The equation r(t)=cos(5t)i + sin(5t)j, 0t≥0 describes the motion of a particle moving along the unit circle. Answer the following questions about the behavior of the particle. a. Does the particle have constant​ speed? If​ so, what is its constant​ speed? b. Is the​ particle's acceleration vector always orthogonal to its velocity​ vector? c. Does the particle move clockwise or counterclockwise around the​ circle? d. Does the particle begin at the point left parenthesis 1 comma 0 right parenthesis(1,0)​?

Answer:

5% or 0.05

Step-by-step explanation:

The null hypothesis will be rejected if p-value is less than significance level.

The null hypothesis can be rejected on 5% and 10% level of significance, but 5% level of significance is a suitable choice because when the 5% significance level is used the confidence level is 95% where as in case of 10% the confidence level is 90%. In short, the significance level indicates the probability of rejection of null hypothesis when the null hypothesis is true and the lesser probability of taking that risk will be better.

So, the scenario indicates the suitable significance level as 0.05.

Final answer:

With a P-value of 0.0278 in a two-tailed 2-sample z-test, the null hypothesis would be rejected at the 0.05 level of significance but not at the stricter 0.01 level.

Explanation:

When you find a P-value of 0.0278 in a two-tailed 2-sample z-test, the level of significance at which you would reject the null hypothesis depends on the predetermined alpha (α) level you have set for your test.

Since the P-value is less than the common significance levels of 0.05 and 0.10, you would reject the null hypothesis at these levels.

However, if your significance level was set at 0.01, which is stricter, you would not reject the null hypothesis because 0.0278 is greater than 0.01.

In summary, you would reject the null hypothesis at the 0.05 level of significance but not at the 0.01 level, given the P-value of 0.0278.

a) When using the binomial distribution for the number of securities that lost value, the following assumptions are made.

b) The probability that all 20 securities lose value is approximately \(0.0008\).

c) The probability that at least 15 of them lose value is the sum of the probabilities of having 15, 16, 17, 18, 19, or 20 securities losing value.

d) Probability that less than 5 of them gain value: Approximately 0.995872.

a) When using the binomial distribution for the number of securities that lost value, the following assumptions are made:

1. Each security in the portfolio has a fixed probability of losing value.

2. The outcomes of different securities are independent of each other.

3. There are only two possible outcomes for each security: it either loses value or it doesn't.

4. The probability of losing value remains constant for each security throughout the evaluation.

b) The probability that all 20 securities lose value can be calculated using the binomial probability formula:

[tex]\[ P(X = 20) = \binom{20}{20} \times 0.7^{20} \times (1 - 0.7)^0 \]\[ P(X = 20) = 1 \times 0.7^{20} \times 1 \]\[ P(X = 20) \approx 0.0008 \][/tex]

c) To find the probability that at least 15 of them lose value, we calculate the cumulative probability from 15 to 20:

[tex]\[ P(X \geq 15) = P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) \]\[ P(X \geq 15) \approx 0.9570 \][/tex]

d) Probability that Less Than 5 of Them Gain Value:

Replace k with the desired number and calculate the probabilities for X < k using the binomial probability formula.

For X < 5:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X < 5) ≈ 0.995872

Answer:

[tex] z<3.95[/tex]

Step-by-step explanation:

Assuming this complete question:

"Suppose a certain species of fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean [tex]\mu =26[/tex] kilograms and standard deviation [tex]\sigma=4.2[/tex] kilograms. Let x be the weight of a fawn in kilograms. Convert the following z interval to a x interval.

[tex] X<42.6[/tex]"

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

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(26,4.2)[/tex]  

Where [tex]\mu=26[/tex] and [tex]\sigma=4.2[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

We know that the Z scale and the normal distribution are equivalent since the Z scales is a linear transformation of the normal distribution.

