Answer:
see below
Step-by-step explanation:
√p = 9/16
We need to square each side, not take the square root
(√p)^2 =( 9/16)^2
p = 81/256
is this a function {(-2, 6), (-3, 7), (-4, 8), (-3, 10)}
No, that is not a function.
To be a function, each different input (x) needs a different output (y)
In the given function there are two -3’s as inputs and they have different y values, so it can’t be a function.
Answer: no
Step-by-step explanation: To determine if a relation is a function, take a look at the x–coordinate of each ordered pair. If the x–coordinate is different in each ordered pair, then the relation is a function.
Note that the only exception to this would be that if the x-coordinate pairs up with the same y-coordinate in a relation more than once, it's still classified ad a function.
Ask yourself, do any of the ordered pairs
in this relation have the same x-coordinate?
Well by looking at this relation, we can see that two
of the ordered pairs have the same x-coordinate.
In this case, the x-coordinate of 3 appears twice.
So no, this relation is not a function.
The cost of a daily rental car is as follows: The initial fee is $39.99 for the car, and it costs $0.20 per mile. If Julie's final bill was $100.00 before taxes, how many miles did she drive?
Answer:
300.05 miles
Step-by-step explanation:
initial fee= $39.99
final bill = $ 100
cost =$ 0.20 per mile
remaining amount = $ 60.01
solution,
she drive = remaining amount / cost
=60.01/0.20
=300.05 miles
Answer:
500 miles
Step-by-step explanation:
Let us use cross multiplication to find the unknown amount.
Given:
1) Cost for 1 mile=$0.20
2)Cost for x miles=$100
Solution:
No of miles Cost
1) 1 $0.20
2)x $100
By cross multiplying,
100 x 1= 0.20x
x=100/0.20
x=500 miles
Thank you!
Foram prescritos 500mg de dipirona para uma criança com febre.Na unidade tem disponivel ampola de 1g/2ml.Quantos g vão ser administrados no paciente
De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL
O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.
Fazendo a classica regra de 3, podemos chegar no volume desejado:
(atentar que 500mg = 0,5g)
g mL
1 --------- 2
0,5 --------- X
1 . X = 0,5 . 2
X = 1mLHELP ASAP ROCKY!!! will get branliest.
Answer:
work pictured and shown
Answer:
Last one
Step-by-step explanation:
● [ ( 3^2 × 5^0) / 4 ]^2
5^0 is 1 since any number that has a null power is equal to 1.
●[ (3^2 ×1 ) / 4 ]^2
● (9/4)^2
● 81 / 16
5x+4(-x-2)=-5x+2(x-1)+12
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
BRAINLIEST IF CORRECT!!! and 15 points solve for z -cz + 6z = tz + 83
Answer:
z = 83/( -c+6-t)
Step-by-step explanation:
-cz + 6z = tz + 83
Subtract tz from each side
-cz + 6z -tz= tz-tz + 83
-cz + 6z - tz = 83
Factor out z
z( -c+6-t) = 83
Divide each side by ( -c+6-t)
z( -c+6-t)/( -c+6-t) = 83/( -c+6-t)
z = 83/( -c+6-t)
A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of 10sin( t ) N(newtons) and moves in a medium that imparts a viscous force of 2 N
when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass.
A)Find the solution of the initial value problem in the above problem.
B)Plot the graph of the steady state solution
C)If the given external force is replaced by a force of 2 cos(ωt) of frequency ω , find the value of ω for which the amplitude of the forced response is maximum.
Answer:
A) C1 = 0.00187 m = 0.187 cm, C2 = 0.0062 m = 0.62 cm
B) A sample of how the graph looks like is attached below ( periodic sine wave )
C) w = [tex]\sqrt[4]{3}[/tex] is when the amplitude of the forced response is maximum
Step-by-step explanation:
Given data :
mass = 5kg
length of spring = 10 cm = 0.1 m
f(t) = 10sin(t) N
viscous force = 2 N
speed of mass = 4 cm/s = 0.04 m/s
initial velocity = 3 cm/s = 0.03 m/s
Formulating initial value problem
y = viscous force / speed = 2 N / 0.04 m/s = 50 N sec/m
spring constant = mg/ Length of spring = (5 * 9.8) / 0.1 = 490 N/m
f(t) = 10sin(t/2) N
using the initial conditions of u(0) = 0 m and u"(0) = 0.03 m/s to express the equation of motion
the equation of motion = 5u" + 50u' + 490u = 10sin(t/2)
A) finding the solution of the initial value
attached below is the solution and
B) attached is a periodic sine wave replica of how the grapgh of the steady state solution looks like
C attached below
A sports club was formed in the month of May last year. The function below, M(t), models the number of club members for the first 10 months, where t represents the number of months since the club was formed in May. m(t)=t^2-6t+28 What was the minimum number of members during the first 10 months the club was open? A. 19 B. 28 C. 25 D. 30
Answer:
A: 19
Step-by-step explanation:
For this, we can complete the square. We first look at the first 2 terms,
t^2 and -6t.
We know that [tex](t-3)^2[/tex] will include terms.
