A stretch of highway that is 12 1 4 12 4 1 ​ 12, start fraction, 1, divided by, 4, end fraction kilometers long has speed limit signs every 7 8 8 7 ​ start fraction, 7, divided by, 8, end fraction of a kilometer. How many speed limit signs are on this stretch of highway?

Answer:

The answer to your question is below

Step-by-step explanation:

pH definition

                         pH = - log [H⁺]

a) For pH = 2.4, solution A

                       2.4 = -log[H⁺]

                      [H⁺] = antilog⁻².⁴

                      [H⁺] = 0.00398

  For pH = 9.4, solution B

                       [H⁺] = antilog⁻⁹.⁴

                       [H⁺] = 3.98 x 10⁻¹⁰

b) Divide hydrogen-ion concentration of solution A by hydrogen-ion concentration of solution B.

                             0.00398 / 3.98 x 10⁻¹⁰

                             10000000 times

c) By 7, because 7 is the number of zeros

Answer:

1740%.

Step-by-step explanation:

We have been given that the pawnbroker loans you ​$600. One month​ later, you get the guitar back by paying the pawnbroker ​$1470.

We will use simple interest formula to solve our given problem.

[tex]A=P(1+rt)[/tex], where,

A = Final amount after t years,

P = Principal amount,

r = Annual interest rate in decimal form,

t = Time in years.

1 month = 1/12 year

[tex]1470=600(1+r*\frac{1}{12})[/tex]

[tex]1470=600+\frac{600}{12}*r[/tex]

[tex]1470=600+50*r[/tex]

[tex]1470-600=600-600+50*r[/tex]

[tex]870=50*r[/tex]

[tex]50r=870=[/tex]

[tex]\frac{50r}{50}=\frac{870}{50}[/tex]

[tex]r=17.4[/tex]

Since our interest rate is in decimal form, so we will convert it into percentage by multiplying by 100 as:

[tex]17.4\times 100\%=1740\%[/tex]

Therefore, you paid an annual interest rate of 1740%.

In a normal distribution, the back-to-knee lengths that separate significant values from others are 24.896 inches (upper percentile) and 19.304 inches (lower percentile) respectively. Using these limits, a back-to-knee length of 24.2 inches is not considered significantly high.

To find the back-to-knee lengths that separate significant values from those that are not, we need to find the values of x for which P(x ≥ some value) ≤ 0.01, and P(x ≤ some value) ≤ 0.01.

These values are known as the upper and lower percentiles, respectively, and can be obtained by transforming to a standard normal distribution (with a mean of 0 and a standard deviation of 1) using z-scores.

Let's use the standard normal table to find z-scores corresponding to 0.01 in the upper side and lower side of the distribution. You would find that the z-score which has an area of 0.01 in the upper tail is approximately 2.33, and in the lower tail it is -2.33.

Now, we can use these z-scores to calculate respective back-to-knee lengths. The formula for a z-score is z = (x - μ) / σ, where μ is the mean, σ is the standard deviation, and x is the observation. Solving for x gives us x = zσ + μ.

Upper percentile, x = (2.33*1.2) + 22.1 = 24.896 inches.

Lower percentile, x = (-2.33*1.2) + 22.1 = 19.304 inches.

So, any back-to-knee length above 24.896 inches or below 19.304 inches can be considered statistically significant. Therefore, a back-to-knee length of 24.2 inches would not be considered significantly high as it is less than 24.896 inches.

https://brainly.com/question/30390016

#SPJ12

In order to find the separating lengths for significant values in a normal distribution, we need to find the values corresponding to z-scores for probabilities 0.01 and 0.99. To know if a back-to-knee length of 24.2 in. is significantly high, compare the probability P(X>=24.2) with 0.01.

In the problem presented, the information is about the normal distribution of the back-to-knee length of a group of adults. A normal distribution graph has its highest point at the mean, which in this case is 22.1 in., and it decrease on either side. The standard deviation, which measures the spread of the values, is given as 1.2 in.