We can convert the corresponding z score for x=42.6 like this:

[tex] z=\frac{42.6-26}{4.2}=3.95[/tex]

So then the corresponding z scale would be:

[tex] z<3.95[/tex]

Answer:

OPTION C:  Sin C - Cos C = s - r

Step-by-step explanation:

ABC is a right angled triangle. ∠A = 90°, from the figure.

Therefore, BC = hypotenuse, say h

Now, we find the length of AB and AC.

We know that:   [tex]$ \textbf{Sin A} = \frac{\textbf{opp}}{\textbf{hyp}} $[/tex]

and    [tex]$ \textbf{Cos A} = \frac{\textbf{adj}}{\textbf{hyp}} $[/tex]

Given, Sin B = r and Cos B = s

⇒    [tex]$ Sin B = r = \frac{opp}{hyp} = \frac{AC}{BC} = \frac{AC}{h} $[/tex]

⇒ [tex]$ \textbf{AC} = \textbf{rh} $[/tex]

Hence, the length of the side AC = rh

Now, to compute the length of AB, we use Cos B.

[tex]$ Cos B = s = \frac{adj}{hyp} = \frac{AB}{BC} = \frac{AB}{h} $[/tex]

⇒  [tex]$ \textbf{AB} = \textbf{sh} $[/tex]

Hence, the length of the side AB = sh

Now, we are asked to compute Sin C - Cos C.

[tex]$ Sin C = \frac{opp}{hyp} $[/tex]

⇒  [tex]$ Sin C = \frac{AB}{BC} $[/tex]

              [tex]$ = \frac{sh}{h} $[/tex]

               = s

Sin C = s

[tex]$ Cos C = \frac{adj}{hyp} $[/tex]

[tex]$ \implies Cos C = \frac{AC}{BC} $[/tex]

⇒ Cos C = [tex]$ \frac{rh}{h} $[/tex]

Therefore, Cos C = r

So, Sin C - Cos C = s - r, which is OPTION C and is the right answer.

Answer:

a) summation of p(x)/n ie. (0.1712+...+0.0051)/6=0.16675

b)1

c).var=summation (x-mean) squared /n ie (0.1712-0.16675)squared +...+(0.0051-0.16675)squared/n=0.027351948

SD =square root of variance =0.16538

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given a set of data sorted from smallest to largest, i.e. arranged in ascending order we are to find out the median, I and III quartiles

We know that the median is the middle entry of data arranged in ascending order

Q1 is the entry below which 25% lie and Q3 is one above which 25% lie

Hence proper definition would be

d. The first quartile is the median of the lower half of the data below the overall median.

The second quartile is the overall median

The third quartile is the median of the upper half of the data above the overall median.

Option b is wrong becuase mean is not necessary here.  Option a is wrong because this has nothing to do with std deviation

Option c is wrong since minimum value cannot be q1

Option e is wrong because we have exactly 25% lie below Q1

Answer:

[tex]m=\frac{332}{182}=1.824[/tex]

[tex]b=\bar y -m \bar x=43.615-(1.824*21)=5.311[/tex]

So the line would be given by:

[tex]y=1.824 x +5.311[/tex]

Step-by-step explanation:

We assume that the data is this one:

x: 15 16 17 18 19 20 21 22 23 24 25 26 27

y: 33 34 33 36 36 47 52 51 41 50 49 50 55

Find the least-squares line appropriate for this data.  

For this case we need to calculate the slope with the following formula:

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]

Where:

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]

So we can find the sums like this:

[tex]\sum_{i=1}^n x_i =273[/tex]

[tex]\sum_{i=1}^n y_i =567[/tex]

[tex]\sum_{i=1}^n x^2_i =5915[/tex]

[tex]\sum_{i=1}^n y^2_i =25547[/tex]

[tex]\sum_{i=1}^n x_i y_i =12239[/tex]

With these we can find the sums:

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=5915-\frac{273^2}{13}=182[/tex]

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=12239-\frac{273*567}{13}=332[/tex]

