Fill in the blanks and explain the pattern.
1, 4, 9, 16, 25,__,__,__,__,100


Fill in the blanks and explain the pattern.
1/2, 1/6, 1/18, 1/54,__,__,__

Answer:

true

Step-by-step explanation:

there is 100 candies. That means we can easily turn the amount of each type of candy into a percent. there was 30 butterscotch which means that is 30 percent. There was 15 strawberry which means that is 15 percent. add that and you get 45. This is a shortcut and i advise you use the way your teacher taught you.

[tex]|\Omega|=100\\|A|=30+15=45\\\\P(A)=\dfrac{45}{100}=0.45[/tex]

So TRUE

Complete Question

On the uploaded image is a similar question that will explain the given question

Answer:

The value of k is  [tex]k = 214285.7[/tex]

The percentage  of the oil that will be cleaned is [tex]x = 80.77\%[/tex]

Step-by-step explanation:

From the question we are told that

   The  cost of cleaning up the spillage is  [tex]C = \frac{ k x }{100 - x }[/tex]  [tex]x \le x \le 100[/tex]

     The  cost of cleaning x =  70% of the oil is  [tex]C = \$500,000[/tex]

Now at  [tex]C = \$500,000[/tex] we have  

       [tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]

       [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]k = 214285.7[/tex]

Now  When  [tex]C = \$900,000[/tex]

       [tex]x = 80.77\%[/tex]

Answer:

Step-by-step explanation:

The determinant of the coefficient matrix is ...

  [tex]D=\left|\begin{array}{ccc}2&5&3\\4&-1&-4\\-5&-2&6\end{array}\right|\\\\=2(-1)(6)+5(-4)(-5)+3(4)(-2)-2(-4)(-2)-5(4)(6)-3(-1)(-5)\\\\=-12+100-24-16-120-15=\boxed{-87}[/tex]

The other determinants are found in similar fashion after substituting the constants on the right for each of the above matrix columns, in turn.

Those determinants are ...

  [tex]D_x=\left|\begin{array}{ccc}21&5&3\\-13&-1&-4\\0&-2&6\end{array}\right|=174[/tex]

  [tex]D_y=\left|\begin{array}{ccc}2&21&3\\4&-13&-4\\-5&0&6\end{array}\right|=-435[/tex]

  [tex]D_z=\left|\begin{array}{ccc}2&5&21\\4&-1&-13\\-5&-2&0\end{array}\right|=0[/tex]

The solutions are ...

  x = 174/-87 = -2

  y = -435/-87 = 5

  z = 0

That is, (x, y, z) = (-2, 5, 0).

Answer:

[tex]x = \frac{1}{3} [/tex]

Step-by-step explanation:

*Move terms to the left and set equal to zero:

4㏒(√x) - ㏒(3x) - ㏒(x²) = 0

*simplify each term:

㏒(x²) - ㏒(3x) - ㏒(x²)

㏒(x²÷x²) -㏒(3x)

㏒(x²÷x² / 3x)

*cancel common factor x²:

㏒([tex]\frac{1}{3x}[/tex])

*rewrite to solve for x :

10⁰ = [tex]\frac{1}{3x}[/tex]

1 = [tex]\frac{1}{3x}[/tex]

1 · x = [tex]\frac{1}{3x}[/tex] · x

1x = [tex]\frac{1}{3}[/tex]

*that would be our answer, however, the convention is to exclude the "1" in front of variables so we are left with:

x = [tex]\frac{1}{3}[/tex]

There will be a total of 4 test scores including the final exam. To get a 70, the 4 tests need to equal 4 x 70 = 280 points , to be 79, they have to equal 4 x 79 = 316 points.

The 3 already done = 61 + 62 + 86 = 209 points.

The final exam needs to be between :

280 -209 = 71

316 -209 = 107. The answer would be between 71 and 100%

Answer:

The answer is D)90

Hope I helped

Answer:

False

Step-by-step explanation:

You Cnat solve it

Answer:

you cannot solve it

Step-by-step explanation:

false

it was all about equating some values

to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.

Let's determine the number of cooks required to bake the required number of cakes during the available time.

We have the following information:

- Each cook can bake either 5 large cakes or 14 small cakes per hour.

- The kitchen is available for 3 hours.

- We need to bake 29 large cakes and 260 cakes in total.

First, let's calculate the number of large cakes that can be baked by one cook in 3 hours:

1 cook can bake 5 large cakes/hour × 3 hours = 15 large cakes.

Next, let's calculate the number of small cakes that can be baked by one cook in 3 hours:

1 cook can bake 14 small cakes/hour × 3 hours = 42 small cakes.

