A micrometer reading for a part is 7.57.5 in. The specifications call for a dimension of 7.59 in.7.59 in. Which is​ larger, the micrometer reading or the​ specification?

Answer:

[tex]f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}....[/tex]

Approximate error = 0.4426

Step-by-step explanation:

f(x)=tanx, a=0

Maclaurin series formula used is given below

[tex]f(x)=\sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)x^{n}}{n!}=f(0)+f'(0)x+\dfrac{f''(0)}{2!}x^{2}+\dfrac{f'''(0)}{3!}x^{3}+....[/tex]

f(x)=tanx

f(0)=tan0=0

[tex]f'(x)=sec^{2}x\\f'(0)=sec^{2}0=1\\f''(x)=2sec^{2}xtanx\\f''(0)=2sec^{2}0tan0=0\\f'''(x)=-4sec^{2}x+6sec^{4}x\\f'''(0)=-4sec^{2}0+6sec^{4}0=-4+6=2\\[/tex]

[tex]f''''(x)=-8(2sec^{2}xtan^{2}x+sec^{4}x)+24(4sec^{4}xtan^{2}x)+sec^{6})\\f''''(0)=-8(0+1)+24(0+1)=-8+24=16\\[/tex]

[tex]f(x)=0+x+0+\dfrac{2x^{3}}{3!}+\dfrac{16x^{4}}{4!}\\[/tex]

[tex]f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}\\[/tex]

Hence, the Taylor series for f(x)=tanx is given by

[tex]f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}....[/tex]

Maclaurin series upper bound error formula used is given as

R_n(x)=|f(x)-T_n(x)|

R_3(x)=|tanx-T_3(x)|

[tex]R_3(x)=|tanx-x-\dfrac{x^{3}}{3}-\dfrac{2x^{4}}{3}|[/tex]

Plugging this value x=1

[tex]R_3(x)=|tan(1)-1-\dfrac{1}{3}-\dfrac{2}{3}|\\[/tex]

R_3(x)=|1.5574-1-0.333-0.666|

R_3(x)=|-0.4426|=0.4426

Hence, upper bound on the error approximation

tan(1)=0.4426

Answer:

A. A straight line intersecting two parallel lines produces corresponding equal angles.

Step-by-step explanation:

If two straight lines are parallel, and a straight line intersects them, then we get alternate angles equal, exterior angle equal to the opposite interior angle on the same side and the interior angles on the same sides equal to two right angles.

Hence, here the rule that is not true is : A. A straight line intersecting two parallel lines produces corresponding equal angles.

Answer: y = v₀tgθx - gx²/2v₀²cos²θ

a = v₀tgθ

b = -g/2v₀²cos²θ

Step-by-step explanation:

x = v₀ₓt

y = v₀y.t - g.t²/2

x = v₀.cosθt → t = x/v₀.cosθ

y = v₀y.t - g.t²/2

v₀y = v₀.senθ

y = v₀senθ.x/v₀cosθ - g/2.(x/v₀cosθ)²

y = v₀.tgθ.x - gx²/2v₀²cos²θ

a = v₀tgθ → constant because v₀ and θ do not change

b = - g/2v₀²cos²θ → constant because v₀, g and θ do not change

Final answer:

The trajectory of a projectile is shown to be parabolic by substituting time from the x-direction motion equation into the y-direction motion equation and rearranging, yielding a parabolic form y = ax + bx², with constants determined by initial velocity and gravity.

Explanation:

To prove that the trajectory of a projectile is parabolic, we start with the equations of motion in the x and y directions for a projectile that starts at the origin. For the x-direction, we have x = v_0x t, where v_0x is the initial velocity component in the x-direction and t is the time.

In the y-direction, the equation is y = v_0y t - (1/2) g t², where v_0y is the initial velocity component in the y-direction and g is the acceleration due to gravity.

To find t from the x equation: t = x / v_0x. Substituting this into the y equation yields: y = v_0y (x / v_0x) - (1/2) g (x / v_0x)^2.

Now, simplifying we get: y = (v_0y / v_0x) x - (g/2 v_0x^2) x^2, which is of the form y = ax + bx². Here, a = v_0y/v_0x and b = -g/2 v_0x²are constants depending on the initial velocity components and acceleration due to gravity. The equation y = ax + bx² represents a parabolic path, confirming that the projectile's trajectory is indeed a parabola.

