7) possible combinations.

Since the thief is trying to guess the correct pin code on the first try, there is only one correct combination out of the total possible combinations.

Therefore,

The probability of guessing the correct pin code on the first try would be 1 out of 4⁷, or approximately 0.00000024, or 0.000024%.

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The probability of randomly guessing a 7-digit PIN from a 4-key keypad is 1 in 16,384. This equals approximately 0.000061 or 0.0061%. Each digit has 4 possible options, and there are 7 digits in total.

The probability of guessing a seven-digit PIN code correctly from a 4-key keypad (where repetition of digits is allowed) can be calculated as follows:

Since each of the 7 digits in the PIN can be any of 4 possible digits (0 through 3), the total number of possible combinations is calculated by raising the number of choices per digit to the power of the number of digits:

→ Total possible combinations = 4^7

= 4⁷

= 16384

Therefore, the probability of guessing the correct PIN on the first try is the reciprocal of the total number of possible combinations:

→ Probability of a correct guess = 1 / 16384

Hence, the probability is approximately 0.000061 or 0.0061%.

Answer: 0.0775

Step-by-step explanation:

Given : Mean : [tex]\mu = 25[/tex]

Standard deviation : [tex]\sigma =15[/tex]

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

Since its normal distribution , then the formula to calculate the z-score is given by :-

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

For x= 28.2 hours

[tex]z=\dfrac{28.2-25}{\dfrac{15}{\sqrt{36}}}=1.28[/tex]

For x= 30 hours

[tex]z=\dfrac{30-25}{\dfrac{15}{\sqrt{36}}}=2[/tex]

The P- value = [tex]P(1.28<z<2)[/tex]

[tex]=P(z<2)-P(z<1.28)= 0.9772498-0.8997274=0.0775224\approx0.0775[/tex]

Hence, the probabiliy that the average time spent stydying for the sampe was between 28.2 and 30 hours = 0.0775

The probability that the average time spent studying for the sample was between 28.2 and 30 hours is calculated as 0.0775 or 75 %.

To calculate the probability that the average time spent studying for the sample was between 28.2 and 30 hours, we use the normal distribution and standardize the sample means to a z-score.

Given the population mean (μ) is 25 hours, the population standard deviation (σ) is 15 hours, and the sample size (n) is 36, the standard error of the mean (SEM) is σ/√n which is 15/6 = 2.5 hours.

The z-scores for 28.2 and 30 hours are calculated as (X - μ)/(SEM).

Z for 28.2 hours = (28.2 - 25)/2.5 = 1.28
Z for 30 hours = (30 - 25)/2.5 = 2

Now we can look up these z-scores in the standard normal distribution table (or use calculator/software) to find the probabilities for these z-scores and then find the probability that lies between them by subtracting the two.

Example: Let's assume the probability corresponding to z=1.28 is 0.8997 and to z=2 is 0.9772.

The probability that the sample mean lies between 28.2 and 30 hours is:

P(1.28 < Z < 2) = P(Z < 2) - P(Z < 1.28)

= 0.9772 - 0.8997
= 0.0775

Hence, there is a 7.75% probability that the sample mean is between 28.2 and 30 hours.

Answer:

They should be able to sell their home for $149706.9

Step-by-step explanation:

Let's first understand the situation.

There is an initial value for the house which is $189000. However, this value varies every 6 years because of a 11% depreciation of the total value.

Because the depreciation is not executed during the 6 years in a constant way, but instead after the whole 6 years have passed, then we can calculate how many depreciations will be applied within the next 15 years:

total years/years needed for depreciation=15years/6years=2.5

The above means that only 2 depreciations are going to be applied. Remember that depreciation is only applied if the whole 6 years have passed.  

