Running times for 400 meters are Normally distributed for young men between 18 and 30 years of age with a mean of 93 seconds and a standard deviation of 16 seconds. How fast does a man have to run to be in the top 1% of runners?

Answer:

When x = 1 and z = 4,   [tex]y=\frac{80}{9}[/tex]

Step-by-step explanation:

The variation described in the problem can be written using a constant of proportionality "b" as:

[tex]y=b\,\,x\,\,z^2[/tex]

The other piece of information is that when x = 5 and z = 1, then y gives 25/9. So we use this info to find the constant "b":

[tex]y=b\,\,x\,\,z^2\\\frac{25}{9} =b\,\,(5)\,\,(1)^2\\\frac{25}{9} =b\,\,(5)\\b=\frac{5}{9}[/tex]

Knowing this constant, we can find the value of y when x=1 and z=4 as:

[tex]y=b\,\,x\,\,z^2\\y=\frac{5}{9} \,\,x\,\,z^2\\y=\frac{5}{9} \,\,(1)\,\,(4)^2\\y=\frac{5*16}{9}\\y=\frac{80}{9}[/tex]

Answer:

c = 24 can i get brainliest

Step-by-step explanation:

Answer:

Can you give the question. Can you post the picture. I can help solve. I will edit this answer once you have given the question/picture.

Answer:

Step-by-step explanation:

(x + k) is a factor of f(x)

x+k=0 => x= -k;    -k is a root of f(x)

=> f(-k)=0

[tex](x + k) is a factor of f(x)x+k=0 = > x= -k; -k is a root of f(x)= > f(-k)=0[/tex]

So the correct option is B.fl-k)=0.

The cube root function is f(x)=3√x f ( x ) = x 3 . A radical function is a function that is defined by a radical expression. The following are examples of rational functions: f(x)=√2x4−5 f ( x ) = 2 x 4 − 5 ; g(x)=3√4x−7 g ( x ) = 4 x − 7 3 ; h(x)=7√−8x2+4 h ( x ) = − 8 x 2 + 4 7 .

The root function is used to find a single solution to a single function with a single unknown. In later sections, we will discuss finding all the solutions to a polynomial function. We will also discuss solving multiple equations with multiple unknowns. For now, we will focus on using the root function.

Learn more about root function here: https://brainly.com/question/13136492

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Answer:

The probability that on a randomly selected day the statistics professor will have five messages is 0.1755.

Step-by-step explanation:

Let the random variable X represent the number of e-mail messages per day a statistics professor receives from students.

The random variable is approximated by the Poisson Distribution with parameter λ = 5.

The probability mass function of X is as follows:

[tex]P(X=x)=\frac{e^{-5}\cdot 5^{x}}{x!};\ x=0,1,2,3...[/tex]

Compute the probability that on a randomly selected day she will have five messages as follows:

[tex]P(X=5)=\frac{e^{-5}\cdot 5^{5}}{5!}[/tex]

               [tex]=\frac{0.006738\times 3125}{120}\\\\=0.17546875\\\\\approx 0.1755[/tex]

Thus, the probability that on a randomly selected day the statistics professor will have five messages is 0.1755.

Answer: 720 ft

Step-by-step explanation: He rides 720 feet.

if 120 feet are in 10 seconds then;

60 seconds are 60/10*120=720 feet

Answer:

720

Step-by-step explanation:

120/10 to find his feet per second which is 12 feet per second

12*60

since there are 60 seconds in a minute

= 720

Answer:

[tex]-5x^{4} -20x^{3} -5x-20[/tex]

Step-by-step explanation:

[tex]-5[(x^{3} +1)(x+4)][/tex]

Use the FOIL method for the last two groups.

[tex]-5(x^{4} +4x^{3} +x+4)[/tex]

Now, distribute the -5 into each term.

[tex]-5x^{4} -20x^{3} -5x-20[/tex]

Answer:

B is the midpoint of line segment AC

Answer:

4/25

Step-by-step explanation:

The probability the first spin lands on gray is 8/10 = 4/5.

The probability the second spin lands on blue is 2/10 = 1/5.

The probability of both events is 4/5 × 1/5 = 4/25.

Answer:

what method exactly r u using ????

