A shed is 4.0 m long and 2.0m wide. A concrete path of constant width is laid all the
way around the shed. If the area of the path is 9.50m? Calculate its width.

The P-value is 0.0107 for the sample statistics n-35 and the coefficient of standard deviation is 82.

α = 0. 05

μ = 1620

size (n)= 35

X_bar=1590

σ=82

From the given sample statistics, the test statistics will be calculated as:

t = (X_bar - μ) / (σ / sqrt(n))

t = (1590 - 1620) / (82 / sqrt(35))

t = (-2.5411)

Using the t-distribution table with 34 degrees of freedom, the critical value will be:

t_critical = -1.6909

Here the calculated test statistic is less than the critical value.

P - value = 2*P(-100< t < -1.9720, when df = 34)

P = tcf (-100,-2.4103,34)

P = 0.0107

Therefore we can conclude that the P-value is 0.0107.

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Jonty is correct as the original cost of paint is £196 which is less than £200 as said by him.

The area of part of cuboid to be painted will be the sum of all the unpainted areas. So, the area remaining to be painted will be = (2 × 3 × 2.5) + (2 × 3 × 12) + (12 × 2.5)

Remaining area = 15 + 72 + 30

Remaining area = 117 m²

Let us assume the original cost of paint be x. So,

x + 10%x = 26.95

110x = 26.95 × 100

110x = 2695

x = £24.5

Now, number of required tins = total unpainted area/area covered by one tin

Number of required tins = 117/15

Number of required tins = 7.8 tins

Taking it as 8 tins.

Previous cost of tins = 8 × 24.5

Previous cost = £196

Since the original cost is less than £200, Jonty is stating truth.

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The complete question is attached in figure.

The measure of <BAE is 130 degree.

We have,

m <C = 65

As, opposite sides are parallel then ABCD is a parallelogram.

Then, opposite angles of a parallelogram are congruent

∠ BAD = ∠ BCD = 65°

∠ ADE = ∠ BAD = 65° ( alternate angles )

Since, DE = AE then Δ ADE is isosceles Triangle.

So, ∠ DAE = ∠ ADE = 65°

We can write,

∠ BAE = ∠ BAD + ∠ DAE = 65° + 65° = 130°

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The probability of exactly one success in the binomial experiment would be 20. 4 %.

The probability that there is one success in a binomial probability which has a chance of success of 5 % can be found by the formula :

P ( X = 1) = (5 choose 1) x ( 0.05 ) x  (0.95 ) ⁴

= ( 0.05 ) x  ( 0. 95 ) ⁴

= 0.05 x 0.8145

= 0.040725

Multiplying both gives:

P(X = 1) = 5 x 0.040725

= 0.203625

In conclusion, the probability of one success is 0.203625 or 20. 4 %.

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To approximate the area of the region defined by[tex]y = √(3x)[/tex] using upper and lower sums, we first divide the interval [0,1] into n subintervals of equal width [tex]Δx = 1/n[/tex]. We then compute the upper and lower sums using the formulae above, and take their average to obtain the approximate area.

To approximate the area of the region defined by the function [tex]y = √(3x)[/tex]using upper and lower sums, we first need to divide the interval of integration [0,1] into subintervals of equal width. Let n be the number of subintervals, then the width of each subinterval is[tex]Δx = 1/n[/tex].

The upper sum is the sum of the areas of rectangles whose heights are taken from the upper endpoints of each subinterval. Specifically, for each i from 1 to n, we compute the height of the rectangle as f(xi), where xi is the upper endpoint of the i-th subinterval.

Upper sum =[tex]Δx [f(x1) + f(x2) + ... + f(xn)], where x1 = 0, x2 = Δx, x3 = 2Δx, ..., xn = (n-1)Δx.[/tex]Similarly, the lower sum is the sum of the areas of rectangles whose heights are taken from the lower endpoints of each subinterval.

Lower sum = [tex]Δx [f(x0) + f(x1) + ... + f(xn-1)][/tex], where[tex]x0 = 0, x1 = Δx, x2 = 2Δx, ..., xn-1 = (n-1)Δx.[/tex] To find the approximate area of the region using upper and lower sums, we simply compute the upper and lower sums using the given number of subintervals, and take their average: Approximate area = (Upper sum + Lower sum)/2.

