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.
Explanation: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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The system of equation, if a b are arbitrary numbers x+2y-3z- a 2x+4y-6z 2a+2 has (A) No solutions regardless of values of a and b (B) Infinitely many solutions regardless of values of a and b (C) a unique solution if a b-0 D) a unique solution regardless of values of a and b
Answer:
(A) No solutions regardless of values of a and b.
Step-by-step explanation:
Asumming that the system of equations is [tex]x+2y-3z=a\\ 2x+4y-6z=2a+2[/tex], the corresponding augmented matrix of the system is [tex]\left[\begin{array}{cccc}1&2&-3&a\\2&4&-6&2a+2\end{array}\right][/tex].
If two time the row 1 is subtracted to row 1, we get the following matrix
[tex]\left[\begin{array}{cccc}1&2&-3&a\\0&0&0&2a+2-2a\end{array}\right][/tex].
Then the system has no solutions regardless of values of a and b.
Consider a fair coin which when tossed results in either heads (H) or tails (T). If the coin is tossed TWO times 1. List all possible outcomes. (Order matters here. So, HT and TH are not the same outcome.) 2. Write the sample space. 3. List ALL possible events and compute the probability of each event, assuming that the probability of each possible outcome from part (a) is equal. (Keep in mind that there should be many more events than outcomes and not all events will have the same probability.)
Answer:
Sample space = {(T,T), (T,H), (HT), (HH)}
Step-by-step explanation:
We are given a fair coin which when tossed one times either gives heads(H) or tails(T).
Now, the same coin is tossed two times.
1) All the possible outcomes
Tails followed by tails
Rails followed by heads
Heads followed by a tail
Heads followed by heads
2) Sample space
{(T,T), (T,H), (HT), (HH)}
3) Formula:
[tex]Probability = \displaystyle\frac{\text{Favourable outcome}}{\text{Total number of outcome}}[/tex]
Using the above formula, we can compute the following probabilities.
Probability((T,T)) =[tex]\frac{1}{4}[/tex]
Probability((T,H)) =[tex]\frac{1}{4}[/tex]
Probability((H,T)) =[tex]\frac{1}{4}[/tex]
Probability((H, H)) =[tex]\frac{1}{4}[/tex]
Probability(Atleast one tails) = [tex]\frac{3}{4}[/tex]
Probability(Atleast one heads) = [tex]\frac{3}{4}[/tex]
Probability(Exactly one tails) = [tex]\frac{2}{4}[/tex]
Probability(Exactly one heads) = [tex]\frac{2}{4}[/tex]
Solve each of the following systems by Gauss-Jordan elimination. (b) X1-2x2+ x3- 4x4=1 X1+3x2 + 7x3 + 2x4=2 -12x2-11x3- 16x4 5 (a) 5x1+2x2 +6x3= 0 -2x1 +x2+3x3 = 0
Answer:
a) The set of solutions is [tex]\{(0,-3x_3,x_3): x_3\; \text{es un real}\}[/tex] y b) the set of solutions is [tex]\{(-6,\frac{-41}{17}-\frac{30}{17}x_4 , \frac{37}{17}+\frac{8}{17} x_4 ,x_4): x_4\;\text{es un real}\}[/tex].
Step-by-step explanation:
a) Let's first find the echelon form of the matrix [tex]\left[\begin{array}{ccc}5&2&6\\-2&1&3\end{array}\right][/tex].
We add [tex]\frac{2}{5}[/tex] from row 1 to row 2 and we obtain the matrix [tex]\left[\begin{array}{ccc}5&2&6\\0&\frac{9}{5} &\frac{27}{5}\end{array}\right][/tex]From the previous matrix, we multiply row 1 by [tex]\frac{1}{5}[/tex] and the row 2 by [tex]\frac{5}{9}[/tex] and we obtain the matrix [tex]\left[\begin{array}{ccc}1&\frac{2}{5} &\frac{6}{5} \\0&1&3\end{array}\right][/tex]. This matrix is the echelon form of the initial matrix.The system has a free variable (x3).
x2+3x3=0, then x2=-3x3 0=x1+[tex]\frac{2}{5}[/tex]x2+[tex]\frac{6}{5}[/tex]x3=x1+[tex]\frac{2}{5}[/tex](-3x3)+[tex]\frac{6}{5}[/tex]x3=
x1-[tex]\frac{6}{5}[/tex]x3+[tex]\frac{6}{5}[/tex]x3
then x1=0.