Answer:

see explanation

Step-by-step explanation:

The n th term of an arithmetic sequence is

[tex]a_{n}[/tex] = a₁ + (n - 1)d

(a)

Given a₁ = - 2 and a₄ = 16, then

a₁ + 3d = 16 , that is

- 2 + 3d = 16 ( add 2 to both sides )

3d = 18 ( divide both sides by 3 )

d = 6

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

(b)

Given

[tex]a_{n}[/tex] = 11998 , then

a₁ + (n - 1)d = 11998 , that is

- 2 + 6(n - 1) = 11998 ( add 2 to both sides )

6(n - 1) = 12000 ( divide both sides by 6 )

n - 1 = 2000 ( add 1 to both sides )

n = 2001

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

Answer:

A

Step-by-step explanation:

Answer:

The answer here is A.

A) A is congruent to D.

A=

Step-by-step explanation:

AP E

Answer:

Bailey: 16%

Coco: 28%

Ginger: 32%

Ruby: 24%

I hope this helps!

Answer:

[tex]y(x)=c_1e^{-3x} cos(2x)+c_2e^{-3x} sin(2x)[/tex]

Step-by-step explanation:

In order to find the general solution of a homogeneous second order differential equation, we need to solve the characteristic equation. This is basically as easy as solving a quadratic.

For a second order differential equation of type:

[tex]ay''+by'+cy=0[/tex]

Has characteristic equation:

[tex]a r^{2} +br+c=0[/tex]

Whose solutions [tex]r_1 , r_2 ,.., r_n[/tex] are the roots from which the general solution can be formed. There are three cases:

Real roots:

[tex]y(x)=c_1e^{r_1 x} +c_2e^{r_2 x}[/tex]

Repeated roots:

[tex]y(x)=c_1e^{r x} +c_1 xe^{r x}[/tex]

Complex roots:

[tex]y(x)=c_1e^{\lambda x}cos(\mu x) +c_2e^{\lambda x}sin(\mu x)\\\\Where:\\\\r_1_,_2=\lambda \pm \mu i[/tex]

Therefore:

The characteristic equation for:

[tex]y''+6y'+13y=0[/tex]

Is:

[tex]r^{2} +6r+13=0[/tex]

Solving for [tex]r[/tex] :

[tex]r_1_,_2= -3 \pm 2i[/tex]

So:

[tex]\mu = 2\\\\and\\\\\lambda=-3[/tex]

Hence, the general solution of the differential equation will be given by:

[tex]y(x)=c_1e^{-3x} cos(2x)+c_2e^{-3x} sin(2x)[/tex]

Answer:

C.

Step-by-step explanation:

Density = Mass / Volume

2.7 = 54 / V

V = 54 / 2.7

V = 20 cubic cm

Answer:

a) P(6) = 0.0097

b) P(More than 3) = 0.1611

Step-by-step explanation:

For each question, there are only two possible outcomes. Either it is guessed correctly, or it is not. Questions are independent of each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

A student takes a multiple-choice test that has 11 questions.

This means that [tex]n = 11[/tex]

Each question has five choices.

This means that [tex]p = \frac{1}{5} = 0.2[/tex]

(a) Find P (6)

This is P(X = 6).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 6) = C_{11,6}.(0.2)^{6}.(0.8)^{5} = 0.0097[/tex]

P(6) = 0.0097

(b) Find P (More than 3).

Either P is 3 or less, or it is more than three. The sum of the probabilities of these outcomes is 1. So

[tex]P(X \leq 3) + P(X > 3) = 1[/tex]

We want P(X > 3). So

[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]

In which

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{11,0}.(0.2)^{0}.(0.8)^{11} = 0.0859[/tex]

[tex]P(X = 1) = C_{11,1}.(0.2)^{1}.(0.8)^{10} = 0.2362[/tex]

[tex]P(X = 2) = C_{11,2}.(0.2)^{2}.(0.8)^{9} = 0.2953[/tex]

[tex]P(X = 3) = C_{11,3}.(0.2)^{3}.(0.8)^{8} = 0.2215[/tex]

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0859 + 0.2362 + 0.2953 + 0.2215 = 0.8389[/tex]

Then

[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.8389 = 0.1611[/tex]

P(More than 3) = 0.1611

Answer:

1/18

Step-by-step explanation:

First you would add 1/2 and 1/2 to get 1 then you would divide it by 18 to get 1/18

Answer:

1/18

Step-by-step explanation:

plz mark me brainliest.