And the slope would be:

[tex]m=\frac{332}{182}=1.824[/tex]

Nowe we can find the means for x and y like this:

[tex]\bar x= \frac{\sum x_i}{n}=\frac{273}{13}=21[/tex]

[tex]\bar y= \frac{\sum y_i}{n}=\frac{567}{13}=43.615[/tex]

And we can find the intercept using this:

[tex]b=\bar y -m \bar x=43.615-(1.824*21)=5.311[/tex]

So the line would be given by:

[tex]y=1.824 x +5.311[/tex]

The equation of the least-squares regression line for predicting beak heat loss from temperature is: Percent heat loss from beak = 0.943(temp) + 16.243

The equation of the least-squares regression line for predicting beak heat loss, as a percent of total body heat loss from all sources, from temperature is:

Percent heat loss from beak = 0.943(temp) + 16.243

This equation can be obtained by performing a linear regression analysis on the given data points, where the temperature is the independent variable and the percent heat loss from the beak is the dependent variable.

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Answer:

i.  the forecast for this year using alpha = 0.5 is 11250

ii.   using alpha = 0.3 is 7350

Step-by-step explanation:

using exponential smoothing,the formula is given as: F(t+1) =αAt + (1-α)Ft, where F(t+1) is the new forecast or required forecast, α is the alpha, At is the each date or observation and Ft is the current trend.

i. using alpha= 0.5, the year forecast = (0.5 x 21000) + (1-0.5) x 1500 = 11250

ii. using alpha = 0.3,the year forecast = (0.3 x 21000) + (1-0.3) x 1500 = 7350.

this year forecast using alpha = 0.5 and 0.3 are 11250 and 7350 respectively

Answer:

(a) No the conclusion is not justified.

b. No

c. Two defective circuits in the sample

Step-by-step explanation:

Ans: (a) No the conclusion is not justified. What is important is the percentage population of defectives;

the sample proportion is only an approximation. The population proportion

for the new process may be more than or less than that of the old process.  We can  decide to pick two hundred samples and discover that the number of defects is greater than the previous process

(b)

.For the defectives, the population proportion for the new process may be 0.12 or more,

although the sample of defectives is just 11 out of 100

(c) Two defective circuits in the sample. This is because the probability of having two defects from the 100n samples is less than having 11 defects

a. The engineer's conclusion is not justified. A hypothesis test is needed to compare the proportions. b. Finding 11 defective circuits would not prove the new process is better. c. Finding 2 defective circuits represents stronger evidence that the new process is better.

a. The engineer's conclusion is not justified. To determine if the new process is better, we need to perform a hypothesis test. We can compare the proportion of defectives in the sample to the proportion of defectives in the known population for the old process using a hypothesis test for a proportion. If the p-value is small (less than the chosen significance level), we can reject the null hypothesis and conclude that the new process is better.

b. No, finding 11 defective circuits in the sample would not prove that the new process is better. We still need to perform a hypothesis test as mentioned in part a. If the p-value is small (less than the chosen significance level), we can reject the null hypothesis and conclude that the new process is better.

c. Finding 2 defective circuits in the sample represents stronger evidence that the new process is better. A smaller proportion of defectives in the sample suggests that the new process is more effective at reducing defects.

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Answer:

It hit the ground at t=2

It reaches its highest point at t=1

Its maximum height =16

Step-by-step explanation:

The graph of the function of the height will be a parapola opening downward where it's x intercepts = the time where the object hit the ground and its maximum point = the point with the maximum point

The x intercepts are 0 when the object was thrown and 2 when it landed

The x coordinate of the vertex is in the middle of the two x intercepts so it will be 1

Substituting the x coordinate of the vertex gives us the y coordinate of the vertex which is the maximum hight 16

The pomegranate hits the ground at time t=2 seconds (tg=2) and reaches its maximum height at time t=1 second (th=1). The maximum height (h) is 16 feet.

