Answer:
m<A = 133 degrees
m<B = 17 degrees
m<C = 30 degrees
Step-by-step explanation:
In a triangle, all the angles add up to 180 degrees.
So, adding all the angles gets us,
39x + 24
This equals 180 degrees so,
39x + 24 = 180
Subtract 24 from both sides,
39x + 24 - 24 = 180 - 24
39x = 156
Divide both sides by 39
x = 4
Now we have x = 4, we use this to plug in each equation of the angles.
m<A = 40(4) - 27 = 160 - 27 = 133
m<B = 25 - 2(4) = 25 - 8 = 17
m<C = 26 + 4 = 30
A square and a rectangle have the same area. If the dimensions of the rectangle are 4 ft by 16 ft, how long is a side of the square?
Answer:
8
Step-by-step explanation:
4×16=64
[tex] \sqrt{64 } = 8[/tex]
The greater than symbols looks like this ____________, and the less than symbol looks like?
Answer:
The greater than symbols looks like this > , and the less than symbol looks like? <
Answer:
Greater than symbol: >
Less than symbol: <
Greater than or equal to symbol: ≥
Less than or equal to symbol: ≤
Equal symbol: =
In this case, you are answering with the greater than symbol as well as the less than symbol.
The greater than symbols looks like this > , and the less than symbol looks like < .
What is the equation of the parabola that has its vertex at (8,-1) and a y-intercept of (0,-17)?
y = a(x + 1.5)^2 - 12.5
y intercept is (0,-8) so:-
-8 = a(0+1.5)^2 - 12.5
-8 = 2.25a - 12.5
a = 4.5/ 2.25 = 2
so we have
y = 2 ( x +1.5)^2 - 12.5
solving for x when y = 0:-
(x + 1.5)^2 = 12.5/2 = 6.25
taking sqrt's x + 1.5 = +/- 2.5
x = -4, 1
so the x intercepts are (-4,0) and (1,0)
Answer:
y = –1∕4(x – 8)^2 – 1
Step-by-step explanation:
I took the exam and got it right.
What is the product of all values of b that satisfy (b-2)(b-8) +4= 0 ?
A) 10
B) 12
C) 16
D) 20
Answer:
D
Step-by-step explanation:
(b-2)(b-8)+4=0
b^2-10b+16+4=0
b^2-10b+20=0
Product of roots is given by c/a=20. So the answer is 20
arthur walks 5/8 mi to school jonathan rides a bus 8 times that far> How far does Jonathan ride to school
Answer:
Step-by-step explanation:
Distance walked by Arthur = 5/8 miles
Distance ride by Jonathan = 8 times that of Arthur
it means that Distance rode on bus by Jonathan is 8 multiplied by Distance walked by Arthur
Distance rode on bus by Jonathan = 8 * Distance walked by Arthur
Distance rode on bus by Jonathan = 8 * 5/8 = 5 Miles Answer
I will mark u brainleist if u help me and 5 stars and a thanks
Answer:
1. Jan checks the weather. It is 27 degrees outside. Jan did chores for two hours. After Jan was done, she checked the weather again. The temperature had decreased 11 degrees.
2. (See screen shot below.)
Step-by-step explanation:
1. It doesn't have to be as complicated as I made it. You can just say that the weather started out with 27 degrees, and decreased later on. Remember, decreased means subtracted and 27+(-11) is the same as 27-11 because when a + and - are together - always wins. So no.1 wants you to say something got subtracted.
2. On the number line, make a dot at 29 because it said it was 29 degrees. Then drag the dot at the number 29 to 13 because it said it decreased 16, so it is 19 minus 16 which is 13.
In the Cash Now lottery game there are 8 finalists who submitted entry tickets on time. From these 8 tickets, three grand prize winners will be drawn. The first prize is one million dollars, the second prize is one hundred thousand dollars, and the third prize is ten thousand dollars. Determine the total number of different ways in which the winners can be drawn. (Assume that the tickets are not replaced after they are drawn.)