Answer:

Area = 8

Step-by-step explanation:

A skecth is given in the attached file, there are two extra lines used to calculate the area with simple geometry:

We must use a double integral to obtain the area:

[tex]\int\limits^2_0 {\int\limits^b_a  \, dy } \, dx[/tex]

Where

b stands for y=2x+1

a stands for y=-x

Carring out the integrals we find the area:

[tex]\int\limits^2_0 {(2x+1 - (-x))} \, dx =  \int\limits^2_0 {3x+1} \, dx = (3x^{2}/2+x) \left \{ {{2} \atop {0}} \right.\\ A =( 3*2^{2}/2) + 2 =8[/tex]

Geometrically we can divide the area bounded by this lines as two triangles and a rectangle from the figure and the intersection of these lines we kno that the three figures have a base of 2. The heigth of the rectangle is 1 and for the triangles we have 4 for the upper triangle and 2 for the lower.

Therefore:

[tex]A = A_{upperT}+ A_{rect}+ A_{lowerT}[/tex]

and

[tex]A_{upperT}=2*4/2=4\\A_{rect}=2*1=2\\ A_{lowerT}=2*2/2=2[/tex]

Summing the four areas we have:

A=8

Greeting!

Answer:

z=10

x1=0

x2=0

x3=3.33

Step-by-step explanation:

First Step convert your constraints in standard equations

so we have

x1 + 2x2 + 3x3+x4 = 10

x1 + x2 +x5= 5

x1 +x6= 1

Now we pass it all to the simplex table

Remember that we choose the column with the most negative value

Pivot Element=3

Divide all elements on Pivot Line by Pivot Element

Line x5= 0*Pivot Line +Line x5

Line x6= 0*Pivot Line+ Line X6

Line Z= 3* Pivot Line + Line Z

We finish when all the elements from the line Z are positive

Hence we have that x3=3.33 and x1=0, x2=0 and the max of z is 10

Answer:

z=10

x1=0

x2=0

x3=3.33

Step-by-step explanation:

Answer:

Answered

Step-by-step explanation:

Online courses are those classes which are delivered entirely online. Students study via web cam,  chat rooms and smart boards. Whereas hybrid learning is a combination of both online learning and  traditional in class learning.  Remote work is working away from the work place at your own comfort and choice of location.

Answer:

Ac = 10100100011

[tex]A \cup B = 11111111110[/tex]

[tex]A \cap B = 00010011000[/tex]

[tex]A - B = 100100111010[/tex]

Step-by-step explanation:

All these operations are bitwise operations.

Ac is the complement of a. So where we have a bit 0, the complement is 1. Where we have a bit 1, the complement is 0. So

A = 01011011100

Ac = 10100100011

The second operation is the union between A and B. This is bitwise(bit 1 of A with bit 1 of B, bit 2 of A with bit 2 of B,...). The union operation is only 0 when it is between two zeros. So:

A = 01011011100

B = 10110111010

[tex]A \cup B = 11111111110[/tex]

The third operation is the intersection between A and B. Again, bitwise. The intersection is only 1 when it is between two bits that are 1. So

A = 01011110100

B = 10110111010

[tex]A \cap B = 00010011000[/tex]

The last operation is the bitwise subtraction between A and B. We start from the least significant bit(the last one). And we have to take care of the borrow operator also, similarly to a decimal subtraction.

We can only borrow from a previous bit 1, and this bit is set to 0

0-0 is 0 with no borrow

1-0 is 1 with no borrow

1 with borrow - 0 is 1 with borrow

1 with borrow - 1 is 1 with no borrow

0-1 is 1 with no borrow

0 with borrow -1 is 0 with no borrow

1-1 is 0 with no borrow

If we arrive at the first bit(the most significant) with a borrow, we must add a 1 at the front of the answer. So

A = 01011110100

B = 10110111010

[tex]A - B = 100100111010[/tex]

The conversion of 1.256 grams is as follows: 1,256,000 micrograms, 1256 milligrams, and 0.001256 kilograms.

To convert 1.256 grams (g) to different units of mass, you use the following conversion factors:

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1.256 grams is equal to 1,256,000 micrograms, 1,256 milligrams, and 0.001256 kilograms. These conversions use basic multiplication and division of the metric system.

To convert 1.256 grams to different units, you should be familiar with the respective conversion factors:

Steps to Convert:

Therefore, the conversions are:

Answer with Step-by-step explanation:

We are given that a number 8881 is not a prime number

We have to prove that it has given a prime  factor that is at most 89.

In order to prove that given number has highest prime factor is 89 we will find the prime factorization of given number.

