A survey was conducted to measure the height of men. In the survey, respondents were grouped by age. In the 20-29 age group, the heights were normally distributed, with a mean of 69.9 inches and a standard deviation of 3.0 inches. a study participant is randomly sleected. what height cuts off the top 5%

Answer:

area of sector = 40.192 unit²

Step-by-step explanation:

Area of a sector = ∅/360 × πr²

where

∅ = angle in degree

r = radius

Area of sector  when ∅ = radian

area of a sector = 1/2r²∅

where

∅ = radian

r = radius

area of sector = 1/2 × 4² × 8/5 × π

area of sector = 1/2 × 16 × 8/5 × π

area of sector = 128/10 × π

area of sector = 12.8 × π

area of sector = 12.8 × 3.14

area of sector = 40.192 unit²

Answer:

a) t₀ =  1.73205 s

b) 1.0 C

Step-by-step explanation:

(A)

The time dilation (t) observed by an observer at rest relative to the time (t₀) measured by observer in motion is;

[tex]t = \frac{t_0}{\sqrt{1 - \frac{V^2}{C^2}}}[/tex]

[tex]t_0 = t \sqrt{1 - \frac{V^2}{C^2}}[/tex] time measured by captain

⇒ [tex]t_0 = 2.0 \sqrt{1 - \frac{0.5^2C^2}{C^2}}[/tex]             V = 0.5 c

⇒ t₀ =  1.73205 s

(B)

Speed of the light never exceeds by its real value. The speed of the light in any frame of reference is constant.

∵ It will be "1.0C" or just "C"

Answer:

Equation:  [tex]F=1500(1.0025)^{12t}[/tex]

The balance after 5 years is:  $1742.43

Step-by-step explanation:

This is a compound growth problem . THe formula is:

[tex]F=P(1+\frac{r}{n})^{nt}[/tex]

Where

F is future amount

P is present amount

r is rate of interest, annually

n is the number of compounding per year

t is the time in years

Given:

P = 1500

r = 0.03

n = 12 (compounded monthly means 12 times a year)

The compound interest formula modelled by the variables is:

[tex]F=1500(1+\frac{0.03}{12})^{12t}\\F=1500(1.0025)^{12t}[/tex]

Now, we want balance after 5 years, so t = 5, substituting, we get:

[tex]F=1500(1.0025)^{12t}\\F=1500(1.0025)^{12*5}\\F=1500(1.0025)^{60}\\F=1742.43[/tex]

The balance after 5 years is:  $1742.43

Answer:

A. $1500

Step-by-step explanation:

We need to find the countertop area, so let's calculate the areas of the rectangles that the problem broke the countertop into:

1. "1 rectangle has a base of 70 inches and height of 20 inches"

The area of a rectangle is denoted by: A = bh, where b is the base and h is the height. Here, b = 70 and h = 20, so the area is: A = 70 * 20 = 1400 inches squared

2. "The other rectangle has a base of 20 inches and height of 30 inches"

Again, use A = bh: b = 20 and h = 30, so A = 20 * 30 = 600 inches squared

Add up these two areas: 1400 + 600 = 2000 inches squared.

The problem says that the cost is $0.75 per square inch, so multiply this by 2000 to get the total cost of 2000 square inches:

2000 * 0.75 = $1500

Thus, the answer is A.

Hope this helps!

Answer:

$1500

Step-by-step explanation:

You can divide this figure into 2 rectangle with dimensions:

1) 70 × 20

2) 20 × (50-20: 20 × 30

Area:

(70×20) + (20×30)

1400 + 600

2000 in²

Cost per in²: 0.75

2000in² cost:

2000 × 0.75

$1500

.24 is equivalent to

24%

24/100

6/25

12/50, and more!

Hope this helped

Answer:

Step-by-step explanation:

This is permutation question

The formula for it:

Finding the answer:

Answer:

B - -5 < y < 5

Step-by-step explanation:

Range is highest and lowest y value the graph goes to.