Answer:
The number of ways is [tex]\left n} \atop {}} \right. P_r = 336[/tex]
Step-by-step explanation:
From the question we are told that
The number of tickets are [tex]n = 8[/tex]
The number of finalist are [tex]r =3[/tex]
Generally the number of way by which this winners can be drawn and arrange in the order of [tex]1^{st} , \ 2nd , \ 3rd[/tex] is mathematically represented as
[tex]\left n} \atop {}} \right. P_r = \frac{n\ !}{(n-r) !}[/tex]
substituting values
[tex]\left n} \atop {}} \right. P_r = \frac{ 8!}{(8-3) !}[/tex]
[tex]\left n} \atop {}} \right. P_r = \frac{ 8* 7*6*5*4*3*2*1}{ 5*4*3*2*1}[/tex]
[tex]\left n} \atop {}} \right. P_r = 336[/tex]
The deck for a card game contains 30 cards. 10 are red, 10 yellow, 5 blue, and 1 green, and 4 are wild cards. Each player is randomly dealt a five-card hand. a) What is the probability that a hand will contain exactly two wild cards? b) What is the probability that a hand will contain two wild cards, two red cards, and one blue cards?
Answer: a) 0.1095 b) 0.0095
Step-by-step explanation:
Given : The deck for a card game contains 30 cards.
10 are red, 10 yellow, 5 blue, and 1 green, and 4 are wild cards.
Each player is randomly dealt a five-card hand.
Number of ways to choose 5 cards out of 30 = [tex]C(30,5)=\dfrac{30!}{5!25!}=142506[/tex]
a) Cards other than wild card = 30-4=26
Number of ways to choose exactly two wild cards = [tex]C(26,3)\timesC(4,2)[/tex]
[tex]=\dfrac{26!}{3!23!}\times\dfrac{4!}{2!2!}\\\\=15600[/tex]
Probability that a hand will contain exactly two wild cards = [tex]\dfrac{15600}{142506}=0.1095[/tex]
b) Number of ways to choose two wild cards, two red cards, and one blue cards = [tex]C(4,2)\times C(10,2)\times C(5,1)[/tex]
[tex]=\dfrac{4!}{2!2!}\times\dfrac{10!}{2!8!}\times5=1350[/tex]
Probability that a hand will contain two wild cards, two red cards, and one blue cards = [tex]\dfrac{1350}{142506}=0.0095[/tex]
A baking scale measures mass to the tenth of a gram, up to 650 grams. A cup of flour is placed on the scale and results in a measure of 121.8 grams. Which of the following statements is not true?
a.The exact mass of the cup of flour must be between 121.7 and 121.9 grams.
b.The cup of flour has a mass of exactly 121.8 grams.
c.Given the limitations of the scale, the measurement has an appropriate level of accuracy.
d.To the nearest gram, the cup of flour has a mass of 122 grams.
Answer
Is it C I may have done my math wrong lol
Step-by-step explanation:
The amount of money spent on textbooks per year for students is approximately normal.
A. To estimate the population mean, 19 students are randomly selected the sample mean was $390 and the standard deviation was $120. Find a 95% confidence for the population meam.
B. If the confidence level in part a changed from 95% 1 to 1999%, would the margin of error for the confidence interval:
1. decrease.
2. stay the same.
3. increase not.
C. If the sample size in part a changed from 19% 10 to 22, would the margin of errot for the confidence interval:
1. decrease.
2. stay the same.
3. increase
D. To estimate the proportion of students who purchase their textbookslused, 500 students were sampled. 210 of these students purchased used textbooks. Find a 99% confidence interval for the proportion of students who purchase used text books.
Answer:
(A) A 95% confidence for the population mean is [$332.16, $447.84] .
(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.
(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.
(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .
Step-by-step explanation:
We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean = $390
s = sample standard deviation = $120
n = sample of students = 19
[tex]\mu[/tex] = population mean
Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.101 < [tex]t_1_8[/tex] < 2.101) = 0.95 {As the critical value of t at 18 degrees of
freedom are -2.101 & 2.101 with P = 2.5%}
P(-2.101 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.101) = 0.95
P( [tex]-2.101 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.101 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.101 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.101 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.101 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.101 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]\$390-2.101 \times {\frac{\$120}{\sqrt{19} } }[/tex] , [tex]\$390+2.101 \times {\frac{\$120}{\sqrt{19} } }[/tex] ]
= [$332.16, $447.84]
(A) Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .
(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is [tex]Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }[/tex] would increase because of an increase in the z value.
(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is [tex]Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }[/tex] would decrease because as denominator increases; the whole fraction decreases.