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If cruz purchased a large pizza for $12.75. It serves 5 people, the cost per serving of the pizza is $2.55. So, correct option is A.

To find the cost per serving of the pizza, we need to divide the total cost of the pizza by the number of servings. In this case, the pizza costs $12.75 and serves 5 people.

Therefore, the cost per serving can be calculated as:

Cost per serving = Total cost of pizza / Number of servings

Cost per serving = $12.75 / 5

Cost per serving = $2.55

So, the cost per serving of the pizza is $2.55.

When working with fractions or dividing quantities, we need to pay attention to the units involved. In this case, the units of the cost and the servings must match for the division to be meaningful.

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The  area of the lateral face of cylinder = 75.4 in²

The  area of the two bases of the cylinder = 25.13 in²

The total surface area of the cylinder =  100.53 in²

We know that the formula for the surface area of cylinder is:

A = 2πrh + 2πr²

where r is the radius of the cylinder

and h is the height of the cylinder

Here, r = 2 in and h = 6 in

The area of the lateral face of cylinder is given by,

A₁ = 2 × π × r × h

A₁ = 2 × π × 2 × 6

A₁ = 24 × π

A₁ = 75.4 sq. in.

And the area of two base is,

A₂ = 2πr²

A₂ = 2 × π × 2²

A₂ = 8 × π

A₂ = 25.13 sq. in.

The total surface area of cylinder would be,

A = A₁ + A₂

A = 75.4 + 25.13

A = 100.53 sq. in.

Therefore, the required area = 100.53 in²

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Answer: Use the table to identify values of p and q that can be used to factor x2 + x – 12 as (x + p)(x + q).A. –2 and 6. B. 3 and –4. C. –3 and 4. D. 2 and –6.

Step-by-step explanation:

The cost of Legos, given that this is based on vowels and consonants would be $ 19.

The vowels and consonants can be arranged such that:

Doll - 1 vowel (o), 3 consonants ( d , l , l ) - $ 17

Kite - 2 vowels ( i, e ) , 2 consonants ( k, t) - $ 14

Skates - 2 vowels ( a, e ), 4 consonants ( s, k , t , s) - $24

Using this, we can solve for vowels and consonants such that cost per vowel is $2, and the cost per consonant is $5.

The cost of Legos is based on 2 vowels (e, o) and 3 consonants (L, g, s)

= ( 2 x 2 ) + ( 5 x 3 )

= $ 19

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Answer:

19

Step-by-step explanation:

Let's call the number we're trying to find "x".

According to the problem:

4x + 25 = 6x - 13

To solve for x, we can start by isolating the x term on one side of the equation. Let's subtract 4x from both sides:

4x + 25 - 4x = 6x - 13 - 4x

25 = 2x - 13

Next, let's add 13 to both sides:

25 + 13 = 2x - 13 + 13

38 = 2x

Finally, we can divide both sides by 2 to solve for x:

38/2 = 2x/2

19 = x

Let 0-1 represent students who participate in sports and let 2-9 represent students who do not participate in sports.

Let 0-9 represent students selected for the sample, with digits 0-8 representing students who participate in sports and digit 9 representing a student who does not participate in sports.

So, an appropriate assignment of digits for this simulation would be: Let 0-1 represent students who participate in sports and let 2-9 represent students who do not participate in sports.

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The school which is a better choice is sea side.

We are given that;

The plot

Now,

If you are interested in a smaller class size, Seaside School is a better choice for you because it has a smaller mean and median class size than Bay Side School. This means that on average and in general, Seaside School has fewer students per class than Bay Side School. Also, Seaside School has a smaller maximum class size than Bay Side School (both have a minimum of zero), so you are less likely to encounter a very large class at Seaside School.

Therefore, by algebra the answer will be sea side.

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In this case, the scientist is studying if students' beliefs about illegal drug use change as they go through college. The null hypothesis (H0) for McNemar's test in this context would state that there is no significant change in students' beliefs about illegal drug use between the time they enter college and four years later. In other words, the proportion of students who change their beliefs about illegal drug use is not significantly different from the proportion who do not change their beliefs.