[tex]8881=83\times 107[/tex]

Therefore, 8881 is not  a prime number and it has two factors 83 and 107.

83 is  a prime factor of 8881 which is less than 89.We have to find  a prime factor of 8881 which is atmost 89.

Therefore, 83 is that prime factor .

Hence, 8881 has a prime factor that is at most 89.

Answer:

9.3125 hours

Step-by-step explanation:

Given:

Number of components consisting in a product = 25

Standard time for work cycle = 7.45 minutes

or

standard time for work cycle = [tex]\frac{\textup{7.45}}{\textup{60}}[/tex] hours

Number of units completed = 75

Now,

The number of standard hours

= Number of units completed × standard time for the work cycle

=  [tex]75\times\frac{\textup{7.45}}{\textup{60}}[/tex] hours

or

The number of standard hours = 9.3125 hours

Answer:

The answer is 9.3125 hours :)

Step-by-step explanation:

Answer: The cube root of 10 is 2.1544 using an Xo value of -0.003723

Step-by-step explanation: The Newton-Raphson is a root finding method and its formula is NR: X=Xo-(f(x)/f'(x). Once you have the equation you also need to find the derivative of that equation before applying the formula. Since the problem stated that X =10, the method was applied to find the best root in order to find the cube root of 10 up to 5 significant figures. The best method is to use a software like Excel that helps you calculate those iterations faster. The root finding for this example was -0.003723.

Answer: There are 90 snack boxes.

Step-by-step explanation:

Given : The number of kinds of fruits = 5

The number of kinds of herbal teas = 3

The number of kinds of flavors of wrap sandwich = 6

Then by using the fundamental principal of counting, the number of possible snack are there will be :_

[tex]5\times3\times6= 90[/tex]

Therefore, the number of possible snacks = 90

Final Answer:

No, it is not possible to find a sequence of m−1 consecutive positive integers none of which is prime for any integer m≥2.

Explanation:

In order to determine whether it is possible to find a sequence of m−1 consecutive positive integers none of which is prime, we need to consider the properties of prime numbers. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Therefore, if we take any m−1 consecutive positive integers, at least one of them will be divisible by a prime number. This means that it is not possible to find a sequence of m−1 consecutive positive integers none of which is prime.

For example, let’s consider the sequence of 4 consecutive positive integers: 10, 11, 12, and 13. Among these numbers, 11 and 13 are prime. Even if we consider larger sequences, we will always encounter at least one prime number within the consecutive positive integers.

In conclusion, due to the nature of prime numbers and their distribution among positive integers, it is not possible to find a sequence of m−1 consecutive positive integers none of which is prime for any integer m≥2.

Step-by-step explanation:

We will prove by mathematical induction that, for every natural [tex]n\geq 4[/tex],  

[tex]5^n\geq 2^{2n+1}+100[/tex]

We will prove our base case, when n=4, to be true.

Base case:

[tex]5^4=625\geq 612=2^{2*4+1}+100[/tex]

Inductive hypothesis:  

Given a natural [tex]n\geq 4[/tex],  

[tex]5^n\geq 2^{2n+1}+100[/tex]

Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.

Inductive step:

[tex]2^{2(n+1)+1}+100=2^{2n+1+2}+100=\\=4*2^{2n+1}+100\leq 4(2^{2n+1}+100)\leq 4*5^n<5^{n+1}[/tex]

With this we have proved our statement to be true for n+1.  

In conlusion, for every natural [tex]n\geq4[/tex].

[tex]5^n\geq 2^{2n+1}+100[/tex]

Answer:

The expended form of the provided expression is: [tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]

Step-by-step explanation:

Consider the provided expression.

[tex](2x-1)^5[/tex]

The binomial theorem:

[tex](a+b)^{n}=\sum _{r=0}^{n}{n \choose r}a^{n-r}b^r[/tex]

Where,

[tex]{n \choose r}= ^nC_r =\frac{n!}{(n-r)!r!}[/tex]

Now by using the above formula.