You can see on the graph that it does not pass 5 and -5

Answer:

A. [tex]\sqrt{64}=8[/tex] miles

Step-by-step explanation:

Given two Cartesian coordinates [tex](x_1,y_1)\&(x_2,y_2)[/tex], the distance between the points is given as:

[tex]d = \sqrt{((x_1-x_2)^2+(y_1-y_2)^2)}[/tex]

Converting to polar coordinates

[tex](x_1,y_1) = (r_1 cos \theta_1, r_1 sin \theta_1)\\(x_2,y_2) = (r_2 cos \theta_2, r_2 sin \theta_2)[/tex]

Substitution into the distance formula gives:

[tex]\sqrt{((r_1 cos\theta_1-r_2 cos \theta_2)^2+(r_1 sin \theta_1-r_2 sin \theta_2)^2}\\=\sqrt{(r_1^2+r_2^2-2r_1r_2(cos \theta_1 cos \theta_2+sin\theta_1 sin \theta_2) }\\= \sqrt{r_1^2+r_2^2-2r_1r_2cos (\theta_1 -\theta_2)}[/tex]

In the given problem,

[tex](r_1,\theta_1)=(8 mi, 63^0) \:and\: (r_2,\theta_2)=(8 mi, 123^0 ).[/tex]

[tex]Distance=\sqrt{8^2+8^2-2(8)(8)cos (63 -123)}\\=\sqrt{128-128cos (-60)}\\=\sqrt{64}=8 mile[/tex]

The closest option is  A. [tex]\sqrt{64}=8[/tex] miles

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let [tex]\bar X[/tex] = sample average time spent studying

The z-score probability distribution for sample mean is given by;

          Z = [tex]\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean hours spent studying = 25 hours

            [tex]\sigma[/tex] = standard deviation = 15 hours

            n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < [tex]\bar X[/tex] < 30 hours)

    P(29 hours < [tex]\bar X[/tex] < 30 hours) = P([tex]\bar X[/tex] < 30 hours) - P([tex]\bar X[/tex] [tex]\leq[/tex] 29 hours)

    P([tex]\bar X[/tex] < 30 hours) = P( [tex]\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }[/tex] < [tex]\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }[/tex] ) = P(Z < 2) = 0.97725

    P([tex]\bar X[/tex] [tex]\leq[/tex] 29 hours) = P( [tex]\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }[/tex] [tex]\leq[/tex] [tex]\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }[/tex] ) = P(Z [tex]\leq[/tex] 1.60) = 0.94520

So, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < [tex]\bar X[/tex] < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Answer:

In the screenshot you have the right answer, it is indeed 1000 times greater

Step-by-step explanation:

Answer:

At critical point in D

a

     [tex](x,y) = (0,0)[/tex]

b

[tex]f(x,y) = f(x) =11 -x^2[/tex]

where [tex]-1 \le x \le 1[/tex]

c

maximum value 11

minimum value  10

Step-by-step explanation:

Given [tex]f(x,y) =10x^2 + 11x^2[/tex]

At critical point

[tex]f'(x,y) = 0[/tex]

 =>  [tex][f'(x,y)]_x = 20x =0[/tex]

=>   [tex]x =0[/tex]

Also

[tex][f'(x,y)]_y = 22y =0[/tex]

=>   [tex]y =0[/tex]

Now considering along the boundary

       [tex]D = 1[/tex]

=>  [tex]x^2 +y^2 = 1[/tex]

=>  [tex]y =\pm \sqrt{1- x^2}[/tex]

Restricting [tex]f(x,y)[/tex] to this boundary

      [tex]f(x,y) = f(x) = 10x^2 +11(1-x^2)^{\frac{2}{1} *\frac{1}{2} }[/tex]

                            [tex]= 11-x^2[/tex]

At boundary point D = 1

Which implies that [tex]x \le 1[/tex]  or [tex]x \ge -1[/tex]

So the range of  x is

                  [tex]-1 \le x \le 1[/tex]

Now along this this boundary the critical point is at

            [tex]f'(x) = 0[/tex]

=>         [tex]f'(x) = -2x =0[/tex]

=>         [tex]x=0[/tex]

Now at maximum point [tex](i.e \ x =0)[/tex]

            [tex]f(0) =11 -(0)[/tex]

                   [tex]= 11[/tex]

For the minimum point x = -1 or x =1

              [tex]f(1) = 11 - 1^2[/tex]

                      [tex]=10[/tex]

              [tex]f(-1) = 11 -(-1)^2[/tex]

                         [tex]=10[/tex]

Answer:

  Its magnitude will be larger than 0.004.