Now, let's calculate the number of large cakes that can be baked by all the cooks in 3 hours:

Total number of large cakes = Number of cooks × Large cakes per cook per 3 hours

We need to bake 29 large cakes, so:

29 = Number of cooks × 15

Number of cooks = 29 / 15 ≈ 1.93

Since we can't have a fraction of a cook, we need to round up to the nearest whole number. Therefore, we need at least 2 cooks to bake the required number of large cakes.

Similarly, let's calculate the number of small cakes that can be baked by all the cooks in 3 hours:

Total number of small cakes = Number of cooks × Small cakes per cook per 3 hours

We need to bake 260 small cakes, so:

260 = Number of cooks × 42

Number of cooks = 260 / 42 ≈ 6.19

Again, rounding up to the nearest whole number, we need at least 7 cooks to bake the required number of small cakes.

Since we need to satisfy both requirements for large and small cakes, we choose the larger number of cooks required, which is 7 cooks.

Therefore, to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.

Learn more about work here

https://brainly.com/question/13245573

#SPJ2

Step-by-step explanation:

utilise the formula a+(n-1)d

a is the first number while d is common difference

Answer:

22

Step-by-step explanation:

Using the formular, Un = a + (n - 1)d

Where n = 10; a = -23; d = 5

U10 = -23 + (9)* 5

U10 = -23 + 45 = 22

Answer:

Question 1 = D) Acute

Question 2 = C)3 feet

Question 3 = D) Obtuse

Question 4 = C)27.73 ft.

Step-by-step explanation:

Question 1: A triangle has sides with lengths 5, 6, and 7. Is the triangle right, acute, or obtuse?

In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem

Where:

If a² + b² = c² = Right angle triangle

If a² +b² > c² = Acute triangle.

If a² +b² < c² = Obtuse triangle.

It is important to note that the length ‘‘c′′ is always the longest.

Therefore, for the above question, we have lengths

5 = a, 6 = b and c = 7

a² + b² = c²

5² + 6² = 7²

25 + 36 = 49

61 = 49

61 ≠ 49, Hence 61 > 49

Therefore, this is an Acute Triangle

Question 2: A 15-foot statue casts a 20-foot shadow. How tall is a person who casts a 4-foot-long shadow?

This is question that deals with proportion.

The formula to solve for this:

Height of the statue/ Length of the shadow of the person = Height of the person/ Length of the shadow of the person

Height of the statue = 15 feet

Length of the shadow of the person = 20 feet

Height of the person = unknown

Length of the shadow of the person = 4

15/ 20 = Height of the person/4

Cross Multiply

15 × 4 = 20 × Height of the person

Height of the person = 15 × 4/20

= 60/20

Height of the person = 3 feet

Therefore, the person is 3 feet tall.

Question 3: A triangle has sides with lengths 17, 12, and 9. Is the triangle right, acute, or obtuse?

In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem

Where:

If a² + b² = c² = Right angle triangle

If a² +b² > c² = Acute triangle.

If a² +b² < c² = Obtuse triangle.

It is important to note that the length ‘‘c′′ is always the longest.

Therefore, for the above question, we have lengths 17, 12, 9

9 = a, 12 = b and c = 17

a² + b² = c²

9² + 12² = 17²

81 + 144 = 289

225 = 289

225 ≠ 289

225 < 289

Hence, This is an Obtuse Triangle.

Question 4: Two friends are standing at opposite corners of a rectangular courtyard. The dimensions of the courtyard are 12 ft. by 25 ft. How far apart are the friends?

To calculate how far apart the two friends are we use the formula

Distance = √ ( Length² + Breadth²)

We are given dimensions: 12ft by 25ft

Length = 12ft

Breadth = 25ft

Distance = √(12ft)² + (25ft)²

Distance = √144ft²+ 625ft²

Distance = √769ft²

Distance = 27.730849248ft

Approximately ≈27.73ft

Therefore, the friends are 27.73ft apart.

Answer:

Mean age: 48

Standard deviation: 4

Step-by-step explanation:

a) Mean

The formula for Mean = Sum of terms/ Number of terms

Number of terms

= 42 + 54 + 50 + 54 + 50 + 42 + 46 + 46 + 48+ 48/ 10

= 480/10

= 48

The mean age is 48

b) Standard deviation

The formula for Standard deviation =

√(x - Mean)²/n

Where n = number of terms

Standard deviation =

√[(42 - 48)² + (54 - 48)² + (50 - 48)² +(54 - 48)² + (50 - 48)² +(42 - 48)² + (46 - 48)² + (46 - 48)² + (48 - 48)² + (48 - 48)² / 10]

= √-6² + 6² + 2² + 6² + 2² + -6² + -2² + -2² + 0² + 0²/10

=√36 + 36 + 4 + 36 + 4 + 36 + 4 + 4 + 0 + 0/ 10

=√160/10

= √16

= 4

The standard deviation of the ages is 4

Answer:

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Answer:

Perimeter = 42 cm

Step-by-step explanation:

A square has all equal sides so you would just add 10.5 + 10.5 + 10.5 + 10.5 to get 42 cm.