Answer:

the area of the circle is 102.11 square metres

Answer:

The number of the class containing the  largest score can be found in frequency  24 and the class is  98 - 103

For the third class 78 - 83 ; the upper limit = 83

The class width for this histogram 5

The number of students that took the exam simply refers to the frequency is  84

The percentage of students that scored higher than 95.5 is 53%

Step-by-step explanation:

The objective of this question is to use the following histogram that shows the exam scores for a Pre-algebra class to answer the question given:

NOW;

The table given in the question can be illustrated as follows:

S/N           Class                Score          Frequency

1                 68  - 73            70.5           0

2                73 - 78             75.5           4

3                78 - 83             80.5           8

4                83 - 88             85.5           12

5                88 - 93            90.5           16

6                93 - 98            95.5           20

7                98 - 103           100.5          24                            

TOTAL:                                                 84

a) The number of the class containing the  largest score can be found in frequency  24 and the class is  98 - 103

b) For the third class 78 - 83 ; the upper limit = 83  ( since the upper limit is derived by addition of  5 to the last number showing  in the highest value specified by the number in the class interval which is 78 ( i.e 78 + 5 = 83))

c)   The class width for this histogram 5 ; since it  is the difference  between the upper and lower boundaries  limit of the given class.

So , from above the difference in any of the class will definitely result into 5

d) The number of students that took the exam simply refers to the frequency ; which is (0+4+8+12+16+20+24) = 84

e) Lastly; the percentage of students that scored higher than 95.5 is ;

⇒[tex]\dfrac{20+24}{84} *100[/tex]

= 0.5238095 × 100

= 52.83

To the nearest percentage ;the percentage of students that scored higher than 95.5 is 53%

Answer:

1. 98-103 (6th class)

2. 88

3. 5

4. 84

5. 52%

Step-by-step explanation:

Find attached the frequency table.

The class of exam scores falls between (1, 2, 3, 4, 5, or 6).

The exam score ranged from 68-103

1) The largest number of exam scores = 24

The largest number of exam scores is in the 6th class = 98 -103

Step 2 of 5:

The upper class limit is the higher number in an interval. Third class interval is 83-88

The upper class limit of the third class 88.

Step 3 of 5:

Class width = upper class limit - lower class limit

We can use any of the class interval to find this as the answer will be the same. Using the interval between 73-78

Class width = 78 - 73

Class width for the histogram = 5

Step 4 of 5:

The total of students that took the test = sum of all the frequency

= 0+4+8+12+16+20+24 = 84

The total of students that took the test = 84

Step 5 of 5:Find the percentage of students that scored higher than 95.5

Number of student that scored higher than 95.5 = 20 + 24 = 44

Percentage of students that scored higher than 95.5 = [(Number of student that scored higher than 95.5)/(total number of students that took the test)] × 100

= (44/84) × 100 = 0.5238 × 100 = 52.38%

Percentage of students that scored higher than 95.5 = 52% (nearest percent)

Answer:

No, because the​ z-score of Z = 1.06 is not unusual since it does not lie within the range of a usual​ event, namely within 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Unusual

If X is more than two standard deviations from the mean, x is considered unusual.

In this question:

[tex]\mu = 511, \sigma = 119, n = 55, s = \frac{119}{\sqrt{55}} = 16.046[/tex]

A random sample of 55 students from this school was​ selected, and the mean math SAT score was 528. Is the high school justified in its​ claim?

If Z is equal or greater than 2, the claim is justified.

Lets find Z.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{528 - 511}{16.046}[/tex]

[tex]Z = 1.06[/tex]

1.06 < 2, so 528 is not unusually high.

The answer is:

No, because the​ z-score of Z = 1.06 is not unusual since it does not lie within the range of a usual​ event, namely within 2 standard deviations of the mean of the sample means.

The statement that could be made regarding the high school about the justification of its claim would be:

- No, because the​ z-score of Z = 1.06 is not unusual since it does not lie within the range of a usual​ event, namely within 2 standard deviations of the mean of the sample means.

Given that,

μ = 511

σ = 119

Sample(n) = 55

and

s = [tex]119/\sqrt{55}[/tex]

[tex]= 16.046[/tex]

As we know,

The claim of the high school could be valid and justified only when

[tex]Z > 2[/tex]

To find,

The value of Z

So,

[tex]Z = (X -[/tex] μ )/σ

by putting the values using Central Limit Theorem,

[tex]Z = (528 - 511)/16.046[/tex]

∵ [tex]Z = 1.06[/tex]

Since [tex]Z < 2[/tex], the claim is not justified.

Learn more about "Standard Deviation" here:

brainly.com/question/12402189

Answer:

The z-score when x=187 is 2.41. The mean is 187. This z-score tells you that x = 187 is 2.41 standard deviations above the mean.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

[tex]\mu = 152, \sigma = 14.5[/tex]

The z-score when x=187 is ...