(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.
Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;
P.Q. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion students who purchase their used textbooks = [tex]\frac{210}{500}[/tex] = 0.42
n = sample of students = 500
p = population proportion
Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions
So, 99% confidence interval for the population proportion, p is ;
P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5%
level of significance are -2.58 & 2.58}
P(-2.58 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 2.58) = 0.99
P( [tex]-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99
P( [tex]\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99
99% confidence interval for p = [ [tex]\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ]
= [ [tex]0.42 -2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }[/tex] , [tex]0.42 +2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }[/tex] ]
= [0.363, 0.477]
Therefore, a 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .
The blueprints of a house have a scale factor of 30. If one side of the house measures 4 inches on the blueprint, how long is the actual side length (in feet)?
A. 7.5 feet
B.10 feet
C. 90 feet
D. 120 feet
If the scale factor is 30, then all you have to do is multiply each measurement by the scale factor. In this case, 4 · 30 = 120.
Given the graph, find an equation for the parabola.
Answer:
[tex]\Large \boxed{\sf \bf \ \ y=\dfrac{1}{16}(a-3)^2-2 \ \ }[/tex]
Step-by-step explanation:
Hello, please consider the following.
When the parabola equation is like
[tex]y=a(x-h)^2+k[/tex]
The vertex is the point (h,k) and the focus is the point (h, k+1/(4a))
As the vertex is (3,-2) we can say that h = 3 and k = -2.
We need to find a.
The focus is (3,2) so we can say.
[tex]2=-2+\dfrac{1}{4a}\\\\\text{*** We add 2. ***}\\\\\dfrac{1}{4a}=2+2=4\\\\\text{*** We multiply by 4a. ***}\\\\16a=1\\\\\text{*** We divide by 16. ***}\\\\a=\dfrac{1}{16}[/tex]
So an equation for the parabola is.
[tex]\large \boxed{\sf y=\dfrac{1}{16}(a-3)^2-2 }[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
Solve for
x
Round to the nearest tenth, if necessary.
9514 1404 393
Answer:
x = 5.0
Step-by-step explanation:
The tangent relation is helpful:
Tan = Opposite/Adjacent
tan(50°) = x/4.2
x = 4.2·tan(50°) ≈ 5.0054 . . . . multiply by 4.2
x ≈ 5.0
Show all work to solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer.
Square root x + 6 - 4 = x
Answer:
-2, (-5 is extraneous)
Step-by-step explanation:
assuming √(x+6) -4 =x
√(x+6) = x+4
square both side
x+6 = (x+4)^2
x+6 = x^2+8x+16
x^2+7x+10=0
(x+2)(x+5) = 0
x = -2, -5
put -5 back into org equation and it is not true
√(-5+6) -4 ≠ -5
a sample from a mummified bull was taken from a certain place. The sample shows that 71% of the carbon-14 still remains. how old is the sample
Answer:
Step-by-step explanation:
Decay of carbon - 14 is exponential in nature . It decays as follows .
[tex]N=N_0e^{-\lambda t}[/tex] λ is called decay constant .
λ = .693 / T where T is half life .
Half life of carbon-14 is 5700 years
λ = .693 / T
= .693 / 5700
= 12.158 x 10⁻⁵ year⁻¹
[tex]N=N_0e^{-\lambda t}[/tex]
N = .71 N₀
[tex].71 N_0 =N_0e^{-\lambda t}[/tex]
[tex].71 =e^{-\lambda t}[/tex]
Taking ln on both sides
ln .71 = - λ t
ln .71 = - 12.158 x 10⁻⁵ t
-0.3425 = - 12.158 x 10⁻⁵ t
t = .3425 / 12.158 x 10⁻⁵
2817 years .
Jamal has two investments, one in Company A, and another in Company B. Jamal purchased 300 shares in Company A at $1.45 per share. Since purchasing the shares, the price per share increased to $1.65 per share, after which Jamal decided to sell, realizing a profit. At the same time, Jamal purchased 200 shares in Company B at $1.20 per share. Since purchasing the shares, the share price fell to $1.10 per share, after which Jamal decided to sell the shares, suffering a loss. Calculate the total profit that Jamal received from his two investments.
Answer:
$20
Step-by-step explanation:
Company A:
Buy 300 shares at $1.45 per share.