The system has infinite solutions of the form (x1,x2,x3)=(0,-3x3,x3), where x3 is a real number.
b) Let's first find the echelon form of the aumented matrix [tex]\left[\begin{array}{ccccc}1&-2&1&-4&1\\1&3&7&2&2\\0&-12&-11&-16&5\end{array}\right][/tex].
To row 2 we subtract row 1 and we obtain the matrix [tex]\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&-12&-11&-16&5\end{array}\right][/tex]From the previous matrix, we add to row 3, [tex]\frac{12}{5}[/tex] of row 2 and we obtain the matrix [tex]\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&0&\frac{17}{5}&\frac{-8}{5}&\frac{37}{5} \end{array}\right][/tex].From the previous matrix, we multiply row 2 by [tex]\frac{1}{5}[/tex] and the row 3 by [tex]\frac{5}{17}[/tex] and we obtain the matrix [tex]\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&1&\frac{6}{5} &\frac{6}{5}&\frac{1}{5}\\0&0&1&\frac{-8}{17}&\frac{37}{17} \end{array}\right][/tex]. This matrix is the echelon form of the initial matrix.The system has a free variable (x4).
x3-[tex]\frac{8}{17}[/tex]x4=[tex]\frac{37}{17}[/tex], then x3=[tex]\frac{37}{17}[/tex]+ [tex]\frac{8}{17}x4.x2+[tex]\frac{6}{5}[/tex]x3+[tex]\frac{6}{5}[/tex]x4=[tex]\frac{1}{5}[/tex], x2+[tex]\frac{6}{5}[/tex]([tex]\frac{37}{17}[/tex]+[tex]\frac{8}{17}x4)+[tex]\frac{6}{5}[/tex]x4=[tex]\frac{1}{5}[/tex], thenx2=[tex]\frac{-41}{17}-\frac{30}{17}[/tex]x4.
x1-2x2+x3-4x4=1, x1+[tex]\frac{82}{17}[/tex]+[tex]\frac{60}{17}[/tex]x4+[tex]\frac{37}{17}[/tex]+[tex]\frac{8}{17}[/tex]x4-4x4=1, then x1=[tex]1-\frac{119}{17}=-6[/tex]The system has infinite solutions of the form (x1,x2,x3,x4)=(-6,[tex]\frac{-41}{17}-\frac{30}{17}[/tex]x4,[tex]\frac{37}{17}[/tex]+ [tex]\frac{8}{17}[/tex]x4,x4), where x4 is a real number.
Ben earns $9 per hour and $6 for each delivery he makes.He wants to earn more than $155 in an 8 hour work day.What is the least number of deliveries he must make to reach his goal?
Answer:
Ben must make at least 14 deliveries to reach his goal.
Step-by-step explanation:
The problem states that Ben earns $9 per hour and $6 for each delivery he makes. So his daily earnings can be modeled by the following function.
[tex]E(h,d) = 9h + 6d[/tex],
in which h is the number of hours he works and d is the number of deliveries he makes.
He wants to earn more than $155 in an 8 hour work day.What is the least number of deliveries he must make to reach his goal?
This question asks what is the value of d, when E = $156 and h = 8. So:
[tex]E(h,d) = 9h + 6d[/tex]
[tex]156 = 9*8 + 6d[/tex]
[tex]156 = 72 + 6d[/tex]
[tex]6d = 84[/tex]
[tex]d = \frac{84}{6}[/tex]
d = 14
Ben must make at least 14 deliveries to reach his goal.
In a NiCd battery, a fully charged cell is composed of Nickelic Hydroxide. Nickel is an element that has multiple oxidation states that is usually found in the following states
Nickel Charge Proportions found
0 0.17
+2 0.35
+3 0.33
+4 0.15
(a) what is the probablity that a cell has at least one of the positive nickel-charged options?
(b) what is the probability that a cell is not composed of a positive nickel charge greater than +3?
Answer:
P(cell has at least one of the positive nickel-charged options) = 0.83.
P(a cell is not composed of a positive nickel charge greater than +3) = 0.85.
Step-by-step explanation:
It is given that the Nickel Charge Proportions found in the battery are:
0 ==> 0.17
.
+2 ==> 0.35
.
+3 ==> 0.33
.
+4 ==> 0.15.