Now, after the first 6 years the depreciation (D) is:

D = 0.11 * $189000 = $20790,

which means that the value of the house will be:

(initial value) - D = $189000 - $20790 = $168210

Now, after the following 6 years, first 12 years, the depreciation (D) is:

D = 0.11 * $168210 = $18503.1,

which means that the value of the house will be:

(initial value) - D = $168210 - $18503.1 = $149706.9

In conclusion, in 15 years from now, they should be able to sell their home for $149706.9

Answer:

Step-by-step explanation:

We can put the given numbers into the given formula and solve for n.

  658.5 = k·3^n

  615.7 = k·4^n

Dividing the first equation by the second, we get ...

  658.5/615.7 = (3/4)^n

The log of this is ...

  log(658.5/615.7) = n·log(3/4)

  n = log(658.5/615.7)/log(3/4) ≈ 0.0291866/-0.124939

  n ≈ -0.233607

Then we can find s from ...

  log(s) = n·log(2)

  s = 2^n

  s ≈ 0.850506

The learning curve helps in estimating the reduction in the time required to produce units as experience is gained in manufacturing. In this problem, you need to solve a system of two equations based on the given work hours for the third and fourth production units, to find the value of exponent n and hence the slope parameter s.

This problem involves understanding of the concept of learning curve used in production and operations management. The learning curve predicts the time required to produce subsequent units given the time consumed by previous units. The initial units take more time to produce due to learning, but as the team gains experience, the time required to produce each subsequent unit decreases.

Above, you've given two data points - the third production unit requires 658.5 work hours and the fourth production unit requires 615.7 work hours. You want to find n in the equation Z_u=K(u^n), where this 'n' is the learning curve exponent which strongly influences how rapidly production time decreases as experience is gained.

We cannot directly calculate n because we do not have a value for K. However, we can set up a system of equations using both data points and solve for n. If you equate the two expressions for Z3 and Z4 and solve for n, you can determine its value and hence find the slope parameter s as well using the relationship s=log⁡s/log⁡2.

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Answer:

Find the slope of the line that passes through the points shown in the table.

The slope of the line that passes through the points in the table is

.

Step-by-step explanation:

Answer: (a) $264.375 ⇒ Amount of Interest

(b) Future Value = $3264.375

Step-by-step explanation:

(a) Principal amount = $3000

Time period = 9 months

Interest rate = 11.75%

Simple interest(SI) = principal amount × rate of interest (i) × time period

                              = 3000 × [tex]\frac{11.75}{100}[/tex] × [tex]\frac{9}{12}[/tex]

                              = 3000 × 0.1175 × 0.75

                               = $264.375 ⇒ Amount of Interest

(b) Future value of loan = principal amount + interest amount

                                        = 3000 + 264.375

                                        = 3264.375

Answer:

The required number of years are 7.52 years.

Step-by-step explanation:

Given : $10000 is deposited in an account earning 4% interest compounded continuously.

To find : How long it takes for the amount in the account to double?

Solution :

Applying Continuous interest formula,

[tex]A=Pe^{rt}[/tex]

Where, P is the principal P=$10000

r is the interest rate r=4%=0.04

t is the time

We have given, Amount in the account to double

i.e. A=2P

Substitute the value in the formula,

[tex]2P=Pe^{rt}[/tex]

[tex]2=e^{0.04t}[/tex]

Taking log both side,

[tex]\log 2=\log (e^{0.04t})[/tex]

[tex]\log 2=0.04t\times log e[/tex]

[tex]t=\frac{\log 2}{0.04}[/tex]

[tex]t=7.52[/tex]

Therefore, The required number of years are 7.52 years.

Answer:

Step-by-step explanation:

First, finding the horizontal asymptote:

[tex]\lim_{t \to \infty} = \frac{200+40t}{1+0.05t} = \frac{\frac{200}{t} 40 }{\frac{1}{t} 0.05} = 800[/tex]

In the context of the problem, the horizontal asymptote speaks about where the population of the fish is headed and capped.