[tex]\frac{5!}{0!\left(5-0\right)!}\left(2x\right)^5\left(-1\right)^0+\frac{5!}{1!\left(5-1\right)!}\left(2x\right)^4\left(-1\right)^1+\frac{5!}{2!\left(5-2\right)!}\left(2x\right)^3\left(-1\right)^2+\frac{5!}{3!\left(5-3\right)!}\left(2x\right)^2\left(-1\right)^3+\frac{5!}{4!\left(5-4\right)!}\left(2x\right)^1\left(-1\right)^4+\frac{5!}{5!\left(5-5\right)!}\left(2x\right)^0\left(-1\right)^5[/tex]

[tex]2^5\cdot \:1\cdot \:1\cdot \:x^5-1\cdot \frac{2^4\cdot \:5x^4}{1!}+1\cdot \frac{2^3\cdot \:20x^3}{2!}-1\cdot \frac{2^2\cdot \:20x^2}{2!}+1\cdot \frac{5\cdot \:2x}{1!}+1\cdot \frac{\left(-1\right)^5}{\left(5-5\right)!}[/tex]

[tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]

Hence, the expended form of the provided expression is: [tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]

Answer: time = 55.48 years

Explanation:

Given:

Principal amount = $10000

Interest rate = 2% p.a

Amount = $30000

We can evaluate the time taken using the following formula:

[tex]Amount=Principal(1+\frac{r}{100})^{t}[/tex]

[tex]30000 = 10000\times(1+\frac{2}{100})^{t}[/tex]

Solving the above equation, we get

[tex]1.02^{t} = 3[/tex]

Now taking log on both sides, we get;

[tex]t\ log(1.02) = log(3)[/tex]

time = 55.48 years

Answer:   100

Step-by-step explanation:

Given : The lodhl diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert.

There are 5 appetizers, 5 soups, 4 main courses, and 5 desserts.

Also, a dessert and a appetizer are not allowed to take together.

By Fundamental counting principal ,

Number of three-course meals with dessert and without appetizer :

[tex]5\times4\times5=100[/tex]      (1)

Number of three-course meals with appetizer and without dessert :

[tex]5\times5\times4=100[/tex]      (2)

Now, the number of meals with either dessert or appetizer :-

[tex]100+100=200[/tex]           [Add (1) and (2)]

Answer:

500

Step-by-step explanation:

5x5x4x5

we do

5x5=25

25x4=100

100x5= 500

Final answer:

To find the length of AC in the right-angled isosceles triangle △ADC with altitude CD and AD equals to BC, we use the Pythagorean theorem, substituting known lengths to solve for AC, which is √11.25 cm.

Explanation:

The student is asking to find the length of side AC in a triangle △ABC where CD is an altitude and AD is equal to BC, with given lengths AB = 3 cm and CD = 3 cm.

To solve for AC, we can use the Pythagorean theorem which states that for a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Since CD is an altitude and AD = BC, △ADC is a right-angled isosceles triangle.

Using the Pythagorean theorem in △ADC, we have:

AD2 + CD2 = AC2

Since AD = BC and CD = 3 cm, let's assume AD = x cm. Then,

x2 + 32 = AC2

x2 + 9 = AC2

But, AB = 3 cm, and AB = AD + DB = x + x = 2x,

Therefore, x = AB/2 = 1.5 cm. Substituting this value back into the previous equation:

(1.5)2 + 9 = AC2 = 2.25 + 9 = 11.25

AC = √11.25 cm

Therefore, the length of AC is √11.25 cm.

Final answer:

The sum of sequence P (YP) is greater than the sum of sequence Q (Q) by 94, with P summing to 2850 and Q summing to 2756.

Explanation:

To determine which sum is greater between P and Q, we need to first understand that P and Q represent the sum of arithmetic sequences. The first sequence, P, begins with 8 and increments by 2, ending at 106. The second sequence, Q, starts at 2 and increments by 2, ending at 104.

To find out the sums, we can use the formula for the sum of an arithmetic sequence: S = n/2 (a1 + an), where S is the sum, n is the number of terms, a1 is the first term, and an is the last term. For the sequence P, the first term a1 is 8, and the last term an is 106. To find n, the number of terms, we can use the formula n = (an - a1) / d + 1, where d is the common difference, which is 2 in this case.

Applying the formula to the sequence P, we get:

The number of terms in P: n(P) = (106 - 8) / 2 + 1 = 50

Sum of P: S(P) = 50/2 x (8 + 106) = 25 x 114 = 2850

For the sequence Q:

The number of terms in Q: n(Q) = (104 - 2) / 2 + 1 = 52

Sum of Q: S(Q) = 52/2 x (2 + 104) = 26 x 106 = 2756

Comparing these sums, we can see that the sum of sequence P, 2850, is greater than the sum of sequence Q, 2756. The difference between the sums is 2850 - 2756 = 94.

Therefore, P is greater than Q by 94.