The numbers associated to the charge are actually the probabilities of the charges because nickel is an element that has multiple oxidation states that is usually found in the above mentioned states.
a) P(cell has at least one of the positive nickel-charged options) = P(a cell has +2 nickel-charged options) + P(a cell has +3 nickel-charged options) + P(a cell has +4 nickel-charged options) = 0.35 + 0.33 + 0.15 = 0.83.
Or:
P(a cell has at least one of the positive nickel-charged options) = 1 - P(a cell has 0 nickel-charged options) = 1 - 0.17 = 0.83.
b) P(a cell is not composed of a positive nickel charge greater than +3) = 1 - P(a cell is composed of a positive nickel charge greater than +3)
= 1 - P(a cell has +4 nickel-charged options) '.' because +4 is only positive nickel charge greater than +3
= 1 - 0.15
= 0.85
To summarize:
P(cell has at least one of the positive nickel-charged options) = 0.83!!!
P(a cell is not composed of a positive nickel charge greater than +3) = 0.85!!!
The probability of a cell having at least one positive nickel-charge is 0.83 and the probability that a cell is not composed of a positive nickel charge greater than +3 is 0.85; This was calculated based on probabilities of Nickel in different charge states.
Explanation:For this problem, you're basically being asked to interpret a probability distribution of Nickel charge proportions, which involves summing probabilities.
(a) The probability that a cell has at least one of the positive nickel-charged options is the sum of the probabilities of Nickel in the +2, +3, and +4 states. From the given data, we simply add: 0.35 (for +2), 0.33 (for +3), and 0.15 (for +4). So, the total probability is 0.83.
(b) The probability that a cell is not composed of a positive nickel charge greater than +3 means we're looking for the probability of Nickel in the 0, +2, and +3 states. Here, we add: 0.17 (for 0 state), 0.35 (for +2 state), and 0.33 (for +3 state) to get a total probability of 0.85.
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You are to give an injection of a drug. The dosage is 0.4 mg per kilogram of bod The concentration of the drug in vial is listed as 500 ug/ml. The patient's chart Hists weight as 168 pounds. How many milliliters (= cc) are you to inject? Patient's weight Concentration of drug Show calculations: mg/ml
Answer:
You inject 60.9628 milliliters of dosage
Step-by-step explanation:
1 pound = 0.453592kg,
Patient's weight in pounds = 168
Patient's weight in kg = [tex]76.2035 kg[/tex]
Now we are given that The dosage is 0.4 mg per kilogram of bod
So, dosage = [tex]0.4 \times 76.2035 mg = 30.4814 mg[/tex]
1 microgram = 0.001 mg
Concentration of drug = [tex]500 micrograms/ml = 500 * 0.001 mg/ml = 0.5 mg/ml[/tex]
Now we are supposed to find How many milliliters (= cc) are you to inject?
So,milliliters of dosage required to inject = [tex]\frac{30.4814}{0.5} = 60.9628[/tex]
Hence you inject 60.9628 milliliters of dosage
A minor league baseball team plays 128 games a seanson. If the tam won 16 more than three times as many games as they lost how many wins and losses did the team have.
The baseball team won 100 games and lost 28 games. We found the number of losses by solving the equation formed by the relationship between wins and losses, and the total number of games played.
To solve the problem, let's denote the number of games the baseball team lost as L, and hence, the games they won would be 3L + 16 as per the condition given. Considering that the team played a total of 128 games, the equation representing the total number of games played is:
L + (3L + 16) = 128
Combining like terms, we get:
4L + 16 = 128
Subtracting 16 from both sides, we have:
4L = 112
Dividing both sides by 4 yields:
L = 28
Now that we have the number of losses, we can calculate the number of wins by substituting L back into 3L + 16:
Wins = 3(28) + 16 = 84 + 16 = 100
Therefore, the team won 100 games and lost 28 games in the season.
Find q, r in \mathbb{Z} so that 105 = 11q + r
with 0 \leq r < 11 as in the division algorithm
Answer:
[tex]q=9\,,\,r=6[/tex]
Step-by-step explanation:
Division Algorithm :
As per division algorithm , for numbers a and b , there exist numbers q and r such that [tex]a=bq+r\,\,,0\leq r< b[/tex]
Here ,
a = Dividend
b = Divisor
q = quotient
r = remainder
Given : 105 = 11q + r such that [tex]0 \leq r < 11[/tex]
Here, clearly a = 105 , b = 11
To find : q and r
Solution : On dividing 105 by 11 , we get [tex]105=11\times 9+6[/tex]
On comparing [tex]105=11\times 9+6[/tex] with [tex]a=bq+r\,\,,0\leq r< b[/tex] , we get [tex]q=9\,,\,r=6[/tex]
Find the greatest common divisor of 252 and 60
Answer:
12
Step-by-step explanation:
The greatest common divisor(gcd) is also known by the name highest common factor(hcf), greatest common factor(gcf).