Null hypothesis (H0): There is no significant difference in students' beliefs about illegal drug use before and after going through college.

The minimum sample size required to estimate an unknown population mean is 148. So, the correct option is D) 148.

To find the minimum sample size required to estimate an unknown population mean with 90% confidence, within 3.4 lb of the population mean, and a population standard deviation of 25 lb, follow these steps:

1. Identify the given values:
  - Confidence level = 90%
  - Margin of error (E) = 3.4 lb
  - Population standard deviation (σ) = 25 lb

2. Find the corresponding z-score for the 90% confidence level. Using a standard normal distribution table or calculator, the z-score is 1.645.

3. Use the formula to find the sample size (n):

  n = (z * σ / E)^2
  n = (1.645 * 25 / 3.4)^2

4. Calculate the sample size:

  n ≈ 147.267

Since we cannot have a fraction of a person, round up to the nearest whole number to ensure the required confidence level.

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The probability that less than 61% of sampled teenagers own smartphones is 0.934 or 93.4%.

To find the probability that less than 61% of sampled teenagers own smartphones, we need to use the standard normal distribution. We first need to standardize the value of 61% using the formula:

z = (x - μ) / σ

where x is the value we want to standardize (in this case, 61%), μ is the mean (0.55), and σ is the standard deviation (0.0397).

Plugging in the values, we get:

z = (0.61 - 0.55) / 0.0397 = 1.511

We can then look up the probability of getting a z-score less than 1.511 in a standard normal distribution table or calculator. The probability is approximately 0.934.

Therefore, the probability that less than 61% of sampled teenagers own smartphones is 0.934 or 93.4%.

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Complete question:

e) Find the probability that less than 61% of sampled teenagers own smartphones.

(a) Find the mean :The mean μ p is 0.55

(b) Find the standard deviation

The standard deviation σp is 0.0397

The lists that contain only rational numbers is 4/3, -12/13 ,9/4 , -5/7 , 3/4

A rational number can be described as the number which can be expressed in the form of p/q where p and q are integers  when writing this number, q  must not equal to 0 .

Examples of rational numbers are , however in mathematics, a rational number i can be seen as one that can be expressed as the quotient or fraction  which can involves  two integers,  wherby one will be the  numerator p and a non-zero  as well as the denominator q.

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The radius of the circle is 4 units

A circle is simply a round shape that has no corners or line segments. The body of a circle is called the circumference and a cut out of circumference is called an arc.

The distance from the centre of a circle to any part of its circumference is called a radius. Twice of a radius is called the diameter.

In the circle, the distance between the center of the circle and it's circumference is ;

4-0 = 4 units

Therefore the radius of the circle is 4 units.

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Therefore, if the person works 54 hours, they would make $1,163.04. Rounded to 2 decimal places, the answer is $1,163.00.

The decimal system employs ten decimal digits, a decimal mark, and a minus sign ("-") for negative quantities when writing numbers. The decimal digits are 0 through 9, with the dot (".") serving as the decimal separator in many (mainly English-speaking) nations and the comma (",") in others.

The fractional portion of the number is represented by the place value that follows the decimal. The number 0.56, for instance, is composed of 5 tenths and 6 hundredths.

We can use proportionality to solve this problem. If the person works 25 hours and makes $539, then their hourly rate is:

$539 ÷ 25 hours = $21.56 per hour

They would make if they work 54 hours, we can multiply their hourly rate by the number of hours worked:

$21.56 per hour × 54 hours = $1,163.04

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The requried relation is, the higher the grade level in school, the more hours spent on homework. Option A is correct.

From the given data, we can see that the hours spent on homework per day increase as the grade level in school increases. This suggests that there is a positive correlation between grade level and homework hours.

In general, higher grade levels in school involve more advanced coursework and greater academic expectations, which can require more time and effort outside of class to complete homework and study. Therefore, it is reasonable to expect that students in higher grade levels will spend more hours on homework per day compared to students in lower grade levels.

Thus, the requried relation is, the higher the grade level in school, the more hours spent on homework. Option A is correct.