For a value to be considered significantly high, the probability P(x or higher) should be less than or equal to 0.01. Similarly, for a value to be significantly low, P(x or lower) should also be less or equal to 0.01.

To find the back-to-knee lengths separating the significant values from non-significant, one can use the z-scores associated with the probabilities 0.01 and 0.99 (since the total probability of a normal distribution is 1). So, the lengths in question will be the ones that correspond to these z-scores.

Provided, a z-score is a value's number of standard deviations from the mean. If z is the z-score, μ is the mean, σ is the standard deviation, and x is the value, the formula is z=(x-μ)/σ.

Regarding the specific measurement of 24.2 in., we would need to calculate the probability P(X>=24.2) using the given mean and standard deviation in Z-score formula. If P(X>=24.2) is less than 0.01, then 24.2 in. can be considered as significantly high.

https://brainly.com/question/31538429

#SPJ2

Answer:

x4 + 12x^3 – 45x^2

Step-by-step explanation:

Answer:

a) 95.99%

b) 4.01%

c) 00.62%

Step-by-step explanation:

Explanation is given in the attachments.

Answer:

Jim's current weight = 119 pounds

Step-by-step explanation:

1 Stone = 14 pounds

[tex]1\frac{1}{2}[/tex] stone = 1.5 stone

1.5 stone = 1.5 (14 pounds) = 21 pounds

Jim lost 21 pounds

Let X be Jim's Original Weight

Y be his present weight

As per given statement in the Question:

After losing 1.5 stones (21 pounds of weight) Jim now weighs Y

Present weight = original weight - 21

Y = X -21                                                        Equation 1

Also Current Weight = 85 % (Original weight)

Y = 85 % (X) =[tex]\frac{85X}{100}[/tex]

Y=[tex]\frac{85X}{100}[/tex]

put in Equation 1

[tex]\frac{85X}{100}[/tex] = X-21

85X = (X-21) 100

85 X = 100 X -2100

or

2100 = 100 X - 85X

2100 = 15X

or

15 X = 2100

[tex]X=\frac{2100}{15}[/tex]

X= 140 pounds ( Original Weight)

Current Weight = Y = Original weight - 21 (From Equation 1)

Y = X -21

Y = 140 -21

Y = 119 pounds (Current Weight)

Answer:

The pipe C alone can fill the tank in 14 hours .

Step-by-step explanation:

Given as :

The three pipes a , b , c can fill the pipes in 6 hours

They work for 2 hours

After that c pipe is close and a , b finish remaining work

Now, According to question

In 1 hour pipes ( a + b + c ) fill [tex]\frac{1}{6}[/tex] of the tank

∴ In 2 hour pipes ( a + b + c ) fill [tex]\frac{2}{6}[/tex] =  [tex]\frac{1}{3}[/tex] of the tank

Remaining ( 1 - [tex]\frac{1}{3}[/tex]  ) = [tex]\frac{2}{3}[/tex]  part is filled by pipes a and b in 7 hours

∴ The whole tank is filled by a and b in 7 × [tex]\frac{3}{2}[/tex] =  [tex]\frac{21}{2}[/tex] hours

∴ In 1 hour pipes A and b fill the tank in [tex]\frac{2}{21}[/tex] hours

∴ In 1 hour pipes C alone can fill the tank in[tex]\frac{1}{6}[/tex] - [tex]\frac{2}{21}[/tex] hours

Or,  In 1 hour pipes C alone can fill the tank in [tex]\frac{9}{126}[/tex] =  [tex]\frac{1}{14}[/tex]

Or, In 1 hour pipes C alone can fill the tank in 14 hours

Hence The pipe C alone can fill the tank in 14 hours . Answer

Option D

Function used to determine the profit the theater group earns from selling "x" tickets is f(x) = 12.50x - 150

Given that a theater group charges $12.50 per ticket for its opening play of the season

Also given that Production costs for the play are $150

To find: function used to determine the profit the theater group earns from selling "x" tickets

So "x" represents the number of tickets sold

Cost per ticket = $ 12.50

[tex]\text { cost of "x" tickets }=" x " \times \text { cost per ticket }[/tex]