Greatest common factor of two number can be defined as the highest integer that divides both the number.
We have to find greatest common divisor of 252 and 60.
The prime factorization of 252 is:
252 = 2×2×3×3×7
The prime factorization of 60 is:.
60 = 2×2×3×5
Common factors are: 2×2×3
Hence, greatest common divisor of 252 and 60 = 2×2×3 = 12
A group of friends goes out for single-scoop ice-cream cones. There are sugar cones, cake cones and waffle cones. But there are only five flavors of ice-cream left (peppermint, horehound, chocolate malt, gingerbread, and squirrel). How many cone/ice cream combinations can be ordered?
Answer: 15
Step-by-step explanation:
Given : The number of kinds of ice-cream cones ( sugar cones, cake cones and waffle cones)=3
The number of flavors of ice-creams =5
By using the fundamental principle of counting , we have
The number of possible cone/ice cream combinations can be ordered will be :-
[tex]5\times3=15[/tex]
Hence, the number of possible cone/ice cream combinations can be ordered =15
What is the answer to (n+4) +7 =
(n+4) +7 remove the parenthesis
n+4+7 add the same number answer is n +11
Find the lengths of the sides of the triangle PQR. P(2, −3, −4), Q(8, 0, 2), R(11, −6, −4) |PQ| = Incorrect: Your answer is incorrect. |QR| = |RP| = Is it a right triangle? Yes No Is it an isosceles triangle? Yes No
Answer:
the length PQ is 9 units,the length QR is 9 units,the length PR is 9.48 units,the triangle is not a right triangle,this is a isosceles triangle
Step-by-step explanation:
Hello, I think I can help you with this
If you know two points, the distance between then its given by:
[tex]P1(x_{1},y_{1},z_{1} ) \\P2(x_{2},y_{2},z_{2})\\\\d=\sqrt{(x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}+(z_{2}-z_{1} )^{2} }[/tex]
Step 1
use the formula to find the length PQ
Let
P1=P=P(2, −3, −4)
P2=Q=Q(8, 0, 2)
[tex]d=\sqrt{(8-2)^{2} +(0-(-3))^{2}+(2-(-4))^{2}} \\ d=\sqrt{(6)^{2} +(3)^{2}+(6 )^{2}}} \\d=\sqrt{36+9+36}\\d=\sqrt{81} \\d=9\\[/tex]
the length PQ is 9 units
Step 2
use the formula to find the length QR
Let
P1=Q=Q(8, 0, 2)
P2=R= R(11, −6, −4)
[tex]d=\sqrt{(11-8)^{2} +(6-0))^{2}+(-4-2 )^{2}} \\\\\\d=\sqrt{(3)^{2} +(6)^{2}+(-6 )^{2}}} \\d=\sqrt{9+36+36}\\d=\sqrt{81} \\d=9\\[/tex]
the length QR is 9 units
Step 3
use the formula to find the length PR
Let
P1=P(2, −3, −4)
P2=R= R(11, −6, −4)
[tex]d=\sqrt{(11-2)^{2} +(-6-(-3)))^{2}+(-4-4 )^{2}} \\\\\\d=\sqrt{(9)^{2} +(-6+3)^{2}+(-4-(-4) )^{2}}} \\d=\sqrt{81+9+0}\\d=\sqrt{90} \\d=9.48\\[/tex]
the length PR is 9.48 units
Step 4
is it a right triangle?
you can check this by using:
[tex]side^{2} +side^{2}=hypotenuse ^{2}[/tex]
Let
side 1=side 2= 9
hypotenuse = 9.48
Put the values into the equation
[tex]9^{2} +9^{2} =9.48^{2}\\ 81+81=90\\162=90,false[/tex]
Hence, the triangle is not a right triangle
Step 5
is it an isosceles triangle?
In geometry, an isosceles triangle is a type of triangle that has two sides of equal length.