Answer:

42 cm

Step-by-step explanation:

Side of square = 10.5 cm (given)

Perimeter of square = Side X 4

                                  = 10.5 X 4

                                  = 42 cm

HOPE THIS HELPED YOU !

:)

Answer:

x=9/2

Step-by-step explanation:

Let's solve your equation step-by-step.

5x+4(−x−2)=−5x+2(x−1)+12

Step 1: Simplify both sides of the equation.

5x+4(−x−2)=−5x+2(x−1)+12

5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)

5x+−4x+−8=−5x+2x+−2+12

(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)

x+−8=−3x+10

x−8=−3x+10

Step 2: Add 3x to both sides.

x−8+3x=−3x+10+3x

4x−8=10

Step 3: Add 8 to both sides.

4x−8+8=10+8

4x=18

Step 4: Divide both sides by 4.

4x/4=18/4

x=9/2

Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]

$a^{2}=5+2 \sqrt{6}$

$a^{3}=11 \sqrt{2}+9 \sqrt{3}$

The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.

Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$

so fits with the other answers.

Answer:

[tex]y^3 -6y-6[/tex]

Answer:

Option (D)

Step-by-step explanation:

Formula to get the area of a regular polygon in a circle will be,

Area = [tex]n[\frac{1}{2}\times (\text{Base})\times (\text{Height})][/tex]

        = [tex]n[\frac{1}{2}\times (\text{s})\times (\text{h})][/tex]

Here 'n' is the number of sides.

If n increases, h approaches r so that 'rh' approaches r².

In other words, if the number of sides of the polygon gets increased, area of the polygon approaches the area of the circle.

Therefore, Option (4) will be the answer.

In this exercise it is necessary to have knowledge about polygons, so we have to:

Letter D

Then using the formula for the area of ​​a regular polygon we find that:

[tex]A=n(1/2*B*H)\\=n(1/2*S*H)[/tex]

So from this way we were not able to identify the option that best corresponds to this alternative.

See more about polygons at  brainly.com/question/17756657

Answer:

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Step-by-step explanation:

Given that:

[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]

[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]

[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]

[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]

[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Option C is the correct option

Step-by-step explanation:

From the question we are told that

   The equation is  [tex]f (x, y , z ) = x^2 +y^2 + z^2[/tex]

    The constraint is  [tex]P(x, y , z) = x + y + z - 24 = 0[/tex]

Now using Lagrange multipliers  we have that  

   [tex]\lambda = \frac{ \delta f }{ \delta x } = 2 x[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta y } = y[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta z } = 2 z[/tex]

=>       [tex]x = \frac{ \lambda }{2}[/tex]

          [tex]y = \frac{ \lambda }{2}[/tex]

         [tex]z = \frac{ \lambda }{2}[/tex]

From the constraint  we have

      [tex]\frac{\lambda }{2} + \frac{\lambda }{2} + \frac{\lambda }{2} = 24[/tex]

=>   [tex]\frac{3 \lambda }{2} = 24[/tex]

=>   [tex]\lambda = 16[/tex]

substituting for x, y, z

=>   x =  8

=>  y =  8

=>   z =  8        

Hence

    [tex]f (8, 8 , 8 ) = 8^2 +8^2 + 8^2[/tex]

    [tex]f (8, 8 , 8 ) = 192[/tex]

Answer:

$3

Step-by-step explanation:

Given

Total Cost Price: $12,000

Unit Cost Price= $6

Total Selling Price = $18,000

Required

Determine the profit on each share

First, we need to determine the units of share bought;

Units = Total cost price / Unit Cost Price

[tex]Units = \frac{\$12000}{\$6}[/tex]

[tex]Units = 2000[/tex]

Next is to determine the selling price of each share; This is calculated as follows;

Unit Selling Price = Total Selling Price / Units Sold

[tex]Unit\ Selling\ Price = \frac{\$18000}{\$2000}[/tex]

[tex]Unit\ Selling\ Price = \$9[/tex]

The profit is the difference between the unit cost price and unit selling price

[tex]Profit = Unit\ Selling\ Price - Unit\ Cost\ Price[/tex]

[tex]Profit = \$9 - \$6[/tex]

[tex]Profit = \$3[/tex]

Answer:

It sold 14 cans boxes of food and 12 cans of food.