[tex]\text { cost of "x" tickets }= x \times 12.50=12.50x[/tex]

Production costs = 150

Then the profit the theater group earns from selling "x" tickets is:

Profit earned from selling "x" tickets = cost of "x" tickets - production cost

Let f(x) denotes profit earned from selling "x" tickets

f(x) = 12.50x - 150

Thus option D is correct

Answer:

[tex]\frac{dh}{dt}[/tex]≅[tex]0.286\frac{ft^{3} }{min}[/tex]

Step-by-step explanation:

[tex]V=\frac{\pi }{3}r^{2}h[/tex]; rate of change [tex]\frac{dV}{dt}=10\frac{ft^{3} }{min}[/tex], we must find the rate of change of the depth [tex]\frac{dh}{dt} =?;h=8ft[/tex]

5h=12r; [tex]V=\frac{\pi }{3}\({\((5h}/12} )} ^{2}h=\frac{\pi }{3}(\frac{25h^{2} }{144})h; V=\frac{25\pi h^{3}}{432}[/tex]; deriving [tex]\frac{dV}{dt} = \frac{25\pi }{432}(3h^{2})\frac{dh}{dt}[/tex] → [tex]10=\frac{25\pi h^{2}}{144} \frac{dh}{dt}[/tex] → h=8 then [tex]\frac{dh}{dt}=\frac{1440}{25\pi 64}=\frac{9}{10\pi}[/tex]≅ 0.286[tex]\frac{ft^{3} }{min}[/tex]

Answer:

[tex]\sin \theta = \dfrac{5}{13}[/tex] and [tex]\sec \theta = -\dfrac{13}{12}[/tex]

Step-by-step explanation:

Assume that the terminal side of thetaθ passes through the point (−12,5).

In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.

Using Pythagoras theorem:

[tex]hypotenuse^2=perpendicular^2+base^2[/tex]

[tex]hypotenuse^2=(5)^2+(12)^2[/tex]

[tex]hypotenuse^2=25+144[/tex]

[tex]hypotenuse^2=169[/tex]

Taking square root on both sides.

[tex]hypotenuse=13[/tex]

In a right angled triangle

[tex]\sin \theta = \dfrac{opposite}{hypotenuse}[/tex]

[tex]\sin \theta = \dfrac{5}{13}[/tex]

[tex]\sec \theta = \dfrac{hypotenuse}{adjacent}[/tex]

[tex]\sec \theta = \dfrac{13}{12}[/tex]

In second quadrant only sine and cosecant are positive.

[tex]\sin \theta = \dfrac{5}{13}[/tex] and [tex]\sec \theta = -\dfrac{13}{12}[/tex]

The inequality is: 5>x≥0

Step-by-step explanation:

We have to write the inequality one by one

The inequality symbols are used to write inequalities.

So,

5 is greater than x will be written as:

5>x

And

0 is less than or equals to x:

0≤x or x≥0

We have to combine the both inequalities so that the variable is not repeated. The inequality symbols have to be written carefully while writing the compound inequalities.

5>x≥0

Keywords: Inequality, Relationships

Learn more about inequality at:

#LearnwithBrainly

The maximum area of the given rectangle will be A = 4(C/3)√(C/3).

The quantity of space enclosing a three-dimensional shape's exterior is its surface area.

In other meaning, if we say side square then it is an area of the square but for a cuboid, there are 6 faces so the surface area will be external to all 6 surfaces area.

As per the given rectangle inscribed in the parabola has been drawn,

The area of rectangle A = (x + x)y

A = 2x(C - x²)

A = 2Cx - 2x³

To find the maximum area, take the first derivative with respect to x.

dA/dx = 2C - 6x² = 0

C - 3x² = 0

x = √(C/3)

Therefore, the area will be as,

A = 2C√(C/3) - 2(√(C/3))³

A = 2C√(C/3) - 2(C/3)√(C/3)

A = 2√(C/3) [C - C/3]

A = 2√(C/3)(2C/3)

A = 4(C/3)√(C/3)

Hence "The maximum area of the given rectangle will be A = 4(C/3)√(C/3)".