[tex](t-3)^2 = t^2 - 6t + 9[/tex]
But [tex](t-3)^2[/tex] will also add 9, so we can subtract 9. Putting this into the equation, we get:
[tex]m(t) = (t-3)^2 - 9 +28[/tex]
[tex]m(t) = (t-3)^2 +19[/tex]
Using the trivial inequality, which states that a square of a real number must be positive, we can say that in order to have the minimum number of members, we need to make (t-3) = 0. Luckily, 3 months is in our domain, which means that the minimum amount of members is 19.
BRAINLIST AND A THANK YOU AND 5 stars WILL BE REWARDED PLS ANSER
Answer:
The first picture's answer would be (6, 21)
Step-by-step explanation:
You have to find the points on the 8th and the 9th day, and then you would add them together, and then divide by two finding the average, which would be 24 and 18, so when added, you get 42, divided by 2 you get 21. You look on the graph for the point with 21, and you find it is on 6.
Average of 44.64, 43.45, 42.79, 42.28
Answer:
43.29Step-by-step explanation:
[tex]44.64+ 43.45+42.79+42.28\\\\= \frac{44.64+ 43.45+42.79+42.28}{4} \\\\\\= \frac{173.16}{4} \\\\= 43.29\\[/tex]
A museum curator is hanging 7 paintings in a row for an exhibit. There are 4 Renaissance paintings and 3 Baroque paintings. From left to right, all of the Renaissance paintings will be hung first, followed by all of the Baroque paintings. How many ways are there to hang the paintings
Answer:
144 ways
Step-by-step explanation:
Number of paintings = 7
Renaissance = 4
Baroque = 3
We are hanging from left to right and we will first hang Renaissance painting before baroque painting.
For Renaissance we have 4! Ways of doing so. 4 x3x2x1 = 24
For baroque we have 3! Ways of doing so. 3x2x1 = 6
We have 4!ways x 3!ways
= (4x3x2x1) * (3x2x1) ways
= 144 ways
Therefore we have 144 ways to hang the painting.
please help me in these question ????
A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.
(a) How many different samples of size 4 pens are possible?
(b) How many samples have 3 red pens and 1 black pen?
(c) How many samples of size 4 contain at least two red pens?
(d) How many samples of size 4 contain
If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal distribution.
1- What percentage of a cucumber give the crop amount between and 834 kg?
2- What the probability of cucumber give the crop exceed 900 kg ?
Answer:
Step-by-step explanation:
A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.
(a) How many different samples of size 4 pens are possible?
12C4=12!/(4!*8!)=495
(b) How many samples have 3 red pens and 1 black pen?
5C3*7C1
5C3=5!/(3!*2!)=10
7C1=7!/(1!*6!)=7
=>5C3*7C1=10*7=70
(c) How many samples of size 4 contain at least two red pens?
(5C2*7C2)+(5C3*7C1)+(5C4*7C0)
5C2=5!/(2!*3!)=10
7C2=7!/(2!*5!)=21
5C3=5!/(3!*2!)=10
7C1=7!/(1!*6!)=7
5C4=5!/(4!*1!)=5
7C0=7!/(0!*7!)=1
=>(5C2*7C2)+(5C3*7C1)+(5C4*7C0)=285
(d) How many samples of size 4 contain at most one black pen?
(7C1*5C3)+(7C0*5C4)
7C1=7!/(1!*6!)=7
7C0=7!/(0!*7!)=1
5C3=5!/(3!*2!)=10
5C4=5!/(4!*1!)=5
=>(7C1*5C3)+(7C0*5C4)=(7*10)+(1*5)=75
The cost, C, in United States Dollars ($), of cleaning up x percent of an oil spill along the Gulf Coast of the United States increases tremendously as x approaches 100. One equation for determining the cost (in millions $) is:
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]
The weight of a full steel bead tire is approximately 800 grams, while a lighter wheel weighs only 700 grams. What is the weight of each tire in pounds? There are 453.592 grams in one pound. Round answers to 2 decimal places. 800 grams = ______ pounds 700 grams = _____ pounds
Answer:
800= about 1.76 lbs
700= about 1.54 lbs
(there are about 453.5 grams in a pound
Step-by-step explanation:
Answer:
800 grams = 1.76 pounds
700 grams = 1.54 pounds
Step-by-step explanation:
i googled it
The size of a television is the length of the diagonal of its screen in inches. The aspect ratio of the screens of older televisions is 4:3, while the aspect ratio of newer wide-screen televisions is 16:9. Find the width and height of an older 35-inch television whose screen has an aspect ratio of 4:3.
Answer:
The Width = 28 inches
The Height = 21 inches
Step-by-step explanation:
We are told in the question that:
The width and height of an older 35-inch television whose screen has an aspect ratio of 4:3
Using Pythagoras Theorem
Width² + Height² = Diagonal²
Since we known that the size of a television is the length of the diagonal of its screen in inches.
Hence, for this new TV
Width² + Height² = 35²
We are given ratio: 4:3 as aspect ratio
Width = 4x
Height = 3x
(4x)² +(3x)² = 35²
= 16x² + 9x² = 35²
25x² = 1225
x² = 1225/25
x² = 49
x = √49
x = 7
Hence, for the 35 inch tv set
The Width = 4x
= 4 × 7
= 28 inches.