The given function f(t) = -16t2 + 32t represents the height of the pomegranate at any time t. The pomegranate hits the ground when its height is zero, so set f(t) = 0 and solve for t.

0 = -16t2 + 32t
0 = t(-16t + 32)
Thus, t = 0 (the initial time) and t = 2 (when it hits the ground). Therefore, tg = 2.

The pomegranate reaches its maximum height when the derivative of f(t) is zero. This will give us the time th.

f'(t) = -32t + 32
Set f'(t) = 0 and solve for t. Then -32t + 32 = 0 implies t = 1. So, th = 1.

The maximum height h is simply f(th) = -16(1)2 + 32(1) = 16 feet.

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Answer:

The answers are shown in the step by step explanation that is attached

Step-by-step explanation:

The step by step calculation is as shown in the attachment below

Answer:

0.082 is the probability that a fish caught will be longer than 18 inch.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 12.3 inch

Standard Deviation, σ = 4.1 inch

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

Formula:

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

a) P( longer than 18 inch)

P(x > 18)

[tex]P( x > 18) = P( z > \displaystyle\frac{18 - 12.3}{4.1}) = P(z > 1.39)[/tex]

[tex]= 1 - P(z \leq 1.39)[/tex]

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

[tex]P(x > 18) = 1-0.918 =0.082[/tex]

0.082 is the probability that a fish caught will be longer than 18 inch.

Answer: the probability that a fish caught there will be longer than 18in is 0.08226

Step-by-step explanation:

Assuming a normal distribution for length of the fishes caught in Lake Amotan, the formula for normal distribution is expressed as

z = (x - µ)/σ

Where

x = length of fishes

µ = mean length

σ = standard deviation

From the information given,

µ = 12.3 in

σ = 4.1 in

We want to find the probability that a fish caught there will be longer than 18in. It is expressed as

P(x > 18) = 1 - P(x ≤ 18)

For x = 18,

z = (18 - 12.3)/4.1 = 1.39

Looking at the normal distribution table, the probability corresponding to the z score is 0.91774

P(x > 18) = 1 - 0.91774 = 0.08226

Answer:

The value of x = 5.

The length of KJ = 29 units.

Step-by-step explanation:

Given L and M are the mid points of the lines.

So, LM becomes the mid segment.

Also, [tex]$ \textbf{LM} = \frac{\textbf{GH + KJ}}{\textbf{2}} $[/tex]

Here, the length of LM = 25 units.

Length of GH = 2x + 11 units.

Length of KJ  = 6x - 1 units.

Therefore, we have: [tex]$ LM = \frac{2x + 11 + 6x - 1}{2} $[/tex]

= [tex]$ \frac{8x + 10}{2} $[/tex]

[tex]$ \implies 25 = 4x + 5 $[/tex]

[tex]$ \implies 4x = 20 $[/tex]

⇒ x = 5

Therefore, KJ = 6(5) - 1

= 29 units.

Hence, the answer.

Answer:

a. The individual's blood pressure is unusually high, compared to others

Step-by-step explanation:

The individual's blood pressure is unusually high, compared to others. Since the z- score is positive, it implies that the individual's blood pressure is higher than the average blood pressure of others. And also given that the z-score is as high as 2.1 (higher than 95% confidence interval which is 1.96) implies that the individual's blood pressure is extremely higher than the average blood pressure of others.

Answer:

[tex] \bar X = \frac{1435040}{80}=17938[/tex]

Step-by-step explanation:

Since we have a groued data for this case we can construct the following table to find the expected value.

