What is bigger 4.17 or 4.085

Answer:

12,500 square millimeters

Step-by-step explanation:

Answer:

c(12,500)

Step-by-step explanation:

Answer:

[tex]95-2.76\frac{6.6}{\sqrt{30}}=91.674[/tex]    

[tex]95+2.76\frac{6.6}{\sqrt{30}}=98.326[/tex]    

We can say at 99% confidence that the true mean is between (91.674;98.326)    

Step-by-step explanation:

Data given

[tex]\bar X=95[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=6.6 represent the sample standard deviation

n=30 represent the sample size  

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

We need to find the critical value [tex]t_{\alpha/2}[/tex] and we need to find first the degrees of freedom, given by:

[tex]df=n-1=30-1=29[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that [tex]t_{\alpha/2}=2.76[/tex]

And replacing we got:

[tex]95-2.76\frac{6.6}{\sqrt{30}}=91.674[/tex]    

[tex]95+2.76\frac{6.6}{\sqrt{30}}=98.326[/tex]    

We can say at 99% confidence that the true mean is between (91.674;98.326)    

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: a) [tex]E(x)=\dfrac{-5945}{(30x+11)(5x+200)}[/tex], b) 0.7975, demand is inelastic, c) increase.

Step-by-step explanation:

Since we have given that

[tex]D(x)=\dfrac{5x+200}{30+11}[/tex]

So, derivative w.r.t x would be

[tex]D'(x)=\dfrac{5(30x+11)-30(5x+200)}{(30x+11)^2}\\\\D'(x)=\dfrac{150x+55-150x-6000}{(30x+11)^2}\\\\D'(x)=\dfrac{5945}{(30x+11)^2}[/tex]

As we know that

[tex]E(x)=\dfrac{-xD'(x)}{D(x)}\\\\\\E(x)=\dfrac{\dfrac{-(-)5945x}{(30x+11)^2}}{\dfrac{5x+200}{30x+11}}\\\\\\E(x)=\dfrac{5945x}{(30x+11)(5x+200)}[/tex]

(b) Find the elasticity when x = 2.

So, we put x = 2, we get that

[tex]E(2)=\dfrac{5945\times 2}{(30(2)+11)((5(2)+200))}\\\\E(2)=\dfrac{11890}{(60+11)(10+200)}\\\\E(2)=\dfrac{11890}{71\times 210}\\\\E(2)=\dfrac{11890}{14910}\\\\E(2)=0.7975[/tex]

Since, 0.7975 < 1, so the demand is inelastic.

(c) At $2 per plant, will a small increase in price cause the total revenue to increase or decrease?

The total revenue will also increase with increase in price.

As total revenue = [tex]price\times quantity[/tex]

Hence, a) [tex]E(x)=\dfrac{-5945}{(30x+11)(5x+200)}[/tex], b) 0.7975, demand is inelastic, c) increase.

This problem involves the calculation of the elasticity of a demand function using the derivative of the function. The elasticity is then used to analyze the effect on the total revenue when the price changes. The elasticity at a specific point is calculated and used for further analysis.

For part (a), to find the elasticity of the demand function, we need to use the formula for the price elasticity of demand, which is E = (dQ/dX) * (X/Q). Here, dQ/dX is the derivative of the demand function concerning X. This needs to be calculated first. The value of E provides us with the measure of elasticity.

For part (b), when x = 2 we substitute this value into the formula for E to get the elasticity at x = 2.

For part (c), based on the concept of elasticity, if E > 1, the demand is said to be elastic and a price decrease will result in an increase in total revenue, and vice versa. If E < 1, the demand is said to be inelastic and a price decrease will result in a decrease in total revenue, and vice versa. So, after calculating E at x = 2, we can use it to determine the effect on total revenue.

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Answer:

The length of the mosquito in the picture is 117.6 mm.

Step-by-step explanation:

This question can be solved using a rule of three.

The real size of the mosquito is 16 mm, which is 100% = 1.

In the picture, the size is magnified 635%, so it is 100+635 = 735% = 7.35. So

16mm - 1

x mm - 7.35

x = 7.35*16 = 117.6

The length of the mosquito in the picture is 117.6 mm.

Final answer:

To find the magnified length of a mosquito in a picture, multiply its actual size (16 mm) by the magnification factor (6.35), resulting in a magnified length of 101.6 millimeters.

Explanation:

The question involves calculating the magnified length of a mosquito in a picture, given that the magnification is 635% of its actual size and the actual size is 16 millimeters. First, understand that 635% magnification means the mosquito's image is 6.35 times its actual size. To find the magnified length, multiply the actual length by the magnification factor.

Magnified length = Actual length × Magnification factor

= 16 mm × 6.35

= 101.6 mm

Therefore, the length of the mosquito in the picture is 101.6 millimeters.

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:

17

Step-by-step explanation:

32+2=34

34/2= 17

Half of the sum of 32 and twice a number 'x' expressed as '2ans' would be calculated by adding 32 and 2x and then dividing by 2, resulting in the expression 16 + x.

The question asks to find half of the sum of 32 and an unspecified number (mentioned as '2ans'). Assuming '2ans' means twice the number 'ans', which can be represented as 2x, where 'x' is the particular value of 'ans'. First, calculate the sum of 32 and 2x, and then divide that sum by two to find half of it.