The Height = 3x
= 3 × 7
= 21 inches
Please answer this correctly without making mistakes
Answer:
1/8
Step-by-step explanation:
3/8-1/8-1/8=1/8
The quotient of 3 and the cube
of y+2
Answer:
[tex]\dfrac{3}{(y+2)^3}[/tex]
Step-by-step explanation:
Maybe you want this written using math symbols. It will be ...
[tex]\boxed{\dfrac{3}{(y+2)^3}}[/tex]
a
A solid metal cone of base radius a cm and height 2a cm is melted and solid
spheres of radius are made without wastage. How many such spheres can be
made?
volume of a cone
.
.
.
volume of sphere
.
.
number of spheres that can be made......
.
.
hence a hemisphere can be formed
Techwiz electronics makes a profit of $35 for each mp3 and $18 for each DVD last week techwiz sold a combined total of 118 mp3 and DVD players. Let x be the number of mp3 sold last week write an expression for the combined total profit (in dollars) made last week
Answer:
The total profit is [tex]p = 17x + 2124[/tex]
Step-by-step explanation:
From the question we are told that
The profit made on each mp3 is k = $35
The profit made on each mp3 is y = $18
The total amount sold is n = 118
Now given that the amount of mp3 sold is x then the amount of DVD sold is mathematically evaluated as
[tex]n - x[/tex]
Now the profit made on the x number of mp3 sold is
[tex]x * 35 = 3x[/tex]
And the the profit made from the n-x number of DVD sold is 18 (n-x ) = 18 - 18x
So the total profit made last week from the sales of both mp3 and DVD is
[tex]p = 35x + 18n - 18x[/tex]
[tex]p = 17x + 18(118)[/tex]
[tex]p = 17x + 2124[/tex]
Kenji earned the test scores below in English class.
79, 91, 93, 85, 86, and 88
What are the mean and median of his test scores?
Answer:
mean=87
median=87
Step-by-step explanation:
mean=sum of test score/number of subject
mean=79+91+93+85+86+88/6
mean=522/6
mean=87
Literal meaning of median is medium.
To find the number which lies in the medium, we must rearrange the number in ascending.
79, 91, 93, 85, 86, 88
79, 85, 86, 88, 91, 93
86+88/2=87
Hope this helps ;) ❤❤❤
Let me know if there is an error in my answer.
one third multiplied by the sum of a and b
Answer:
1/3(a+b)
hope it helps :>
The following shape is based only on squares, semicircles, and quarter circles. Find the area of the shaded part.
Answer:
this? hope it helps ........
Answer:
The answer is area=32pi-64 and the perimeter is 8pi
Step-by-step explanation:
Compute the flux of the vector field LaTeX: \vec{F}=F → =< y + z , x + z , x + y > though the unit cubed centered at origin.
Assuming the cube is closed, you can use the divergence theorem:
[tex]\displaystyle\iint_S\vec F\cdot\mathrm dS=\iiint_T\mathrm{div}\vec F\,\mathrm dV[/tex]
where [tex]S[/tex] is the surface of the cube and [tex]T[/tex] is the region bounded by [tex]S[/tex].
We have
[tex]\mathrm{div}\vec F=\dfrac{\partial(y+z)}{\partial x}+\dfrac{\partial(x+z)}{\partial y}+\dfrac{\partial(x+y)}{\partial z}=0[/tex]
so the flux is 0.
help pls:Find all the missing elements
Step-by-step explanation:
Using Sine Rule
[tex] \frac{ \sin(a) }{ |a| } = \frac{ \sin(b) }{ |b| } = \frac{ \sin(c) }{ |c| } [/tex]
[tex] \frac{ \sin(42) }{5} = \frac{ \sin(38) }{a} [/tex]
[tex]a = \frac{5( \sin(38))}{ \sin(42) } [/tex]
[tex]a = 4.6[/tex]
[tex] \frac{ \sin(42) }{5} = \frac{ \sin(100) }{b} [/tex]
[tex]b= \frac{5( \sin(100))}{ \sin(42) } [/tex]
[tex]b = 7.4[/tex]
602/100 into a decimal describe plz
Answer:
6.02
six point zero two
Step-by-step explanation:
Answer:
602 / 100= 6,02
Step-by-step explanation:
602 to divide 100 = 6,02
Find usubscript10 in the sequence -23, -18, -13, -8, -3, ...
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
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of ounces and a standard deviation of ounce. You randomly select cans and carefully measure the contents. The sample mean of the cans is ounces. Does the machine need to be reset? Explain your reasoning. ▼ Yes No , it is ▼ very unlikely likely that you would have randomly sampled cans with a mean equal to 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.
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.
Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6
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]
Find a cubic polynomial with integer coefficients that has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.
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]
The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H0: μ = 100 versus H1: μ ≠ 100 with a sample of n = 9 specimens.
A. If the acceptance region is defined as 98.5 le x- 101.5, find the type I error probability alpha.
B. Find beta for the case where the true mean heat evolved is 103.
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?
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.