Step-by-step explanation:

The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.

Let the no. of sets of food boxes be x.

According to the question,

6x+7x=26

13x=26

x=26/13

x=2

No. of food cans =6x=6×2=12 cans

No. of food boxes=7x=7×2=14 boxes

Please mark brainliest ,if it is truly the best ! Thank you!

Answer:

  x = 1.00995066776

  x = 2.52925492433

Step-by-step explanation:

This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...

  [tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]

The attached graph shows zeros at

  x = 1.00995066776 and 2.52925492433

_____

Comment on the equation

Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068

Comment on the answer refinement

We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.

Answer:

The upper bound of the confidence interval is 0.34

Step-by-step explanation:

Here in this question, we want to calculate the upper bound of the confidence interval.

We start by calculating the margin of error.

Mathematically, the margin of error = 0.29 -0.24 = 0.05

So to get the upper bound of the confidence interval, we simply add this margin of error to the mean

That would be 0.05 + 0.29 = 0.34

Answer:

50.93

Step-by-step explanation:

Add up the frequencies:

2 + 5 + 14 + 15 + 21 + 18 + 15 + 9 + 2 = 101

Divide by 2: 101/2 = 50.5

So the median is the 51st number, with 50 below and 50 above.

Add up the frequencies until you find the interval that contains the 51st number.

2 + 5 + 14 + 15 = 36

2 + 5 + 14 + 15 + 21 = 57

So the median is in the group 49.5 − 51.5.  To estimate the median, we use interpolation.  Find the slope of the line from (36, 49.5) to (57, 51.5).

m = (51.5 − 49.5) / (57 − 36)

m = 2/21

So at x = 51:

2/21 = (y − 49.5) / (51 − 36)

y = 50.93

Answer:

1/3(a+b)

hope it helps :>

Complete question is;

A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.

(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.

Answer:

Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

We are given;

Mean: μ = 128

Standard deviation; σ = 0.2

n = 35

Now, formula for standard error of mean is given as;

se = σ/√n

se = 0.2/√35

se = 0.0338

Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;

μ ± 2se = 128 ± 0.0338

This gives; 127.9662, 128.0338

So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Answer: $2750

Step-by-step explanation:

Formula to calculate interest : I = Prt , where P = Principal amount , r = rate of interest ( in decimal) , t= time.

Given:  I= $495

t= 3 years

r= 6% = 0.06

Then, according to the above formula:

[tex]495 = P (0.06\times3)\\\\\Rightarrow\ P=\dfrac{495}{0.18}\\\\\Rightarrow\ P=2750[/tex]

Hence, the principal invested = $2750

Answer:

1048576

Step-by-step explanation:

Given the following :

Pizza order :

Size = small, medium, large, or extra large = 4 possible sizes

Toppings = any combination of 8 possible toppings (getting no toppings is allowed, as is getting all 8).

Combination of Toppings = 2^8

Four different sizes of pizza = 4

Number of possibilities in ordering for a single pizza :

(4 * 2^8) = 4 * 256 = 1024

Number of possibilities in ordering two pizzas :

(4 * 2^8)^2

(2^2 * 2^8)^2

From indices :

[2^(2+8)]^2

[2^(10)]^2

2^(10*2)

2^20

= 1048576

Answer:

Solution : Option B

Step-by-Step Explanation:

We have the following system of equations at hand here.

{ x = 5 cot(t), y = - 3csc(t) + 4 }

Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,

x = 5 cot(t) ⇒ x - 5 = cot(t),

y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)

Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations  as well. --- Step #2

( y - 4 / - 3 )² = (csc(t))²

- ( x - 5 / 1 )² = (cot(t))²  

___________________

(y - 4)² / 9 - x² / 25 = 1

And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.

Answer: y=(4/3)x+2/3

Step-by-step explanation:

Slope-intercept form is expressed as y=mx+b

First, find the slope (m):

m= rise/run or vertical/horizontal or y/x (found between the highlighted points)

m = 4/3

Second, find b:

Use one of the highlighted points for (x, y)

2=4/3(1)+b

6/3=4/3+b

2/3=b

b=2/3

Plug it into the equation:

You get y=(4/3)x+2/3 :)

Answer:

A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu = 100}[/tex]

[tex]\mathtt{H_1: \mu \neq 100}[/tex]

A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .

Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]

∴

[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]

[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]

when  [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]

[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]

From the standard normal distribution tables

[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]

[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]

[tex]\mathbf{\alpha = 0.0244 }[/tex]

Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]

Thus;

β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )

∴

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 103[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]

[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 105[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]

[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.