 Interval               Frequency(fi)      Midpoint(xi)           xi*fi

5001-10000              16                       7500.5            120008

10001-15000             14                      12500.5           175007

15001-20000            15                      17500.5           262507.5

20001-25000           17                      22500.5          382508.5

25001-30000           18                      27500.5          495009

Total                          80                                              1435040

And we can calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^n f_i x_i}{n}[/tex]

Where [tex] n=\sum_{i=1}^n f_i = 80[/tex]

And if we replace we got:

[tex] \bar X = \frac{1435040}{80}=17938[/tex]

Answer:

Zero

Step-by-step explanation:

The possible outcomes from rolling a die are 1,2,3,4,5 and 6. The probability of getting one of these numbers when the die is rolled

=1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6

=1

The probability of not getting one of the number plus the probability of getting one of them = 1

Therefore;

probability of not getting one of the number + 1 = 1

probability of not getting one of the number = 0

3.37 lb

Step-by-step explanation:

The question requires you to convert weight in Newtons to weight in pounds.

Given 15.0 N to convert to pounds, remember the conversion rate where;

1 Newton = 0.224809 pound-force

1 N= 0.224809 lb

15 N= ?

Perform cross-product

=15*0.224809

=3.37 lb

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Since the population is first divided into various groups and the sampling was done later.  

It is not the example of a simple random sampling as it does not allow for selecting more than 1 adult from each group that would have been possible in all simple random sampling.

Learn more about the sampling units described in part.

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Answer:

A. The annual growth rate in Egypt in 1990 was 31 - 9 = 22 per 1,000 people. The estimated annual growth rate in 2010 is 25 - 5 = 20 per 1,000 people.

B. I feel like Egypt is in the third stage (Late Developing) in the Demographic transition model because the birth and death rate was higher but then as time progressed, they both started to level out. In order for Egypt to advance in the model, they will have to have a developed country (urbanization) and lower their birth/death rates through the use of birth controls and healthcare.

Step-by-step explanation:

Answer:

In 80% of the tests the number of points scored was less than 49 and in 20% of the tests the number of points scored was higher than 49.

Step-by-step explanation:

When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.

On a 60 point written assignment, the 80th percentile for the number of points earned was 49. Interpret the 80th percentile in the context of this situation.

The interpretation is that in 80% of the tests the number of points scored was less than 49 and in 20% of the tests the number of points scored was higher than 49.

Final answer:

The 80th percentile indicates a student scored higher than 80 percent of peers, with a score of 49 out of 60 on the assignment.

Explanation:

The 80th percentile for a 60-point written assignment, where the score was 49, means that 80 percent of the students earned 49 points or less on the assignment. Conversely, it also means that 20 percent of the students earned more than 49 points. Therefore, a student who scored 49 points on this assignment performed better than 80 percent of their peers.

The value of intercept will be 34.

The equation of line in point-slope form passing through the points

(x₁ , y₁) and (x₂, y₂) with slope m is defined as;

⇒ y - y₁ = m (x - x₁)

Where, m = (y₂ - y₁) / (x₂ - x₁)

Given that;

In a trend line based on five observations,

The average of Y = 100

And, The slope of line = 22

Now,

Since, The equation of line is,

⇒ Y = mx + c

Where, m is slope and c is y - intercept.

When x = 1;

⇒ Y = m + c

When x = 2;

⇒ Y = 2m + c

When x = 3;

⇒ Y = 3m + c

When x = 4;

⇒ Y = 4m + c

When x = 5;

⇒ Y = 5m + c

Here, The average of Y is 100.

So, We get;

⇒ (m + c) + (2m + c) + (3m + c) + (4m + c) + (5m + c) / 5 = 100

⇒ 15m + 5c / 5 = 100

⇒ 3m + c = 100

Substitute m = 22;

⇒ 3 × 22 + c = 100

⇒ 66 + c = 100

⇒ c = 100 - 66

⇒ c = 34

Thus, The value of intercept is,

⇒ c = 34

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Answer:

Step-by-step explanation:

Binomial distribution is to be used here due to following reasons.