Answer:

Step-by-step explanation:

First you would find the area of the lot:

50 x 36 = 1800 m^2

Then you would find the area of the house:

31 x 9 = 279 m^2

Then you subtract the are of the lot minus the area of the house:

1800 - 279 = 1521 m^2

And the final answer is:

1521 square meters is left

The area of space which is left over is [tex]1521m^{2} [/tex]

Required steps shown below,

Area of lot [tex]=length*width[/tex]

                  [tex]=50*36=1800m^{2} [/tex]

Area of house[tex]=length*width[/tex]

                       [tex]=31*9=279m^{2} [/tex]

To calculate area of space is left over, subtract area of house from area of lot.

The space left over is,

                                [tex]1800-279=1521m^{2} [/tex]

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Answer:

Yes, rational numbers are closed under addition.

Step-by-step explanation:

Rational numbers are number that can be expressed in the form of fraction [tex]\frac{x}{y}[/tex], where x and y are integers and y ≠ 0.

Now, the closure property of addition of rational number states that if we add two rational number, then the sum of these two rational number will also be a rational number.

Let a and b be two rational number, then,

a+b = c, where c is the sum of and b

c is also a rational number.

Thus, rational numbers are closed under addition.

This can be explained with the help of a example.

[tex]\frac{1}{7} + \frac{2}{7}[/tex] = [tex]\frac{2}{7}[/tex]

It is clear that [tex]\frac{1}{7}, \frac{2}{7}, \frac{3}{7}[/tex] are rational numers.

Thus, rational numbers are closed under addition.

Answer:

12,000.

Step-by-step explanation:

Let x be the number of people that were enrolled last year.

We have been given that enrollment at ELAC decreased by 5%, or 600 people, the year. We are asked to find the number of people that were enrolled last year.

We can set as equation such that 5% of x equals 600.

[tex]\frac{5}{100}\cdot x=600[/tex]

[tex]0.05x=600[/tex]

[tex]\frac{0.05x}{0.05}=\frac{600}{0.05}[/tex]

[tex]x=12,000[/tex]

Therefore, 12,000 people were enrolled last year.

Answer: a. You must pay $11.19 for the item.

b.The price you would expect to pay would be $36.59

Step-by-step explanation:

Hi, for first the question you need to calculate the 30 percent of the price of the item, and then subtract that result to the original price.

So: $15.99 × 0,30 = $4.797

$15.99 - $4.797 = $11.19

You must pay $11.19 for the item.

It´s a similar resolution, first you calculate the 40 percent of the retail price, and then subtract that result to the retail price.

So:

$60.99 × 0,40= $24.396

$60.99 - $24.396 = $36.59

The price you would expect to pay would be $36.59

In summary, the raw data can only be retrieved from a dot plot, not a histogram, as the dot plot represents individual data points, whereas a histogram represents data in grouped intervals.

When comparing a dot plot to a histogram, there are several key points to consider:

A frequency table or a frequency polygon can also be useful tools for data representation in complement to histograms and dot plots, especially when comparing distributions.

The total volume of the bird feeder is 47.1 cubic inches.

The total volume of the bird feeder.

Given:

So, the total volume of the bird feeder is 47.1 cubic inches.

Answer:

Venom C is  0.625 times as potent as venom A.

Step-by-step explanation:

We are given that venom  A is four times as potent as venom B.

Venom C is 2.5 times as potent as Venom B.

We have to find that how many times as potent is venom C is compared to venom A.

Let Venom B =x

Then Venom A=[tex]4\times x=4x[/tex]

Venom C=[tex]2.5\times x=2.5 x[/tex]

[tex]\frac{Venom\;C}{venom\;A}=\frac{2.5x}{4x}=\frac{25}{40}=\frac{5}{8}=0.625[/tex]

Hence, venom C is  0.625 times as potent as venom A.

In relative terms, venom C is 0.625 times as potent as venom A, using venom B as a common comparison base.

To solve this, we first need to establish a common comparison base. If venom A is four times as potent as venom B, let's assign a relative potency value to venom B, let's say 1. Therefore, the potency of venom A is 4. Similarly, if venom C is 2.5 times as potent as venom B, then the potency of venom C is 2.5.

To find out how much more potent venom C is compared to venom A, we divide the potency of venom C by the potency of venom A. Mathematically, it looks like this: Potency of C ÷ Potency of A = 2.5 ÷ 4 = 0.625. Therefore, venom C is 0.625 times as potent as venom A.

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Answer:  Option 'c' is correct.