To learn more about surface area,

https://brainly.com/question/2835293

#SPJ2

Answer:

Since we FAIL to reject the null hypothesis, then if we construct an interval of 95% of confidence, the 0 should be included, because on the test hypothesis we conclude that there would be no significant difference between the means of the two groups analyzed, and the results obtained on the hypothesis test needs to be consistent with the confidence interval.

Step-by-step explanation:

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

[tex]\bar x_{meat}=111.7[/tex] represent the sample mean of calories for the meat hot dogs

[tex]\bar x_{beef}=135.4[/tex] represent the sample mean of calories for the beef hot dogs

The system of hypothesis on this case would be:

Null hypothesis: [tex]\mu_{meat}-\mu_{beef}=0[/tex]

Alternative hypothesis: [tex]\mu_{meat}-\mu_{beef}\neq 0[/tex]

On this case we have the p value obtained, after calculate the statistic and we got that [tex]p_v =0.124[/tex] if we select a 5% significance level [tex]\alpha=0.05[/tex] we see that [tex]p_v >\alpha[/tex] and on this case we can FAIL to rejec the null hypothesis, so there is not a significant difference between the mean of the two tpes of hot dogs analyzed at 5% of significance.

And since we FAIL to reject the null hypothesis, then if we construct an interval of 95% of confidence, the 0 should be included, because on the test hypothesis we conclude that there would be no significant difference between the means of the two groups analyzed, and the results obtained on the hypothesis test needs to be consistent with the confidence interval.

A 95% confidence interval for μMeat - μBeef including 0 indicates no significant difference in calorie content between meat and beef hot dogs.

A 95% confidence interval for μMeat - μBeef that includes 0 suggests that there is no significant difference in the mean calorie content between meat and beef hot dogs. In this case, the difference in mean calorie content between the two types of hot dogs is not statistically significant, as the null hypothesis is not rejected. The P-value of 0.124 suggests that there is a 12.4% chance of observing a difference in mean calorie content as extreme as the one observed if the null hypothesis were true. Therefore, we cannot conclude that there is a significant difference in the mean calorie content between meat and beef hot dogs.

https://brainly.com/question/34700241

#SPJ11

Answer:

a) Frequency table:

Category       Frequency

 Dog                     4

 Cat                       3

 Fish                      2

 Hamster               1

b) Relative frequencies of each animal type

c) Popularity

Explanation:

a. Make a frequency table for the results.

There are four kind of pets: dog, cat, hamster, and fish.

A frequency table shows the number of items for each category (kind of pets).

Count the number of each kind of pet:

With that you build your frequency table.

Frequency table:

Category       Frequency

 Dog                     4

 Cat                       3

 Fish                      2

 Hamster               1

b. Relative frequencies of each animal type.

The relative frequency is how often an outcome appears divided by the total number of outmcomes.

Here the total number of outcomes is 10 (the ten pets).

So, calculate each relative frequency:

An important feature of the relative frequency is that they must add up 1. Check:

c. Using the relative frequencies explain which animals are most and least popular.

Popularity is determined by the frequency with each outcome is repeated. The most popular is the most repeated. The least popular is the least repeated.

Trigonometry can be used to model projectile motion, such as the flight of a baseball. Given the angle at which the ball leaves the bat and the initial velocity, you can determine the distance the ball will travel.  

Answer:

Trigonometry can be used to model projectile motion, such as the flight of a baseball. Given the angle at which the ball leaves the bat and the initial velocity, you can determine the distance the ball will travel.