Answer: [tex]x_1=6.73\\\\x_2=3.26[/tex]

Step-by-step explanation:  

Given the following expression:

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

And knowing that:

[tex](a\±b)^2=a^2-2ab+b^2[/tex]

We get:

[tex]x^2-2(x)(5)+5^2=3\\\\x^2-10x+25=3[/tex]

Move the 3 to the left side of the equation:

[tex]x^2-10x+25-3=0\\\\x^2-10x+22=0[/tex]

Apply the Quadratic formula:

[tex]x=\frac{-b\±\sqrt{b^2-4ac} }{2a}[/tex]

In this case:

[tex]a=1\\b=-10\\c=22[/tex]

Substituting values into the Quadratic formula, we get:

[tex]x=\frac{-(-10)\±\sqrt{(-10)^2-4(1)(22)} }{2(1)}\\\\\\x_1=6.73\\\\x_2=3.26[/tex]

Answer:  The required probability that a randomly selected day in November will be snowy if it is cloudy​ is 86.79%.

Step-by-step explanation:  Given that for the month of November in a certain city, 53​% of the days are cloudy. Also in the month of November in the same​ city, 46​% of the days are cloudy and snowy.

We are to find the probability that a randomly selected day in November will be snowy if it is cloudy​.

Let A denote the event that the day is cloudy and B denote the event that the day is snowy.

Then, according to the given information, we have

[tex]P(A)=53\%=0.53,\\\\P(A\cap B)=46\%=0.46.[/tex]

Now, we need to find the conditional probability of event B given that the event A has already happened.

That is, P(B/A).

We know that

[tex]P(B/A)=\dfrac{P(B\cap A)}{P(A)}=\dfrac{0.46}{0.53}=0.87=87.79\%.[/tex]

Thus, the required probability that a randomly selected day in November will be snowy if it is cloudy​ is 87.79%.

Final answer:

The probability that a randomly selected cloudy day in November will be snowy is calculated using conditional probability. The result is approximately 86.79%.

Explanation:

To determine the probability that a randomly selected day in November will be snowy if it is cloudy, we use the given information: 53% of days are cloudy and 46% of days are both cloudy and snowy. The probability we are looking for is the conditional probability of it being snowy given that it is cloudy, which can be calculated by dividing the probability of it being both cloudy and snowy by the probability of it being cloudy, which is P(Snowy | Cloudy) = P(Cloudy and Snowy) / P(Cloudy).

So the calculation would be:

P(Snowy | Cloudy) = (0.46) / (0.53)
= 0.8679 (or 86.79%).

Therefore, there is an 86.79% chance that it will be snowy on a day that is cloudy in that city in November.

Answer:

Not a probability distribution

Step-by-step explanation:

The given table doesn't describe a probability distribution as in order for the given distribution to be a probability distribution the sum of probabilities is required to be equal to one.

Here,

Sum of probabilities = 0.001+0.025+0.101+0.246+0.503 = 0.876

The sum of probabilities is not equal to one.

Therefore, the given distribution is not a probability distribution ..

Answer:

A) w > -7

B) 1.5 < t

C) -3.5 makes A) true and B) false

    10 makes both inequalities true

Step-by-step explanation:

The idea of these exercises is to clear our variable, we need it to be alone on one side of the inequality

A) 2w + 17 ˃ -4w -25

First, we will put together on one side the terms with a w and on the other the terms without w.

For that, we have to add 4w - 17 on both sides

2w + 17 + 4w - 17 ˃ -4w -25 + 4w - 17 (Notice that 17-17=0 and -4w+4w=0, so we don't have to write them below)

2w + 4w > -25 - 17

Now we can sum the terms (we didn't do it before because we can't sum a term with a w with one without it)

6w > -42

We divide by 6 on both sides and we have

6/6w > -42/6

w > -7

B) 2.3 + 0.6t ˂ 2 + 0.8t

We start as before; in this case we have to put together the terms with a t (our variable changes name but the idea is the same)

We will add -2 - 0.6t on both sides

2.3 + 0.6t -2 - 0.6t ˂ 2 + 0.8t -2 - 0.6t  

2.3 - 2 < 0.8t - 0.6t

Now we sum the terms  

0.3 < 0.2t

We divide by 0.2 on both sides and we have

0.3/0.2 < 0.2/0.2t

1.5 < t

C) Let's check -3.5 on both inequalities:

We have to replace the variable by -3.5:

2*(-3.5) + 17 ˃ -4*(-3.5) -25 (remember that if there is no sign between a number and a variable, it means that is a multiplication)

Now we just solve the calculation

-7 + 17 > 14 -25

10 > -11

That's true, so -3.5 makes the inequality true.