Step-by-step explanation:

When a divisor is less than 1, the quotient will be greater than the dividend.

When the divisor is "almost zero", the quotient will be much greater than the dividend. Here, the dividend may be considered to be "almost zero", so we cannot say anything about the actual quotient except to say its magnitude will be greater than the dividend.

_____

The dividend is positive, so the quotient will have the same sign as the divisor. (Negative divisors can be "almost zero," too.)

Answer:

2695 miles

Step-by-step explanation:

The car can travel 2,695 miles on 55 gallons of gas.

To determine how many miles a car can go on 55 gallons of gas if it can go 490 miles on 10 gallons, we need to find the car's miles per gallon (mpg) and then use that to calculate the distance for 55 gallons.

First, calculate the miles per gallon (mpg):

mpg = 490 miles / 10 gallons = 49 miles per gallon

Now, use the mpg to find the distance the car can travel on 55 gallons:

Distance = 49 miles per gallon * 55 gallons = 2,695 miles

Therefore, the car can go 2,695 miles on 55 gallons of gas.

Answer:

B

Step-by-step explanation:

Answer:

Explanation:

The amount of animal waste one zoo is diposing daily is approximately normal with:

The proportion of waste over 350 lbs may be found using the table for the area under the curve for the cumulative normal standard probability.

First, find the z-score for 350 lbs:

      [tex]z-score=\dfrac{X-\mu}{\sigma}[/tex]

      [tex]z-score=\dfrac{350lbs-348.5lbs}{38.2lbs}\approx0.04[/tex]

There are tables for the cumulative areas (probabilities) to the left and for the cumulative areas to the right of the z-score.

You want the proportion of the days when the z-score is more than 0.04; then, you can use the table for the values to the rigth of z = 0.04.

From such table, the area or probability is 0.4840.

The attached image shows a portion of the table with that value: it is the cell highlighted in yellow.

Hence, the answer is the option (A) 0.484.

By calculating the z-score for 350 pounds of waste and consulting the standard normal distribution, the proportion of days the zoo can expect to have enough animal waste to power the entire aquarium is approximately 0.484.

To determine the proportion of days the zoo can expect to generate enough animal waste to run the entire aquarium, we can use z-scores in a normal distribution. Given the mean (μ = 348.5) and standard deviation (σ = 38.2), we want to find the proportion of the data that is above 350 pounds.

First, we calculate the z-score for 350 pounds:

z = (X - μ) / σ = (350 - 348.5) / 38.2 ≈ 0.04

Now we need to find the probability that the z-score is greater than 0.04. Consulting a standard normal distribution table or using a calculator, this gives us a probability of approximately 0.484.

Therefore, the proportion of the days the zoo can expect to obtain enough waste to cover the energy demands for the entire aquarium is 0.484.

Answer:

–11 and 2

Step-by-step explanation:

observe

x² + 9x – 22 = 0

(x + 11)(x – 2) = 0

x = –11 or x = 2

Answer:

a) [tex]\delta = 1.4\,\%[/tex], b) [tex]\delta_{max} = 1.35\,\%[/tex]

Step-by-step explanation:

a) The area formula for a square is:

[tex]A =l^{2}[/tex]

The total differential for the area is:

[tex]\Delta A = \frac{\partial A}{\partial l}\cdot \Delta l[/tex]

[tex]\Delta A = 2\cdot l \cdot \Delta l[/tex]

The absolute error for the area of the square is:

[tex]\Delta A = 2\cdot (10\,cm)\cdot (0.07\,cm)[/tex]

[tex]\Delta A = 1.4\,cm^{2}[/tex]

Thus, the relative error is:

[tex]\delta = \frac{\Delta A}{A}\times 100\,\%[/tex]

[tex]\delta = \frac{1.4\,cm^{2}}{100\,cm^{2}} \times 100\,\%[/tex]

[tex]\delta = 1.4\,\%[/tex]

b) The maximum allowable absolute error for the area of the square is:

[tex]\Delta A_{max} = \left(\frac{\delta}{100} \right)\cdot A[/tex]

[tex]\Delta A_{max} = \left(\frac{2.7}{100} \right)\cdot (100\,cm^{2})[/tex]

[tex]\Delta A_{max} = 2.7\,cm^{2}[/tex]

The maximum allowable absolute error for the length of a side of the square is:

[tex]\Delta l_{max}= \frac{\Delta A_{max}}{2\cdot l}[/tex]

[tex]\Delta l_{max} = \frac{2.7\,cm^{2}}{2\cdot (10\,cm)}[/tex]

[tex]\Delta l_{max} = 0.135\,cm[/tex]

Lastly, the maximum allowable relative error is:

[tex]\delta_{max} = \frac{\Delta l_{max}}{l}\times 100\,\%[/tex]

[tex]\delta_{max} = \frac{0.135\,cm}{10\,cm} \times 100\,\%[/tex]

[tex]\delta_{max} = 1.35\,\%[/tex]

Answer:

a) [tex] P(-1<Z<1)= P(Z<1) -P(Z<-1)= 0.841-0.159= 0.682[/tex]

b) [tex] P(-2<Z<2)= P(Z<2) -P(Z<-2)= 0.977-0.0228= 0.954[/tex]

c) [tex] P(-3<Z<3)= P(Z<3) -P(Z<-3)= 0.999-0.0013= 0.998[/tex]

The results are on the fogure attached.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Part a

For this case we want to find this probability:

[tex] P(-1<Z<1)[/tex]

And we can find this probability with this difference:

[tex] P(-1<Z<1)= P(Z<1) -P(Z<-1)[/tex]

And if we find the probability using the normal standard distribution or excel we got:

[tex] P(-1<Z<1)= P(Z<1) -P(Z<-1)= 0.841-0.159= 0.682[/tex]

Part b

For this case we want to find this probability:

[tex] P(-2<Z<2)[/tex]

And we can find this probability with this difference:

[tex] P(-2<Z<2)= P(Z<2) -P(Z<-2)[/tex]

And if we find the probability using the normal standard distribution or excel we got:

[tex] P(-2<Z<2)= P(Z<2) -P(Z<-2)= 0.977-0.0228= 0.954[/tex]

Part c

For this case we want to find this probability:

[tex] P(-3<Z<3)[/tex]

And we can find this probability with this difference:

[tex] P(-3<Z<3)= P(Z<3) -P(Z<-3)[/tex]

And if we find the probability using the normal standard distribution or excel we got:

[tex] P(-3<Z<3)= P(Z<3) -P(Z<-3)= 0.999-0.0013= 0.998[/tex]

Final answer:

The question asks to find the area under the standard normal curve for specific z-score ranges. Using the empirical rule, we conclude that respective areas for those ranges are approximately 68%, 95%, and 99.7%. The exact areas can be found using a Z-table.

Explanation:

The question involves finding the area under the standard normal curve between specified z-scores. This is a fundamental concept in statistics, often used to find probabilities related to normally distributed data.

To find the exact areas based on the z-scores, we can refer to the Z-table of Standard Normal Distribution. This table lists the cumulative probabilities from the mean up to a certain z-score. By looking up the area to the left of each positive z-score and doubling it, we can get the approximate area between the negative and positive z-scores mentioned above.