Step-by-step explanation:

Answer:

-12

Step-by-step explanation:

Answer:

The first One

Step-by-step explanation:

I just got 100% on My Quiz

Answer: [tex]S_n=5(1-\dfrac{1}{n+1})[/tex] ; 5

Step-by-step explanation:

Given series : [tex][\dfrac{5}{1\cdot2}]+[\dfrac{5}{2\cdot3}]+[\dfrac{5}{3\cdot4}]+....+[\dfrac{5}{n\cdot(n+1)}][/tex]

Sum of series = [tex]S_n=\sum^{\infty}_{1}\ [\dfrac{5}{n\cdot(n+1)}]=5[\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}][/tex]

Consider [tex]\dfrac{1}{n\cdot(n+1)}=\dfrac{n+1-n}{n(n+1)}[/tex]

[tex]=\dfrac{1}{n}-\dfrac{1}{n+1}[/tex]

⇒ [tex]S_n=5\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}=5\sum^{\infty}_{1}[\dfrac{1}{n}-\dfrac{1}{n+1}][/tex]

Put values of n= 1,2,3,4,5,.....n

⇒ [tex]S_n=5(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......-\dfrac{1}{n}+\dfrac{1}{n}-\dfrac{1}{n+1})[/tex]

All terms get cancel but First and last terms left behind.

⇒ [tex]S_n=5(1-\dfrac{1}{n+1})[/tex]

Formula for the nth partial sum of the series :

[tex]S_n=5(1-\dfrac{1}{n+1})[/tex]

Also, [tex]\lim_{n \to \infty} S_n = 5(1-\dfrac{1}{n+1})[/tex]

[tex]=5(1-\dfrac{1}{\infty})\\\\=5(1-0)=5[/tex]

Answer:

$824

Step-by-step explanation:

3% (rate of interest) of 800= 24

800 + 24= 824

Answer: option B is the correct answer

Step-by-step explanation:

The diagram of the tree is shown in the attached photo. The triangle ABC formed is not a right angle triangle. The last angle, angle C is gotten by subtracting the sum of angle A and angle B from 180(sum of angles) in a triangle is 180). It becomes

C = 180-(98+24)= 180 -122

C = 58 degrees

To find the height of the tree, we would apply the sine rule

a/sinA = b/sin B = c/ sinC

We would apply b/sin B = c/ sinC

b/sin24 = 27/sin58

b/0.4067 = 27/0.8480

Cross-multiplying,

27 ×0.4067 = b × 0.8480

10.9809 = 0.8480b

b = 10.9809/0.8480 = 12.949

Approximately 13 meters

The bug crawls 13 meters from the base to the top of the tree

Answer:

The dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

Step-by-step explanation:

Let the length of garden be x

Let the breadth of garden be y

Area of Rectangular garden = [tex]Length \times Breadth = xy[/tex]

We are given that the area of the garden is 122 square feet

So, [tex]xy=122[/tex] ---A

A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft

So, cost of brick along length x = 20 x

On the other three sides by a metal fence costing $10/ft.

So, Other three side s = x+2y

So, cost of brick along the other three sides= 10(x+2y)

So, Total cost = 20x+10(x+2y)=20x+10x+20y=30x+20y

Total cost = 30x+20y

Substitute the value of y from A

Total cost = [tex]30x+20(\frac{122}{x})[/tex]

Total cost = [tex]\frac{2440}{x}+30x[/tex]

Now take the derivative to minimize the cost

[tex]f(x)=\frac{2440}{x}+30x[/tex]

[tex]f'(x)=-\frac{2440}{x^2}+30[/tex]

Equate it equal to 0

[tex]0=-\frac{2440}{x^2}+30[/tex]

[tex]\frac{2440}{x^2}=30[/tex]

[tex]\sqrt{\frac{2440}{30}}=x[/tex]

[tex]9.018 =x[/tex]

Now check whether it is minimum or not

take second derivative

[tex]f'(x)=-\frac{2440}{x^2}+30[/tex]

[tex]f''(x)=-(-2)\frac{2440}{x^3}[/tex]

Substitute the value of x

[tex]f''(x)=-(-2)\frac{2440}{(9.018)^3}[/tex]

[tex]f''(x)=6.6540[/tex]

Since it is positive ,So the x is minimum

Now find y

Substitute the value of x in A

[tex](9.018)y=122[/tex]

[tex]y=\frac{122}{9.018}[/tex]

[tex]y=13.528[/tex]

Hence the dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

Answer:

0.8802

Step-by-step explanation:

given that the Police estimate that 84​% of drivers wear their seatbelts.

when they stop 140 cars, no of trials = no of cars checked = 140

Each car is independent of the other

Hence X no of cars with drivers wearing seat belts is binomial with p = 0.85

Required probability =

the probability they find at least 27 drivers not wearing their​ seatbelts

Since normal approximation is required we can approximate to

X is Normal with mean = np = [tex]140(0.84)\\=117.6[/tex]

std dev = [tex]\sqrt{npq} =4.338[/tex]

Required probability =atelast 27 drivers not wearing their​ seatbelts

= P(X>(140-27))

= P(X>113)

[tex]=P(X>112.5)\\=1- 0.1198\\=0.8802[/tex]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

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

a) P(calls last between 4.7 and 5.5 minutes)

[tex]P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z <0)\\= 0.9452 - 0.5000 = 0.4452 = 44.52\%[/tex]

[tex]P(4.7 \leq x \leq 5.5) = 44.52\%[/tex]

b) P(calls last more than 5.5 minutes)

[tex]P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)[/tex]

Calculating the value from the standard normal table we have,

[tex]1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%[/tex]

c) P( calls last between 5.5 and 6 minutes)

[tex]P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z <1.6)\\= 0.9953 - 0.9452 = 0.0501 = 5.01\%[/tex]

[tex]P(5.5 \leq x \leq 6) = 5.01\%[/tex]

d) P( calls last between 4 and 6 minutes)

[tex]P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z <-1.4)\\= 0.9953 - 0.0808 = 0.9145 = 91.45\%[/tex]

[tex]P(4 \leq x \leq 6) = 91.45\%[/tex]

e) We have to find the value of x such that the probability is 0.03.

P(X > x)  

[tex]P( X > x) = P( z > \displaystyle\frac{x - 4.7}{0.50})=0.03[/tex]  

[tex]= 1 -P( z \leq \displaystyle\frac{x - 4.7}{0.50})=0.03 [/tex]  

[tex]=P( z \leq \displaystyle\frac{x - 4.7}{0.50})=0.997 [/tex]  

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

P(z < 2.75) = 0.997

[tex]\displaystyle\frac{x - 4.7}{0.50} = 2.75\\x = 6.075 \approx 6.08[/tex]  

Hence, the call lengths must be 6.08 minutes or greater for them to lie in the highest 3%.

The fraction of calls that last between 4.7 and 5.5 minutes is 0.4452, the fraction of calls that last more than 5.5 minutes is 0.0548, the fraction of calls that last between 5.5 and 6 minutes is 0.0501, the fraction of calls that last between 4 and 6 minutes is 0.9145, and the length of the longest 3% of the calls is 5.85 minutes.

To solve this problem, we can use the standard normal distribution table by converting the given values into z-scores. The z-score formula is: z = (x - μ) / σ where x is the given value, μ is the mean, and σ is the standard deviation. Let's calculate the fractions for each part of the question:

(a) Between 4.7 and 5.5 minutes:

(b) More than 5.5 minutes:

(c) Between 5.5 and 6 minutes:

(d) Between 4 and 6 minutes:

(e) The longest 3% of the calls:

So, the time for the longest 3% of the calls is 5.85 minutes.

https://brainly.com/question/34741155

#SPJ11

Answer:

f(t) = 250[tex]e^{-0.007752t}[/tex]

Step-by-step explanation:

Let f(t) = [tex]\alpha[/tex][tex]e^{\beta t }[/tex]

where f is the amount of radioactive substance in grams

and t is the time in minutes

initially (at t=0), f = 250 grams

⇒f(0) = 250 grams

⇒[tex]\alpha[/tex][tex]e^{0\beta}[/tex] = 250

⇒[tex]\alpha[/tex][tex]e^{0}[/tex] = 250

⇒[tex]\alpha[/tex] = 250 grams {∵[tex]e^{0} = 1[/tex]}

⇒f(t) = 250[tex]e^{\beta t }[/tex]