Now, in the other inequality, we replace the t by -3.5 and solve as before

2.3 + 0.6*(-3.5) ˂ 2 + 0.8*(-3.5)

2.3 - 2.1 < 2 - 2.8

0.2 < -0.8

That's false because we are saying that a negative number is bigger than a positive one, so -3.5 makes the inequality not true.

  Now we do the same with 10 in both inequalities:

   2*10 + 17 ˃ -4*10 -25

   20 + 17 > -40 -25

   37 > - 65

   It's true!

   2.3 + 0.6*10 ˂ 2 + 0.8*10

   2.3 + 6 < 2 + 8

   8.3 < 10

   It's true!

To solve the problem, first calculate all possible combinations of selecting 3 apples from 11. Then calculate the favorable combinations, which include selecting 1 red apple (from 6 available) and 2 yellow apples (from 5 available). Divide these values to get the probability.

The topic at hand is one of probability, more specifically, it's a problem of combinations in probability. The bag contains a total of 11 apples (6 red and 5 yellow). When 3 apples are chosen, we want to find the probability that 1 is red and 2 are yellow.

First, calculate the total number of ways to choose 3 apples from 11, which is denoted as '11 choose 3', using combination formula C(n,r) = n! / [r!(n - r)!]. Then, consider the number of favorable outcomes: choosing 1 red apple from 6 (denoted as '6 choose 1') and 2 yellow apples from 5 (denoted as '5 choose 2'). Multiply these two results because we choose '1 red' and '2 yellow', using the rule of product. Calculate these individual results and then divide the favorable outcomes by the total outcomes to get the required probability.

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Answer:

Length: 11.95

Width: 9.95

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Algebraic Equations.

To begin solving this question we need to create a formula to then solve. According to the information given in the question we can create 2 formulas.

[tex]P = 2L+2W[/tex]

[tex]L = 2+W[/tex]

Where:

Now we can replace L for the L in the P formula and solve for W, Like so....

[tex]43.8 = 2(2+W)+2W[/tex]

[tex]43.8 = 4+2W+2W[/tex]

[tex]39.8 = 4W[/tex]

[tex]9.95 = W[/tex]

Now that we have the value of W we can plug that into the L formula and find L

[tex]L = 2+9.95[/tex]

[tex]L = 11.95[/tex]

Finally, we can see that the value of the Length is 11.95 in and the Width is 9.95 in

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

The vector pointing from (2, 1) to (1, 3) points in the same direction as the vector [tex]\vec u=(1,3)-(2,1)=(-1,2)[/tex]. The derivative of [tex]f[/tex] at (2, 1) in the direction of [tex]\vec u[/tex] is

[tex]D_{\vec u}f(2,1)=\nabla f(2,1)\cdot\dfrac{\vec u}{\|\vec u\|}[/tex]

We have

[tex]\|\vec u\|=\sqrt{(-1)^2+2^2}=\sqrt5[/tex]

Then

[tex]D_{\vec u}f(2,1)=(f_x(2,1),f_y(2,1))\cdot\dfrac{(-1,2)}{\sqrt5}=\dfrac{-f_x(2,1)+2f_y(2,1)}{\sqrt5}=-\dfrac2{\sqrt5}[/tex]

[tex]\implies f_x(2,1)-2f_y(2,1)=2[/tex]

The vector pointing from (2, 1) to (5, 5) has the same direction as the vector [tex]\vec v=(5,5)-(2,1)=(3,4)[/tex]. The derivative of [tex]f[/tex] at (2, 1) in the direction of [tex]\vec v[/tex] is