Answer:

.5

Step-by-step explanation:

-1.9+4=2.1

2.1+(-1.6)=.5

Step-by-step explanation:

Bringing like terms on one side

2 + 7 - 5 = 2x - 4x

9 - 5 = - 2x

4 = - 2x

4/ - 2 = x

- 2 = x

Answer:

$362.50

Profit: $217.50

Step-by-step explanation:

Answer:

The equation is correct

Step-by-step explanation:

The equation, written as:

[log_2 (10)][log_4 (8)][log_10 (4)] = 3

Consider the change of base formula:

log_a (x) = [log_10 (x)]/ [log_10 (a)]

Applying the change of base formula to change the expressions in base 2 and base 4 to base 10.

(1)

log_2 (10) = [log_10 (10)]/[log_10 (2)]

= 1/[log_10 (2)]

(Because log_10 (10) = 1)

(2)

log_4 (8)  = [log_10 (8)]/[log_10 (4)]

Now putting the values of these new logs in base 10 into the left-hand side of original equation to verify if we have 3, we have:

[log_10 (2)][log_8 (4)][log_10 (4)]

= [1/ log_10 (2)][log_10 (8) / log_10 (4)][log_10 (4)]

= [1/log_10 (2)] [log_10 (8)]

= [log_10 (8)]/[log_10 (2)]

= [log_10 (2³)]/[log_10 (2)]

Since log_b (a^x) = xlog_b (a)

= 3[log_10 (2)]/[log_10 (2)]

= 3 as required

Therefore, the left hand side of the equation is equal to the right hand side of the equation.

Answer:

B on E2020.

Step-by-step explanation:

Answer:

Step-by-step explanatio n: ummmoirnd iehcn

Answer:

Step-by-step explanation:

quadratic equation: ax² + bx + c =0

x' = [-b+√(b²-4ac)]/2a   and x" =  [-b-√(b²-4ac)]/2a  

6 = x² – 10x ; x² - 10x -6 =0

(a=1, b= - 10 and c = - 6

x' = [10+√(10²+4(1)(-6)]/2(1)  and x" = [10-√(10²+4(1)(-6)]/2(1)

x' =5+√31  and x' = 5-√31

Answer:

12,500 square millimeters

Step-by-step explanation:

Answer:

c(12,500)

Step-by-step explanation:

Answer:

Hence total of 10 students are not able to complete the exam.

Step-by-step explanation:

Given:

Mean for completing exam =80 min

standard deviation =10 min.

To find:

how much student will not complete the  exam?

Solution:

using the Z-table score we can calculate the required probability.

Z=(Required time -mean)/standard deviation.

A standard on an avg class contains:

60 students.

consider for 70 mins and then 90 mins (generally calculate ±  standard deviation of mean)(80-10 and 80+10).

1)70 min

Z=(70-80)/10

Z=-1

Now corresponding p will be

P(z=-1)

=0.1587

therefore

Now for required 90 min will be

Z=(90-80)/10

=10/10

z=1

So corresponding value of p is

P(z<1)=0.8413

this means 0.8413 of 60 students are able to complete the exam.

0.8413*60

=50.47

which approximate 50 students,

total number =60

and total number student will able to complete =50

Total number of student will not complete =60-50

=10.

About 15.87% of college students are expected not to finish the final examination within the 90-minute time limit, based on the properties of the normal distribution with a mean of 80 minutes and a standard deviation of 10 minutes.

The student's question involves using the properties of the normal distribution to determine the percentage of students who will not finish a final examination in the given time frame.

To compute this, we need to calculate the z-score that corresponds to the 90-minute time limit. The z-score formula is:

Z = (X - μ) / σ

where X is the value of interest, μ (mu) is the mean, and σ (sigma) is the standard deviation. Plugging in the numbers:

Z = (90 - 80) / 10 = 1

A z-score of 1 corresponds to a percentile of approximately 84.13%, meaning about 84.13% of students will finish within 90 minutes. To find the percentage that will not finish in time, we subtract this from 100%:

100% - 84.13% = 15.87%

Therefore, approximately 15.87% of the class will not finish the exam in time.