At t = 250 minutes, f = 36 grams

⇒f(250) = 36 grams

⇒250[tex]e^{250\beta}[/tex] = 36

⇒[tex]e^{250\beta}[/tex] = [tex]\frac{36}{250}[/tex] = 0.144

⇒250[tex]\beta[/tex] = ㏑ 0.144 = -1.938

⇒[tex]\beta[/tex] = -[tex]\frac{1.938}{250}[/tex] = -0.007752 [tex]min^{-1}[/tex]

∴f(t) = 250[tex]e^{-0.007752t}[/tex]

Answer:

475 billion dollars

Step-by-step explanation:

Let P be the customer credit

At the end of year X, 36% of P is gotten from automobile installment credit

57 billion credit is 1/3 of the automobile installment credit. This means that the total automobile installment credit 57*3 = 171 billion dollars

36% * P = 171

36/100 * P = 171

36P = 171 *100

P = 17100/36

P= 475 billion dollars

Answer:

Step-by-step explanation:

Total amount of money to be shared among Neymar, Rohit Sharma and Nadal is $250

Neymar gets two parts. This means that Neymar gets 1/2 × total amount of money.

Neymar gets 1/2 × 250 = 125

The percentage will be amount that Neymar gets divided by the total amount and multiplied by 100. It becomes

125/250 × 100 = 50%

Rohit gets three parts. This means that Rohit gets 1/3 × total amount of money.

Rohit gets 1/3 × 250 = 83.33

The percentage will be amount that Rohit gets divided by the total amount and multiplied by 100. It becomes

83.33/250 × 100 = 33.33%

Nadal gets five parts. This means that Nadal gets 1/5 × total amount of money.

Nadal gets 1/5 × 250 = 50

The percentage will be amount that Nadal gets divided by the total amount and multiplied by 100. It becomes

50/250 × 100 = 20%

Answer:

Step-by-step explanation:

A total of 10 parts are allocated to Neymar, Rohit, and Nadal, so each part represents 1/10 of the amount, or 10%.

Neymar gets 2 parts, or 20% of $250, so gets $50.

Rohit gets 3 parts, or 30% of $250, so gets $75.

Nadal gets 5 parts, or 50% of $250, so gets $125.

_____

Comment on the question

The question seems incomplete in that there are 4 names, but only 3 allocations.

Answer:

Both ratios will increase where the accounts payable balance is paid off.

Step-by-step explanation:

The current ratio is given as

Current ratio = Current asset / current liabilities

Where the current assets are asset that can be converted into cash easily ( including cash and cash equivalents) while the current liabilities are liabilities to be settled in a short term, say 1 year.

Acid test ratio is given as

Acid test ratio = (Current asset -  Inventories) / current liabilities

Here, the current assets excludes the assets that are not so easily converted to cash.

From the two formulas stated above, where the accounts payable balance which is an element of the current liabilities is paid off, the current liabilities balance reduces thus resulting in an increase in both ratio.

Hence, current and the acid-test ratios will increase where the accounts payable balance is paid off.

Paying off accounts payable increases both the current and acid-test ratios, assuming they are originally greater than 1, indicating positive financial stability to investors and creditors.

The effect of paying off an accounts payable balance on both the current and the acid-test ratios, if they are greater than 1, would be an increase. The current ratio is calculated as current assets divided by current liabilities. When accounts payable (a current liability) is paid off, the denominator of the ratio decreases, leading to an increase in the ratio. Similarly, for the acid-test (or quick) ratio, once again we see a decrease in the denominator after pay off, leading to an increase in the ratio.

Although the net change in an entity’s financial position may seem neutral (decrease in an asset offset by a decrease in liabilities), these ratio increases can be viewed positively by investors and creditors as they imply a greater degree of short-term solvency.

https://brainly.com/question/31531442

#SPJ3

Answer:

  $91.76

Step-by-step explanation:

Interest the first month was ...

  I = Prt

  I = $15000(0.0387)(1/12) = $48.375 ≈ $48.38

Interest in the second month was ...

  I = $15000(0.0347)(1/12) = $43.375 ≈ $43.38

So the total interest amount is ...