(a)

Probability of sending a box to an experienced inspector

= Probability of getting non-zero unacceptable cans

= 1 - Probability of getting zero unacceptable cans

=1- P(X = 0) = 1 - 10.08^0 *(1 – 0.08)^(4-0) = 0.283607

(b)

Expected cost per inspector in an hour in case of 8% unacceptable cans

= {(1-0.283607)*16+0.283607*22} = $ 17.70164

If groceries reduce their unacceptable rate to 4% then X - Bin(4, 0.04) .

In this scenario,

Probability of sending a box to an experienced inspector

= Probability of getting non-zero unacceptable cans

= 1 - Probability of getting zero unacceptable cans

=1- P(X = 0) = 1 - 0.04^0 * (1 – 0.04)*(4-0) = 0.150653

Expected cost per inspector in an hour in case of 4% unacceptable cans

= {(1-0.150653)*16+0.150653*22} = $ 16.90392

Percentage of savings realized = (17.70164-16.90392)/17.70164*100% = 4.506475%

Answer:

a) [tex] \mu -3\sigma = 336-3*6=318[/tex]

[tex] \mu+-3\sigma = 336+3*6=354[/tex]

b) For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%

And then the % above would be 100-84= 16%

Step-by-step explanation:

Assuming this question : "Bigger animals tend to carry their young longer before birth. The  length of horse pregnancies from conception to birth varies according to a roughly normal distribution with  mean 336 days and standard deviation 6 days. Use the 68-95-99.7 rule to answer the following questions. "

(a) Almost all (99.7%) horse pregnancies fall in what range of lengths?

First we need to remember the concept of empirical rule.

From this case we assume that [tex] X\sim N(\mu = 336. \sigma =6)[/tex] where X represent the random variable "length of horse pregnancies from conception to birth"

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

From the empirical rule we know that we have 99.7% of the data within 3 deviations from the mean so then we can find the limits for this case with this:

[tex] \mu -3\sigma = 336-3*6=318[/tex]

[tex] \mu+-3\sigma = 336+3*6=354[/tex]

(b) What percent of horse pregnancies are longer than  342 days?

For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%

And then the % above would be 100-84= 16%

The problem is applying the concept of normal distribution in statistics to describe the length of horse pregnancies, which are said to follow a normal distribution with a mean of 336 days and a standard deviation of 6 days.

The student is being asked to deal with a problem that relates to the normal distribution concept in statistics applied to horse pregnancies. If X represents the length of horse pregnancies, it's stated that it follows a normal distribution with a mean (average) of 336 days and a standard deviation of 6 days.

The normal distribution, also known as the Gaussian or bell curve, is a function that describes the probability distribution of many kinds of data, in this case, the horse gestation period. The distribution is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

In practical terms, it means that most horses will have a gestation period near the 336 days (mean value), with few horses having gestation periods significantly shorter or longer. The standard deviation (in this case, 6 days) gives an indication of how much the gestation period is expected to vary from the mean.

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Answer:

The equation of line is [tex]P(t)=26.9t+1241[/tex]

Step-by-step explanation:

Consider the provided information.

World grain production was 1241 million tons in 1975 and 2048 million tons in 2005,

We need to find the linear function.

The difference of the year is: 2005-1975=30

The function represents the world grain production at time t years after 1975, Thus, the first points is (0,1241) and the second points is (30,2048)

Find the slope of the line by using the formula: [tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Substitute the respective values in the above formula.

[tex]m=\dfrac{2048-1241}{30-0}\\\\m=\dfrac{807}{30}\\\\m=26.9[/tex]

The slope of the linear function is 26.9

The slope intercept form is: [tex]y=mx+b[/tex]

Where b is the y intercept. y intercept is a point where the value of x is 0.

Therefore y intercept of the linear function is 1241 because the first points is (0,1241).

Hence, the equation of line is [tex]P(t)=26.9t+1241[/tex]

Given the data, the approximated rate of increase in world grain production from 1975 to 2005 is about 32.28 million tons per year. Increases might be driven by population growth and rising food demands. Future food supply forecasts point to significant increases by 2050.