The steps to solve this are:

Calculate the sum: 32 + 2x.

To find half of the sum, divide by 2: (32 + 2x) / 2.

Simplify the expression: 16 + x.

Therefore, half of the sum of 32 and twice a number 'x' (2x) is 16 + x.

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:

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) [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:

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:

$362.50

Profit: $217.50

Step-by-step explanation:

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:

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

Step-by-step explanation:

H*W*L

280=7*W*(6+W)

280=42W+7W^2

40=6w+w^2

W^2+6W-40=0

(W-4)(W+10)=0

W=4

We can only use positive results

L=6+W

L=4+6=10

Given that, the height of the cardboard box=10 inches and volume = 540 cubic inches.

Let the width of the cardboard box be x, then the length will be 3x.

The formula to find the volume of a rectangular prism is Volume=Length×Width×Height.

Now, 540=10×x×3x

⇒x²=18

⇒x=3√2 inches.

Therefore, the width of a cardboard box is 3√2 inches.

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Answer:

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

Step-by-step explanation:

Answer:

Step-by-step explanation:

Since x is a normally distributed random variable, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = the random variable

µ = mean

σ = standard deviation

From the information given,

µ = 16

σ = 2

a.​ P(x ≥ 17.5​) = 1 - (x < 17.5)

For x < 17.5

z = (17.5 - 16)/2 = 0.75

Looking at the normal distribution table, the probability corresponding to the z score is 0.77

P(x ≥ 17.5​) = 1 - 0.77 = 0.23

b.​ P(x ≤ 12​)

z = (12 - 16)/2 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.023

P(x ≤ 12​) = 0.023

c) P(16.78 ≤ x ≤ 20.46)

For x = 16.78,

z = (16.78 - 16)/2 = 0.39

Looking at the normal distribution table, the probability corresponding to the z score is 0.65

For x = 20.46,

z = (20.46 - 16)/2 = 2.23

Looking at the normal distribution table, the probability corresponding to the z score is 0.987

Therefore,

P(16.78 ≤ x ≤ 20.46) = 0.987 - 0.65 = 0.337

d) P(11.48 ≤ x ≤ 19.06)

For x = 11.48,

z = (11.48 - 16)/2 = - 2.26

Looking at the normal distribution table, the probability corresponding to the z score is 0.012

For x = 19.06,

z = (19.06 - 16)/2 = 1.53

Looking at the normal distribution table, the probability corresponding to the z score is 0.94

Therefore,

P(11.48 ≤ x ≤ 19.06) = 0.94 - 0.012 = 0.928

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:

B

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:

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:

Step-by-step explanatio n: ummmoirnd iehcn

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:

Yes, if I'm assuming that it was written as [tex]119*10^-3[/tex], then this is a basis for a scientific notation.

The mathematical statement translates to an inequality 20 ≤ 4C - 8, and solving this results in C ≥ 7.

The question provides a mathematical statement which can be translated into an equation. The statement 'The number 20 is no less the difference of 4 times the number C and 8' translates into the equation 20 ≤ 4C - 8. In this equation, 4 times a number C minus 8 can result in a value that is at least 20. To find the value for C, you would need to isolate C. This is done by first adding 8 to both sides to get 28 ≤ 4C, then dividing by 4 to get C ≥ 7. C, therefore, could be any value that is 7 or greater.

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Answer:

A. 5 * 1/7

Step-by-step explanation:

When solving division, you can also multiply by the reciprocal of the second number to get the same answer.

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.

Answer:

16*pi=50.24

Step-by-step explanation:

Answer:

16* pi = 50.24

Step-by-step explanation:

The standard deviation of the sampling distribution of the sample​ mean is 4.587 and this can be determined by using the formula of the standard deviation of the sample mean.

Given :

The standard deviation of a random variable X is 20 and a random sample of size n equals 19.

The formula of the standard deviation of the sample mean can be used to determine the standard deviation of the sampling distribution of the sample​ mean.

The standard deviation of the sample mean is given by:

[tex]\sigma_m=\dfrac{\sigma}{\sqrt{n} }[/tex]   --- (1)

Now put the value of [tex]\sigma[/tex] that is 20 and the value n that is 19 in the equation (1).

[tex]\sigma_m = \dfrac{20}{\sqrt{19} }[/tex]

[tex]\sigma_m = \dfrac{20}{4.36}[/tex]

[tex]\sigma_m = 4.587[/tex]

So, the standard deviation of the sampling distribution of the sample​ mean is 4.587.

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The standard deviation of the sampling distribution of the sample mean, also referred to as the standard error, for this scenario is approximately 4.58.

The standard deviation of a sampling distribution of the sample mean, also known as the standard error, can be calculated using the formula Standard Error = σ/√n, where σ is the population standard deviation and n is the sample size. In your case, the standard deviation (σ) of the random variable X is 20, and your sample size (n) is 19.

Plugging these values into the formula, we get Standard Error = 20/√19, which can be approximated as 4.58.

Therefore, the standard deviation of the sampling distribution of the sample mean for this scenario is approximately 4.58.

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