  $48.38 +43.38 = $91.76

Marcella earned $91.76 in two months.

_____

Comment on rounding

We have assumed that Marcella's account statement will report the interest rounded to 2 decimal places (cents). Hence she obtains the benefit from rounding for both months.

If there is no statement, so that rounding is not required until the end of the second month, then she may not have that extra penny in her account.

Answer:

Step-by-step explanation:

a³-b³=(a-b)(a²+ab+b²)

6³-1=6³-1³=(6-1)(6²+6*1+1²)=5×43

so 6³-1 is divisible by 5.

[tex]\displaystyle 6^3 - 1 = 216 - 1 = 215[/tex]

The number 215 is divisible by 5 because it ends in 5, according to the divisibility rules.

I am joyous to assist you anytime.

Answer:

[tex]6^{14}[/tex]

Step-by-step explanation:

Use the "Power Law" of exponents that tells us that when you have a base to a power "n" and all that raised to a power "m", it is the same as writing the original base to the single exponent which is the product of n time m:

[tex](b^n)^m=b^{n*m}[/tex]

therefore in your case, the base "b" is 6, the exponent "n" is 2, and the exponent "m" is 7. Then:

[tex](6^2)^7=6^{2*7}=6^{14}[/tex]

Final answer:

To simplify the expression [tex](6^2)^7[/tex] using a single exponent, multiply the inner exponent 2 by the outer exponent 7, which yields 6¹⁴.

Explanation:

To write the expression [tex](6^2)^7[/tex] using a single exponent, you need to apply the rule for raising a power to a power. This rule states that you multiply the exponents together. So for our expression, we have the base number 6 raised to the power of 2, and this result is then raised to the power of 7. To combine them into a single exponent, you multiply 2 by 7, which gives you 14.

Therefore,[tex](6^2)^7[/tex] can be simplified to 6¹⁴. This is because when you raise a power to another power, the powers are multiplied: for example, (a^b)^c = a^(b*c).

Answer:

Area of Δ ABC = 21.86 units square

Perimeter of Δ ABC = 24.59 units

Step-by-step explanation:

Given:

In Δ ABC

∠A=45°

∠C=30°

Height of triangle = 4 units.

To find area and perimeter of triangle we need to find the sides of the triangle.

Naming the end point of altitude as 'D'

Given [tex]BD\perp AC[/tex]

For Δ ABD

Since its a right triangle with one angle 45°, it means it is a special 45-45-90 triangle.

The sides of 45-45-90 triangle is given as:

Leg1 [tex]=x[/tex]

Leg2 [tex]=x[/tex]

Hypotenuse [tex]=x\sqrt2[/tex]

where [tex]x[/tex] is any positive number

We are given BD(Leg 1)=4

∴ AD(Leg2)=4

∴ AB (hypotenuse) [tex]=4\sqrt2=5.66 [/tex]  

For Δ CBD

Since its a right triangle with one angle 30°, it means it is a special 30-60-90 triangle.

The sides of 30-60-90 triangle is given as:

Leg1(side opposite 30° angle) [tex]=x[/tex]

Leg2(side opposite 60° angle) [tex]=x\sqrt3[/tex]

Hypotenuse [tex]=2x[/tex]

where [tex]x[/tex] is any positive number

We are given BD(Leg 1)=4

∴ CD(Leg2) [tex]=4\sqrt3=6.93[/tex]

∴ BC (hypotenuse) [tex]=2\times 4=8 [/tex]  

Length of side AC is given as sum of segments AD and CD

[tex]AC=AD+CD=4+6.93=10.93[/tex]

Perimeter of Δ ABC= Sum of sides of triangle

⇒ AB+BC+AC

⇒ [tex]5.66+8+10.93[/tex]

⇒ [tex]24.59[/tex] units

Area of Δ ABC = [tex]\frac{1}{2}\times base\times height[/tex]

⇒  [tex]\frac{1}{2}\times 10.93\times 4[/tex]

⇒ [tex]21.86[/tex] units square