Step-by-step explanation:

Since we have given that

The salary earned for a month stored in = A35

Rate of tax reduces by = 22%

We need to remove the % sign by dividing 22 by 100 and it becomes 0.22.

So, the dollar amount of taxes is given by

[tex]Taxes=A35\times \dfrac{22}{100}\\\\Taxes=A35\times 0.22[/tex]

Hence, Option 'c' is correct.

Answer:

int t(int n){

     if(n == 1)

          return 1;

     else

          return n*n*t(n-1);

}

Step-by-step explanation:

A recursive function is a function that calls itself.

I am going to give you an example of this algorithm in the C language of programming.

int t(int n){

     if(n == 1)

          return 1;

     else

          return n*n*t(n-1);

}

The function is named t. In the else clause, the function calls itself, so it is recursive.

Answer:

a) False b) True c) False d ) False e) True f) False g) False h) False

Step-by-step explanation:

a) False

x² × x³ = [tex]x^5[/tex]

When we multiply two exponential number with same base, their powers add up.

b) True

2π + 5π = π(2+5) = 7π

The coefficients of π are added together.

c) False

x² ≥ 0. For any value of x, negative or positive x² is always positive. But for x = 0, x² = 0×0 = 0

d) False

315 - 8 = 307, which is clearly an odd number.

e)True

The sum of all integers is always even.

Let m and n be two odd integers.

Thus, they can be expressed as m = 2r + 1 and n = 2s +1, where r and s are even integers.

m + n = 2r +2s + 2, which is clearly even.

f) False

Since √2 is an irrational number. It cannot belong to z, which is collection of all integer number.

√2 ∉ z

g) False

Since -1 is a negative integer, it cannot belong to [tex]z^+[/tex], as it is collection of all positive integers.

h) False

π cannot belong to Q because Q is a collection of all rational numbers and π is not a rational number. The decimal expansion of π is non- terminating that is it does not end.

Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

a)

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

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

where

[tex]\bar x=mean\; of\;the \;sample[/tex]

[tex]\mu=mean\; established\; in\; H_0[/tex]

[tex]\sigma=standard \; deviation[/tex]

N = size of the sample.

So,

[tex]z=\frac{185-175}{20/\sqrt {10}}=1.5811[/tex]

[tex]\boxed {z=1.5811}[/tex]

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of  1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of  1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is  

[tex]\boxed {p=0.057}[/tex]

b)

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

Step 1

Compute [tex]\bar x_{critical}[/tex]

[tex]1.64=z_{critical}=\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}[/tex]

[tex]\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}=\frac{\bar x_{critical}-175}{6.3245}=1.64\Rightarrow \bar x_{critical}=185.3721[/tex]

So we would make a Type II error if our sample mean is less than 185.3721.  

Step 2

Compute the probability that your sample mean is less than 185.3711  

[tex]P(\bar x < 185.3711)=P(z< \frac{185.3711-185}{6.3245})=P(z<0.0586)=0.5234[/tex]

So, the probability of making a Type II error is 0.5234 = 52.34%

c)

The power of a hypothesis test is 1 minus the probability of a Type II error. So, the power of the test is

1 - 0.5234 = 0.4766

The p-value is 0.014, indicating strong evidence against the null hypothesis. The probability of a type II error is 0.0907, and the power of the test is 0.9093.

To find the p-value, we need to determine the probability of observing a sample average of 185 or higher, given that the true mean foam height is 175. Since the sample size is small, we'll use the t-distribution instead of the normal distribution. With a sample size of 10, the degrees of freedom is 9. Using the t-distribution table or a calculator, we find the p-value to be 0.014 (rounded to 3 decimal places).

The probability of a type II error is the probability of failing to reject the null hypothesis when it is actually false. In this case, the null hypothesis is μ = 175, but the true mean foam height is 200. We assume α = 0.05, so the critical value is 1.645 (from the t-distribution table for a one-tailed test). Using the formula for the standard error of the sample mean, σ/√n, we can calculate the standard deviation of the sample mean to be 20/√10 = 6.32 (rounded to 2 decimal places). The difference between the critical value and the true mean is (200 - 175)/6.32 = 3.96 (rounded to 2 decimal places). Using a t-distribution table or calculator to find the area to the right of 3.96 with 9 degrees of freedom, we find the probability of a type II error to be 0.0907 (rounded to 4 decimal places).

The power of the test is 1 minus the probability of a type II error. So the power of the test is 1 - 0.0907 = 0.9093 (rounded to 4 decimal places).

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