Sell 300 shares at $1.65 per share.
Profit: ($1.65 - $1.45) * 300 = $60
Company B:
Buy 200 shares at $1.20 per share.
Sell 200 shares at $1.10 per share.
Loss: ($1.20 - $1.10) * 200 = $20
Net profit:
$40 - $20 = $20
Answer:
Step-by-step explanation:
Profit on Company A =$(1.65−1.45)×300=$60.
Loss on Company B =$(1.20−1.10)×200=$20.
Therefore the total profit Jamal achieved was $60−$20=$40.
In order to earn an A in her math course,
Bernadette must have an average of at
least 90 on her exam scores. She has
grades of 83, 97, 89, and 82 on her first 4
exams. What is the minimum she can
score on the final exam to earn an A in the
course?
Step-by-step explanation:
Let minimum score on the final exam to earn an A be X
[tex]mean \: = \frac{sum \: of \: observation}{number \: of \: observation} [/tex]
[tex]90 = \frac{83 + 97 + 89 + 82 + x}{5} [/tex]
Further solving :
X = 99 marks
how do i solve this, i’m really confused
Answer:
x = 10
Step-by-step explanation:
[tex]x = 3 \times 2 + 4[/tex]
[tex]x = 6 + 4[/tex]
[tex]x = 10[/tex]
The product of a number and 3 is equal to 15 minutes twice the number, find the number.
Answer:
The answer is 3Step-by-step explanation:
Let the number to be found be x
The product of a number and 3 is written as
3 × x = 3x15 minus twice the number is written as
15 - 2xNow equate the two statements
That's
3x = 15 - 2x
Group like terms
3x + 2x = 15
5x = 15
Divide both sides by 5
the final answer is
x = 3Hope this helps you
How many solutions does this system of equations have?
y = -1/3x+7
y = -2x3 + 5x2 + x - 2
O A.
no solution
OB
1 solution
O C.
2 solutions
OD
3 solutions
Answer:
1 solution
Step-by-step explanation:
the two equations only cross once
The system of equations may have either 2 or 1 solution, but we cannot determine the exact number of solutions without further information or analysis.
We have,
The given system of equations consists of two equations in two variables:
y = (-1/3)x + 7 _______(1)
y = -2x³ + 5x² + x - 2 _______(2)
To find the solutions of the system, we need to find the values of x and y that satisfy both equations simultaneously.
Since both equations are in the form y = ...,
we can set the expressions on the right-hand sides of the equations equal to each other:
(-1/3)x + 7 = -2x³ + 5x² + x - 2
Rearranging and simplifying, we get:
2x³ - 5x² - (2/3)x + 9 = 0
To solve for x, we can use factoring or the rational root theorem.
However, the polynomial on the left-hand side of the equation does not appear to have any rational roots or easily factorable forms.
Therefore, to determine the number of solutions,
we can use Descartes' rule of signs to find the possible number of positive and negative roots of the polynomial:
The number of positive roots is either equal to the number of sign changes in the coefficients or less than it by an even integer.
There are two sign changes in the coefficients, so there can be either 2 or 0 positive roots.
The number of negative roots is either equal to the number of sign changes in the coefficients or less than it by an even integer.
There is one sign change in the coefficients, so there can be either 1 or 0 negative roots.
Since the sum of the possible number of positive and negative roots is either 2 or 1, the total number of solutions is either 2 or 1.
Therefore,
The system of equations may have either 2 or 1 solution, but we cannot determine the exact number of solutions without further information or analysis.
Learn more about solutions of equations here:
https://brainly.com/question/545403
#SPJ7
find the least number that can be divided 9and 12 without leaving remainder
Answer:
By finding LCM of 9 and 12 the answer is 36
Step-by-step explanation:
If you like my answer than please mark me brainliest thanks
Suppose your weekly local lottery has a winning chance of 1/106. You buy lottery from them for x weeks in a row. What is the probability that you never win?