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As per the unitary method, it would cost $33 for Kelly and her two friends to bowl one game before 4 pm and one game after 4 pm each at Super Strike Bowling.

Bowling is a popular recreational activity enjoyed by many people around the world. Super Strike Bowling charges different rates for games played at different times of the day. To model the cost for one game, we can use a step function, where the value of x represents the number of hours after 12 pm. This function is defined as follows:

Cost per game (C) =

$5 if 1 pm ≤ x < 4 pm

$6 if 4 pm ≤ x < 8 pm

$8 if 8 pm ≤ x ≤ 12 am

Now, let's apply this step function to the scenario of Kelly and her two friends bowling. They each played one game before 4 pm, which cost $5 per game, and one game after 4 pm, which cost $6 per game. Therefore, the total cost for one person to play two games is:

Total cost = ($5 per game) + ($6 per game)

= $11

Since there were three people bowling, we can multiply the total cost by 3 to get the cost for all three to bowl:

Cost for all three to bowl = 3 × $11

= $33

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While it is not very likely that there will be exactly 14 births on any given day, it is still possible, and the probability of it happening is about 8.3%.

Let's start by calculating the mean or average number of births per day. To do this, we divide the total number of births in a year (4126) by the number of days in a year. Since there are 365 days in a year, the mean number of births per day is:

4126 / 365 = 11.3

This means that on average, there are about 11 to 12 births per day in this hospital.

In this case, the average rate of occurrence is 11.3 births per day. Using the Poisson distribution formula, we can calculate the probability of having 14 births in a day as follows:

P(X=14) = (e⁻¹¹°³) x (11.3¹⁴) / 14!

where e is the mathematical constant approximately equal to 2.71828, X is the random variable representing the number of births in a day, and ! represents the factorial function.

Using a calculator or a software tool, we get:

P(X=14) = 0.083

This means that the probability of having exactly 14 births in a day is about 8.3%.

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Answer:Step 1:

Enter the expression you want to simplify into the editor.

The simplification calculator allows you to take a simple or complex expression and simplify and reduce the expression to it's simplest form. The calculator works for both numbers and expressions containing variables.

Step-by-step explanation: I really hope this helps

There was a significant reduction in peoples over estimation of the line length, t = XXX with df = XXX, p < 0.01 two-tailed, with M = 10%, n = 25, s = 15%, Cohen's d = XXX , M = 10%, 95% CI [XXX, XXX].

8a. A. was
8b. B. t = XXX with df = XXX
8c. A. < 0.01 two-tailed
8d. C. M = 10%, n = 25, s = 15%, Cohen's d = XXX , M = 10%, 95% CI [XXX, XXX].

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The required solution for the given right angle triangle is given below.

Blank #1: We can use the Pythagorean theorem to find AB:

[tex]AB = \sqrt{AC^2 + CB^2} \\= \sqrt{12^2 + 9^2}\\ =15[/tex]

Therefore, AB = 15.

Blank #2: We can use the inverse tangent function to find the angle A:

tan(A) = opposite / adjacent = CB / AC = 9 / 12

[tex]A = tan^{-1}(9/12) = 36.86^o[/tex]

Therefore, angle A ≈ 36 degrees.

Blank #3: We can use the inverse cosine function to find the angle C:

cos(C) = adjacent / hypotenuse = CB / AB = 9 / 15

C = arccos(9/15) ≈ 53.14

Therefore, angle C ≈ 53.14 degrees.

Blank #4: We can use the fact that the sum of angles in a triangle is 180 degrees to find angle B:

B = 180 - A - C ≈

B= 90

Therefore, angle B ≈ 90

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We can expect a difference of about 6.67 between the two sample means.

To calculate how much difference to expect between two sample means when the population means are equal, we need to compute the standard error of the difference between means (SED).

The formula for SED in the independent-measures t-test is:

SED = sqrt((s1^2/n1) + (s2^2/n2))

where s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.

a) For the first situation, we have:

s1^2 = SS1/(n1-1) = 500/(6-1) = 100

s2^2 = SS2/(n2-1) = 524/(12-1) = 49.45

Plugging these values into the formula, we get:

SED = sqrt((100/6) + (49.45/12)) = 5.76

Therefore, we can expect a difference of about 5.76 between the two sample means.

b) For the second situation, we have:

s1^2 = SS1/(n1-1) = 600/(6-1) = 120

s2^2 = SS2/(n2-1) = 696/(12-1) = 69.6

Plugging these values into the formula, we get:

SED = sqrt((120/6) + (69.6/12)) = 6.67

Therefore, we can expect a difference of about 6.67 between the two sample means.