The subject of your question appears to be related to linear growth over time, specifically in the context of world grain production. Given the data provided, we can calculate the approximate rate of increase in grain production by taking the difference in amounts (2048 million tons in 2005 minus 1241 million tons in 1975) and divide it by the difference in years (2005 minus 1975). This calculation results in approximately 32.28 million tons increase per year.

It's also noteworthy to consider that grain production increases might be driven by the increasing global population and rising food demands. Additionally, future projections for food supply such as milk and meat production, estimated to face significant increases by 2050, might also apply to grain production, further emphasizing the importance of understanding and managing these growth rates for global food security.

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Answer: The margin of error E = 0.0104

Step-by-step explanation:

The formula to find the margin of error that corresponds to the given statistics and confidence level for population proportion is given by :-

[tex]E=z*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] , where

n= Sample size

[tex]\hat{p}[/tex] = Sample proportion

z* = critical value.

As per given , we have

n= 3200

[tex]\hat{p}=0.15[/tex]

Confidence level : 90%

The critical z-value for 90% confidence is z* = =1.645[By z-table]

Substitute all values in the formula , we get

[tex]E=(1.645)\sqrt{\dfrac{0.15(1-0.15)}{3200}}[/tex]

[tex]E=(1.645)\sqrt{0.00003984375}[/tex]

[tex]E=(1.645)(0.00631219058648)=0.0103835535148\approx0.0104[/tex]

Hence, the margin of error E = 0.0104

Answer: The intersection of the line through (0, 1) and (4.3, 2) and the line through (2.1, 3) and (5.3, 0) is (3.392, 1.789).

Step-by-step explanation:

We know that the equation of a line that passes through two points (a,b) and (c,d) is given by :-

[tex](y-b)=\dfrac{d-b}{c-a}(x-a)[/tex]

Similarly , the equation of line that passes through (0, 1) and (4.3, 2)  would  be:

[tex](y-1)=\dfrac{2-1}{4.3-0}(x-0)[/tex]

[tex](y-1)=\dfrac{1}{4.3-}(x)[/tex]

[tex]4.3(y-1)=x[/tex]

[tex]4.3y-4.3=x-----(1)[/tex]

Equation of line that passes through (2.1, 3) and (5.3, 0) would  be:

[tex](y-0)=\dfrac{3-0}{2.1-5.3}(x-5.3)[/tex]

[tex]y=\dfrac{3(x-5.3)}{-3.2}-----(2)[/tex]

To find the intersection point (x,y) , we substitute the value of y from (2)in (1) , we get

[tex]4.3(\dfrac{3(x-5.3)}{-3.2})-4.3=x[/tex]

[tex]-4.03125(x-5.3)-4.3=x[/tex]

[tex]-4.03125x+21.365625-4.3=x[/tex]

[tex]-4.03125x+17.065625=x[/tex]

[tex]x+4.03125x=17.065625[/tex]

[tex]5.03125x=17.065625[/tex]

[tex]x=\dfrac{17.065625}{5.03125}\approx3.392[/tex]

Put value of x in (2) , we get

[tex]y=\dfrac{3(3.392-5.3)}{-3.2}[/tex]

[tex]y=\dfrac{3(-1.908)}{-3.2}\approx1.789[/tex]

Hence, the intersection of the line through (0, 1) and (4.3, 2) and the line through (2.1, 3) and (5.3, 0) is (3.392, 1.789).

This solution involves finding the equations of the two lines using the slope and y-intercept, setting the equations equal to each other to find the x-coordinate of the intersection, and substituting the x-value into one of the equations to find the corresponding y-coordinate of the intersection.

To find the intersection of the two lines, first we need to find the equations of these lines. We can use the formula y = mx + c, where m is the slope and c is the y-intercept.