Answer:
The probability mass function that you never win [tex]^xC_o[/tex] = [tex](\dfrac{999999}{1000000})^x[/tex]
Step-by-step explanation:
Given that;
the winning chance of a weekly local lottery = [tex]\dfrac{1}{10^6}[/tex]
= [tex]\dfrac{1}{1000000}[/tex]
The probability of losing = 1 - probability of winning (winning chance)
The probability of losing = [tex]1- \dfrac{1}{1000000}[/tex]
The probability of losing =[tex]\dfrac{999999}{1000000}[/tex]
The probability mass function that you never win [tex]^xC_o[/tex] = [tex](\dfrac{1}{10^6} )^0 ( \dfrac{999999}{1000000})^x[/tex]
The probability mass function that you never win [tex]^xC_o[/tex] = [tex](\dfrac{999999}{1000000})^x[/tex]
A model for the average price of a pound of white sugar in a certain country from August 1993 to August 2003 is given by the function
S(t) = −0.00003237t5 + 0.0009037t4 − 0.008956t3 + 0.03629t2 − 0.04547t + 0.4778
where t is measured in years since August of 1993. Estimate the times when sugar was cheapest and most expensive during the period 1993-2003. (Round your answers to three decimal places.)
t= __________________________ (cheapest)
t=__________________________ (most expensive)
Answer:
[tex]t = 0.811\,s[/tex] contains the cheapest reference to sugar; [tex]t = 4.511\,s[/tex] contains the most expensive reference to sugar.
Step-by-step explanation:
Let be [tex]s(t) = -0.00003237\cdot t^{5} + 0.0009037\cdot t^{4}-0.008956\cdot t^{3}+0.03629\cdot t^{2}-0.04547\cdot t + 0.4778[/tex], the times when sugar is the cheapest and the most expensive (absolute minimum and maximum) are determined with the help of first and second derivatives of this function (First and Second Derivative Tests):
First Derivative Test
[tex]s'(t) = -0.00016185\cdot t^{4}+0.0036148\cdot t^{3}-0.026868\cdot t^{2}+0.07258\cdot t - 0.04547[/tex]
Let equalize the polynomial to zero and solve the resulting expression:
[tex]-0.00016185\cdot t^{4}+0.0036148\cdot t^{3}-0.026868\cdot t^{2}+0.07258\cdot t - 0.04547 = 0[/tex]
[tex]t_{1} \approx 9.511\,s[/tex], [tex]t_{2}\approx 7.431\,s[/tex], [tex]t_{3}\approx 4.511\,s[/tex] and [tex]t_{4}\approx 0.881\,s[/tex]
Second Derivative Test
[tex]s''(t) = -0.0006474\cdot t^{3}+0.0108444\cdot t^{2}-0.053736\cdot t+0.07258[/tex]
This function is now evaluated at each root found in the First Derivative section:
[tex]s''(9.511\,s) = -0.0006474\cdot (9.511\,s)^{3}+0.0108444\cdot (9.511\,s)^{2}-0.053736\cdot (9.511\,s)+0.07258[/tex]
[tex]s''(9.511\,s) = -0.015[/tex] (A maximum)
[tex]s''(7.431\,s) = -0.0006474\cdot (7.431\,s)^{3}+0.0108444\cdot (7.431\,s)^{2}-0.053736\cdot (7.431\,s)+0.07258[/tex]
[tex]s''(7.431\,s) = 6.440\times 10^{-3}[/tex] (A minimum)
[tex]s''(4.511\,s) = -0.0006474\cdot (4.511\,s)^{3}+0.0108444\cdot (4.511\,s)^{2}-0.053736\cdot (4.511\,s)+0.07258[/tex]
[tex]s''(4.511\,s) = -8.577\times 10^{-3}[/tex] (A maximum)
[tex]s''(0.811\,s) = -0.0006474\cdot (0.811\,s)^{3}+0.0108444\cdot (0.811\,s)^{2}-0.053736\cdot (0.811\,s)+0.07258[/tex]
[tex]s''(0.811\,s) = 0.036[/tex] (A minimum)
Each value is evaluated in order to determine when sugar was the cheapest and the most expensive:
Cheapest (Absolute minimum)
[tex]s(0.811\,s) = -0.00003237\cdot (0.811\,s)^{5}+0.0009037\cdot (0.811\,s)^{4}-0.008956\cdot (0.811\,s)^{3}+0.03629\cdot (0.811\,s)^{2}-0.04547\cdot (0.811\,s)+0.4778[/tex]
[tex]s(0.811\,s) = 0.460[/tex]
[tex]s(7.431\,s) = -0.00003237\cdot (7.431\,s)^{5}+0.0009037\cdot (7.431\,s)^{4}-0.008956\cdot (7.431\,s)^{3}+0.03629\cdot (7.431\,s)^{2}-0.04547\cdot (7.431\,s)+0.4778[/tex]
[tex]s(7.431\,s) = 0.491[/tex]
[tex]t = 0.811\,s[/tex] contains the cheapest reference to sugar.