[tex]D_{\vec v}f(2,1)=\nabla f(2,1)\cdot\dfrac{\vec v}{\|\vec v\|}[/tex]

[tex]\|\vec v\|=\sqrt{3^2+4^2}=5[/tex]

so that

[tex](f_x(2,1),f_y(2,1))\cdot\dfrac{(3,4)}5=1[/tex]

[tex]\implies3f_x(2,1)+4f_y(2,1)=5[/tex]

Solving the remaining system gives [tex]f_x(2,1)=\dfrac95[/tex] and [tex]f_y(2,1)=-\dfrac1{10}[/tex].

Answer:

3 students

Step-by-step explanation:

If everyone in the class has either a pierced nose or ear, we just simply have to add up the total number of hands raised and minus the number of students in the class.

35+8=43

43-40=3

3 students have both a pierced nose and pierced ear.

Answer:

The answer is option C : 5%.

Step-by-step explanation:

Jack typed 80 words per minute when he enrolled in a typing course.

His typing speed increased by 3% two weeks into the course.

At the end of the course, Jack was able to type his entire 1, 680 word document In 20 minutes.

Hence, the percent of increase in his typing speed from the beginning of the course to the end is given by:

[tex]1680/20=84[/tex]

[tex](84-80)/80[/tex] =5%

Therefore, the answer is option C : 5%.

To find the percent increase in Jack's typing speed, compare his initial and final speeds. The percent increase is 5%.

To find the percent increase in Jack's typing speed, we need to compare his initial speed to his final speed. Let's start by calculating his initial typing speed:

80 words per minute

To find his final typing speed, we need to determine how many words he typed in 20 minutes:

1,680 words / 20 minutes = 84 words per minute

Now we can find the percent increase:

(Final speed - Initial speed) / Initial speed * 100

(84 - 80) / 80 * 100 = 5%

Therefore, the percent of increase in Jack's typing speed from the beginning of the course to the end is 5%.

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Answer: 15.06

Step-by-step explanation:

Given : The number of individuals participated in a lifetime study :[tex]n=913[/tex]

The probability of individuals with certain gene of contracting a certain disease :[tex]p= 0.46[/tex]

Now, the standard deviation of the number of people who eventually contract the disease is given by :_

[tex]\sigma =\sqrt{np(1-p)}\\\\=\sqrt{913\times0.46(1-0.46)}=15.059521904\approx15.06[/tex]

Hence, the  the standard deviation of the number of people who eventually contract the disease = 15.06

Answer: a) -6, b) 32, c) -48, d) 9, e) -12

Step-by-step explanation:

Since we have given that

A and B are 4 × 4 matrices.

Here,

det (A) = -3

det (B) = 2

We need to find the respective parts:

a) det (AB)

[tex]\mid AB\mid=\mid A\mid.\mid B\mid\\\\\mid AB\mid=-3\times 2=-6[/tex]

b) det (B⁵ )

[tex]\mid B^5\mid=\mid B\mid ^5=2^5=32[/tex]

c) det (2A)

Since we know that

[tex]\mid kA\mid =k^n\mid A\mid[/tex]

so, it becomes,

[tex]\mid 2A\mid =2^4\mid A\mid=16\times -3=-48[/tex]

d) [tex]\bold{det(A^TA)}[/tex]

Since we know that

[tex]\mid A^T\mid=\mid A\mid[/tex]

so, it becomes,

[tex]\mid A^TA\mid=\mid A^T\mid \times \mid A\mid=-3\times -3=9[/tex]

e) det (B⁻¹AB)

As we know that

[tex]\mid B^{-1}\mid =\mid B\mid[/tex]

so, it becomes,

[tex]\mid B^{-1}AB}\mid =\mid B^{-1}.\mid \mid A\mid.\mid B\mid=2\times -3\times 2=-12[/tex]

Hence, a) -6, b) 32, c) -48, d) 9, e) -12