When the samples have larger variability (bigger SS values), the standard error for the sample mean difference will increase. This is because larger variability means that the scores are more spread out around their respective means, which increases the amount of variability in the difference between the two sample means. In contrast, when the variability is smaller, the scores are more tightly clustered around their means, and the standard error for the sample mean difference will be smaller.

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The missing figures in the diagrams are:      88km                    616km²,   37.7km                      113.14km respectively

A circle is a shape consisting of all points in a plane that are at a given distance from a given point, the center. Equivalently, it is the curve traced out by a point that moves in a plane so that its distance from a given point is constant.

                Radis          diameter            circumference            area

S/N           r                        2r                         2πr                           πr²

Applying the above formulae in each of the questions we have as follows:

1             3                  2*3=6             2*3.14*3=18.84        3.14*3*3=28.26       2        3.5                   7                 2*3.14*3.5=21.98    3.14*3.5*3.5=38.47      

3       7.5                 15               2*3.14*7.5=47.1ft      3.14*7.5*7.5=176.25

4       14km           28km         2*3.14*14=87.92km    3.14*14*14=615.44km²

5        5mi           10mi        2*3.14*5=31.4mi               3.14*5*5= 78.5mi²

6       2.5cm         5cm         2*3.14*2.5=15.7cm          3.14*2.5*2.5=19.63cm²

7      

14                     14*2 28             2*22/7*14                    22/7*14*14  

                                                     88km                    616km²

6                             12                2*22/7*6                     22/7*6*6

                                                    37.7km                      113.14km

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(a) The statement is false because we have found a case where n - m² is even.

(b) The statement holds true.

(c) The statement is false because we have found a case where s^2 is rational despite s being irrational.

(d) It is not possible to find two odd integers n and m such that n²m² - 1 is odd. Thus, the statement is false.

(a) The statement "for any integers n and m, if both n and m are odd, then n - m² is even" is incorrect. Let's consider a counterexample:

Take n = 3 and m = 1. Both n and m are odd.

n - m² = 3 - 1² = 3 - 1 = 2, which is an even number.

Therefore, the statement is false because we have found a case where n - m² is even.

(b) The statement "for any prime number p, p-2 is not prime" is generally true. Let's consider the cases:

If p is an odd prime greater than 2, then p-2 is an even number, and the only even prime number is 2. Therefore, p-2 cannot be prime in this case.

If p = 2, then p-2 = 0, which is not considered a prime number.

In both cases, p-2 is not a prime number. Therefore, the statement holds true.

(c) The statement "for any real number s, if s is irrational, then s^2 is irrational" is incorrect. Let's consider a counterexample:

Take s = √2. √2 is an irrational number.

s^2 = (√2)^2 = 2, which is a rational number.

Therefore, the statement is false because we have found a case where s^2 is rational despite s being irrational.

(d) The statement "There are two odd integers n and m such that n²m² - 1 is odd" is true. Let's consider the following example:

Take n = 1 and m = 1. Both n and m are odd.

n²m² - 1 = 1² * 1² - 1 = 1 * 1 - 1 = 0, which is an even number.

However, if we take n = 3 and m = 1, both n and m are still odd.

n²m² - 1 = 3² * 1² - 1 = 9 * 1 - 1 = 9 - 1 = 8, which is an even number.

Therefore, it is not possible to find two odd integers n and m such that n²m² - 1 is odd. Thus, the statement is false.

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1. The discriminant is positive, it has two real solutions

2. The discriminant is zero, it has a real solution

3. The discriminant is negative, it has no real solution

The discriminant of a graph is expressed as the part of the quadratic formula that is found under the square root symbol: b²-4ac.

It describes and gives information on whether there are two solutions, one solution, or no solutions.

It is important to note the following about discriminants;

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