For the line passing through (0, 1) and (4.3, 2), we find the slope (m) first: m = (2-1) / (4.3-0) = 1/4.3. The line passes through the y-axis at (0,1), so c = 1. Thus, the equation is y = (1/4.3)x + 1.

For the line passing through (2.1, 3) and (5.3, 0), the slope m = (0-3) / (5.3-2.1) = -3/3.2. This line does not pass through the y-axis, so c is not 0. Substituting one of the points into the equation y = mx + c, we can find c: 3 = -3/3.2*2.1 + c, yields c = 3.984375. So, the equation is y = -3/3.2*x + 3.984375.

Now, we set these two equations equal to each other and solve for x: (1/4.3)x + 1 = -3/3.2*x + 3.984375. This will yield the x-coordinate of the intersection. Substitute the x-coordinate into any of the line equations to get the y-coordinate. These coordinates give us the intersection point of the two lines.

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Step-by-step explanation:

All fractions have the same denominator = 5.

The larger the counter, the greater the fraction.

Therefore

[tex]\dfrac{1}{5}<\dfrac{2}{5}<\dfrac{3}{5}<\dfrac{4}{5}[/tex]

Answer:

a) it is neither even nor odd

b) it is an even signal

c) it is an odd signal

Step-by-step explanation:

A function f(x) or a signal is said to be even if its satisfies the condition of f(-x) = f(x). this implies that the graph of such a function or signal has a symmetrical relationship with respect to the y-axis.

A function f(x) or a signal is said to be odd if its satisfies the condition of f(-x) = - f(x). this implies that the graph of such a function or a signal has a skew-symmetrical relationship with respect to the y-axis.

from the first option ; a) x[n] = u[n] - u [n-6], from the conditions attached to even and odd functions, it can be inferred that the first option does not satisfy the conditions for even and odd functions hence, it is neither even nor odd.

The attachements below gives a detailed explanation of the second and third option.

Answer:

There are no values in the data that is two standard deviations below the mean.

Step-by-step explanation:

We are given the following data set in question:

165, 171, 174, 180, 182, 188, 189, 192, 198, 202, 202, 225, 228, 235, 240

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{2971}{15} = 198.06[/tex]

The mean of sample is 198.06 pounds.

Sum of squares of differences = 7984.93

[tex]S.D = \sqrt{\dfrac{7984.93}{14}} = 23.88[/tex]

The sample standard deviation is 23.88 pounds.

We have to find the weight that is 2 standard deviations below the mean.

[tex]x < \bar{x}- 2s\\x < 198.06 -2(23.88)\\x < 150.3[/tex]

Thus, we have to find a value less than 150.3.

Sorted data: 165, 171, 174, 180, 182, 188, 189, 192, 198, 202, 202, 225, 228, 235, 240

There are no values in the data that is less than 150.3

Answer:

0.0769

Step-by-step explanation:

Let x be the number of student that offer both subjects

15 - x + x + 13 - x + 9 = 26

-x + 37 = 26

-x = -11

x = 11

Number of student that offer english but not math = 13 - 11

                                                                                        = 2

The probability of english but not math = 2/26

The probability that a student is chosen at random likes English but not maths is [tex]\dfrac{1}{13}[/tex].

Given information:

In a class of 26 students, 15 of them like maths, 13 of them like English and 9 of them like neither.

Now, the number of students who like either maths or English will be,

[tex]26-9=17[/tex]

Now, out of 17, 15 students like maths and 13 students like English.

So, the number of students who like both the subjects will be,

[tex]E\cap M=13+15-17\\=11[/tex]

Now, 11 students like both the subjects and 13 students like English.

So, the number of students who like English but not maths will be,

[tex]13-11=2[/tex]

Thus, the probability that a student chosen at random likes English but not maths will be calculated as,

[tex]P=\dfrac{2}{26}\\=\dfrac{1}{13}[/tex]

Therefore, the probability that a student is chosen at random likes English but not maths is [tex]\dfrac{1}{13}[/tex].

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