Most expensive (Absolute maximum)
[tex]s(4.511\,s) = -0.00003237\cdot (4.511\,s)^{5}+0.0009037\cdot (4.511\,s)^{4}-0.008956\cdot (4.511\,s)^{3}+0.03629\cdot (4.511\,s)^{2}-0.04547\cdot (4.511\,s)+0.4778[/tex]
[tex]s(4.511\,s) = 0.503[/tex]
[tex]s(9.511\,s) = -0.00003237\cdot (9.511\,s)^{5}+0.0009037\cdot (9.511\,s)^{4}-0.008956\cdot (9.511\,s)^{3}+0.03629\cdot (9.511\,s)^{2}-0.04547\cdot (9.511\,s)+0.4778[/tex]
[tex]s(9.511\,s) = 0.498[/tex]
[tex]t = 4.511\,s[/tex] contains the most expensive reference to sugar.
The required values are,
[tex]t=0.881199[/tex] at the cheapest.
[tex]t=4.51081[/tex] at the most expensive.
Minimum or Maximum:A high point is called a maximum (plural maxima ). A low point is called a minimum (plural minima ).
Given equation is,
[tex]S(t) = -0.00003237t^5 + 0.0009037t^4- 0.008956t^3 + 0.03629t^2-0.04547t + 0.4778[/tex]
Differentiating the given equation we get,
[tex]S'(t)=-0.00003237\times 5t^4+0.0009037\times 4t^3-0.008956\times 3t^2+0.03629\times 2t-0.04547+0\\S'(t)=0\\-0.00003237\times 5t^4+0.0009037\times 4t^3-0.008956\times 3t^2+0.03629\times 2t-0.04547+0=0\\t=0.881199\\t=4.51081\\t=7.43087\\t=9.51137\\[/tex]
Now we can directly plug those fours values of t into given function S(t) to find which one gives max or minimum or you can also use the 2nd derivative test. Although that is not compulsory
[tex]t=0.881199,S(t)=0.46031095\\t=4.51081, S(t)=0.50278423\\t=7.43087, S(t)=0.49096762\\t=9.51137, S(t)=0.49832202\\[/tex]
We see that sugar is cheapest at [tex]t=0.881199[/tex] which is approx 1 and corresponds to the year [tex]1993+1=1994[/tex]
Similarly sugar is most expensive at [tex]t=4.51081[/tex] which is approx 5 and corresponds to year [tex]1993+5=1998[/tex]
Learn more about the topic Minimum or Maximum:
https://brainly.com/question/10359210
5
Select the correct answer.
What is this expression in simplified form?
5/2 . 9/6
Answer:
Bro 1st expression is in simplest form and 2nd in simplest is 3/2
Step-by-step explanation:
If you like my answer than please mark me brainliest
Answer:
(5/2)*(9/6)
three can go into the fraction(9/6) giving(3/2)
(5/2)*(3/2)=15/4
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PLEASE HELP 30 POINTS
How long will it take in hours for a car traveling from Tucson to Phoenix (120 km)
to reach Phoenix at a rate of 10km/hr.? How long would it take that car to circle the Earth
at the equator? (c= 2 nr) rof earth is 6,378 km.
Answer:
1. It would take the car to get from Tucson to Phoenix 12 hours.
2. for the car to go around the equator it would take 637 hours if it is still travelling at 10km/hr.
hope this helps
Step-by-step explanation:
1. 120 km divided by 10 = 12 hours
graph 3x-y-2=0 using the x- and y-intercepts
Step-by-step explanation:
I used an app called DESMOS It Is usually super helpful!!!
What is the rate of change from x = 0 to x = pi over 2 ? (6 points) trig graph with points at (0, -4) and (pi over 2, 0) and (pi, 4) and (3 pi over 2, 0) and (2 pi, -4)
Answer: [tex]\dfrac{8}{\pi}[/tex] .