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{187 - 152}{14.5}[/tex]

[tex]Z = 2.41[/tex]

The z-score when x=187 is 2.41. The mean is 187. This z-score tells you that x = 187 is 2.41 standard deviations above the mean.

Answer:an=20(-1/2)^n-1

Step-by-step explanation:

Answer:

50 = p + 42

Step-by-step explanation:

The unknown part of this equation is the variable p, the number of people that left. So you want to add p to 42 and that will give you the total number of football players, which is 50. In order to get p, you need to get it by itself and make it equal something. Subtract 42 from both sides and you are stuck with 50-42 = p

p = 8

Answer:

50-p=42

Step-by-step explanation:

Answer:

2.5

Step-by-step explanation:

(-5/6 ) / (-1/3)

multiply the numerator and denominator by the same number -3 gives:

(-5 * -3 /6 ) / (-1* -3/3)

(15/6 ) / (3/3)

(15/6 ) / 1

(15/6 )

12/6 + 3/6

2 3/6

2 1/2

2.5

Answer: decimal

Step-by-step explanation:

because i did this quiz

Answer: see below

Step-by-step explanation:

To find a coterminal angle, add or subtract 360° to the given angle as many times as needed to get a positive or negative angle.

I should mention that there are an infinite number of answers!

4) 400°

I can subtract 360° to get a positive angle of 40°

I can subtract another 360° to get a negative angle of -320°

5) -360°

I can subtract 360° to get a negative angle of -720°

I can add 360° twice to get a positive angle of 360°

6) -1010°

I can add 360° to get a negative angle of -650°

I can add 360° another 3 times to get a positive angle of 720°

7) 567°

I can subtract 360° to get a positive angle of 207°

I can subtract another 360° to get a negative angle of -153°

8) -164°

I can subtract 360° to get a negative angle of -524°

I can add 360° to get a positive angle of 194°

9) 358°

I can subtract 360° to get a negative angle of -2°

I can add 360° to get a positive angle of 718°

Answer:

The area of the triangle is 18.70 sq.units.

Step-by-step explanation:

It is provided that a triangle is bounded by the y-axis, the line [tex]f(x)=y=9-\frac{2}{3}x[/tex].

The slope of the line is: [tex]m_{1}=-\frac{2}{3}[/tex]

A perpendicular line passes through the origin to the line f (x).

The slope of this perpendicular line is:[tex]m_{2}=-\frac{1}{m_{1}}=\frac{3}{2}[/tex]

The equation of perpendicular line passing through origin is:

[tex]y=\frac{3}{2}x[/tex]

Compute the intersecting point between the lines as follows:

[tex]y=9-\frac{2}{3}x\\\\\frac{3}{2}x=9-\frac{2}{3}x\\\\\frac{3}{2}x+\frac{2}{3}x=9\\\\\frac{13}{6}x=9\\\\x=\frac{54}{13}[/tex]

The value of y is:

[tex]y=\frac{3}{2}x=\frac{3}{2}\times\frac{54}{13}=\frac{81}{13}[/tex]

The intersecting point is [tex](\frac{54}{13},\ \frac{81}{13})[/tex].

The y-intercept of the line f (x) is, 9, i.e. the point is (0, 9).

So, the triangle is  bounded by the points:

(0, 0), (0, 9) and [tex](\frac{54}{13},\ \frac{81}{13})[/tex]

Consider the diagram attached.

Compute the area of the triangle as follows:

[tex]\text{Area}=\frac{1}{2}\times 9\times \frac{54}{13}=18.69231\approx 18.70[/tex]

Thus, the area of the triangle is 18.70 sq.units.

Answer:

The answer would be 1,853,614,297

Answer:

The standard error = 5

Step-by-step explanation:

Explanation:-

Given sample size 'n' = 100

Given mean reading test score μ =  850

Given standard deviation of the population 'σ' = 50

The standard error is determined by

 Standard error =   [tex]\frac{S.D}{\sqrt{n} }[/tex]

                    S.E = σ/√n

[tex]S.E = \frac{S.D}{\sqrt{n} } = \frac{50}{\sqrt{100} } = 5[/tex]

Final answer:-

The standard error ( S.E) = 5

Answer:

Step-by-step explanation:

all numbers except y = 1

because (f*g)(x) = 1+1/x

and 1/x cannot be equal to 0