Step-by-step explanation:
We know that the rate of change of function f(x) from x=a to x= b is given by :-
[tex]k=\dfrac{f(b)-f(a)}{b-a}[/tex]
The given points on graph : (0, -4) and (pi over 2, 0) and (pi, 4) and (3 pi over 2, 0) and (2 pi, -4).
The rate of change from x = 0 to x = pi over 2 will be :-
[tex]\dfrac{0-(-4)}{\dfrac{\pi}{2}-0}=\dfrac{4}{\dfrac{\pi}{2}}[/tex] [By using points (0, -4) and (pi over 2, 0) ]
[tex]=\dfrac{8}{\pi}[/tex]
Hence, the rate of change from x = 0 to x = pi over 2 is [tex]\dfrac{8}{\pi}[/tex] .
which expression is equivalent to x^-5/3
Answer:
B
Step-by-step explanation:
Since the power is negative, you automatically know it has to be a or b, because the only way it would be negative is if it was brought from the denominator to the numerator.
The answer is B, because the numerator of the power, is what is inside the square root, while the denominator is what is outside the square root.
A market researcher believes that brand perception of one of the company's products may vary between different groups. After interviewing 307 persons, the following data was compiled. Can we conclude that brand perception is dependent on age?
Age Favorable Unfavorable Neutral Total
18-30 67 24 20 111
30-45 50 14 16 80
Over 45 69 41 26 116
Total 186 59 62 307
Find the value of the test statistic.
Answer:
The value for the Chi -square test statistics = 1.149
Step-by-step explanation:
The observed value Table can be shown better as:
Observed Value
Age Favorable Unfavorable Neutral Total
18-30 67 24 20 111
30-45 50 14 16 80
Over 45 69 21 26 116
Total 186 59 62 307
NOTE: when computing the question, in the third row and the second column, there is a mistake , the value is supposed to be 21 and not 41 because :
69 +21+ 26 will eventually give = 116
69 + 41 + 26 = 136
With that error being fix , let's get started.
Expected Value
The expected value can be determined by using the formula:
[tex]Expected \ Value = \dfrac{ row \ total \times column \ total }{grand \ total }[/tex]
For 67; (111 * 186)/307 = 67.251
For 24 : (111 * 59)/307 = 21.332
For 20 : (111 * 62)/307 = 22.417
For 50 :(80*186)/307 = 48.469
For 14 : (80* 59)/307 = 15.375
For 16 : ( 0 * 62)/307 = 16.156
For 69 : (116 * 186)/307 = 70.280
For 21 : (116* 59)/307 = 22.293
For 26 : (116*62)/307 = 23.427
Expected Value :
Age Favorable Unfavorable Neutral Total
18-30 67.251 21.332 22.417 111
30-45 48.469 15.375 16.156 80
Over 45 70.280 22.293 23.427 116
Total 186 59 62 307
The Chi - square test statistics = [tex]\dfrac{(observed \ value - Expected \ value)^2}{Expected \ value}[/tex]
For 67.251 : ( 67 - 67.251)²/67.251 = 0.0009
For 21.332 : ( 24 - 21.332)²/21.332 = 0.3337
For 22.417 : ( 20 - 22.417)²/ 22.417 = 0.2606
For 48.469 : ( 50 - 48.469)²/ 48.469 = 0.0484
For 15.375 : ( 14 - 15.375)²/ 15.375 = 0.1230
For 16.156 : ( 16 - 16.156)²/ 16.156 = 0.0015
For 70.280 : ( 69 - 70.280)²/ 70.280 = 0.0233
For 22.293 : ( 21 - 22.293)²/ 22.293 = 0.0750
For 23.427 : ( 26 - 23.427)²/ 23.427 = 0.2826
The chi square table is as follows:
Age Favorable Unfavorable Neutral Total
18-30 0.0009 0.3337 0.2606 0.5952
30-45 0.0484 0.1230 0.0015 0.1729
Over 45 0.0233 0.0750 0.2826 0.3809
Total 0.0726 0.5317 0.5447 1.149
The value for the Chi -square test statistics = 1.149
radical 16 * redical 12
[tex]\sqrt{16}\times\sqrt{12}[/tex]
$=\sqrt{4^2}\times\sqrt{2^2\cdot3}$
$=4\times2\sqrt3=8\sqrt3$