Now side PQ=QR, so this is a isosceles triangle
Have a great day
Suppose that you draw two cards from a deck. After drawing the first card, you do not put the first card back in the deck. What is the probability (rounded to the nearest ten thousandth) that both cards are diamonds?
(A) 0.0543
(B) 0.0588
(C) 0.0625
(D) 0.0643
(E) None of the above
Answer:
(B) 0.0588
Step-by-step explanation:
The probability is calculated as a division between the number of possibilities that satisfy a condition and the number of total possibilities. Then, the probability that the first card is diamonds is:
[tex]P_1=\frac{13}{52}[/tex]
Because the deck has 52 cards and 13 of them are diamonds.
Then, if the first card was diamonds, the probability that the second card is also diamond is:
[tex]P_2=\frac{12}{51}[/tex]
Because now, we just have 51 cards and 12 of them are diamonds.
Therefore, the probability that both cards are diamonds is calculated as a multiplication between [tex]P_1[/tex] and [tex]P_2[/tex]. This is:
[tex]P=\frac{13}{52}*\frac{12}{51}=\frac{1}{17}=0.0588[/tex]
what is the answer of 2.8 plus 7.2
Answer:
10.022
Step-by-step explanation:
1. 49/9
2. 106/25
3. 10.022
4. When you add two rational numbers, each number can be written as a :
fraction
5. The sum of two fractions can always be written as a : fraction
6. Therefore, the sum of two rational numbers will always be : rational
Suppose you are planning to sample cat owners to determine the average number of cans of cat food they purchase monthly. The following standards have been set: a confidence level of 99 percent and an error of less than 5 units. Past research has indicated that the standard deviation should be 6 units. What is the final sample required?
Answer: 10
Step-by-step explanation:
The formula to find the sample size is given by :-
[tex]n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2[/tex]
Given : Significance level : [tex]\alpha=1-0.99=0.1[/tex]
Critical z-value=[tex]z_{\alpha/2}=2.576[/tex]
Margin of error : E=5
Standard deviation : [tex]\sigma=6[/tex]
Now, the required sample size will be :_
[tex]n=(\dfrac{(2.576)\ 6}{5})^2=9.55551744\approx10[/tex]
Hence, the final sample required to be of 10 .
Prove that for all integer m and n, if m-n is even then m^3-n^3 is even.
Let [tex]m,n[/tex] be any two integers, and assume [tex]m-n[/tex] is even. (This would mean either both [tex]m,n[/tex] are even or odd, but that's not important.)
We have
[tex]m^3-n^3=(m-n)(m^2+mn+n^2)[/tex]
and the parity of [tex]m-n[/tex] tells us [tex]m^3-n^3[/tex] must also be even. QED
In European roulette, the wheel is divided into 37 compartments numbered 1 through 36 and 0. (In American roulette there are 38 compartments numbered 1 through 36, 0, and 00.) One-half of the numbers 1 through 36 are red, the other half are black, and the number 0 is green. Find the expected value of the winnings on a $7 bet placed on black in European roulette. (Round your answer to three decimal places.)
Answer:
The expectation is -$0.189.
Step-by-step explanation:
Consider the provided information.
In European roulette, the wheel is divided into 37 compartments numbered 1 through 36 and 0.
One-half of the numbers 1 through 36 are red, the other half are black, and the number 0 is green.
We need to find the expected value of the winnings on a $7 bet placed on black in European roulette.
Here the half of 36 is 18.
That means 18 compartments are red and 18 are black.
The probability of getting black in European roulette is 18/37
The probability of not getting black in European roulette is 19/37. Because 18 are red and 1 is green.
If the ball lands on a black number, the player wins the amount of his bet.
The bet is ball will land on a black number.
The favorable outcomes are 18/37 and unfavorable are 19/37.
Let S be possible numerical outcomes of an experiment and P(S) be the probability.
The expectation can be calculated as:
E(x) = sum of S × P(S)
For[tex] S_1 = 7[/tex]
[tex]P(S_1) = \frac{18}{37}[/tex]
For [tex]S_2 = -7[/tex](negative sign represents the loss)
[tex]P(S_2) = \frac{19}{37}[/tex]
Now, use the above formula.
[tex]E(x) = 7\times \frac{18}{37}-7\times \frac{19}{37}\\E(x) = -0.189[/tex]
Hence, the expectation is -$0.189.
2. The cost of a cell phone is $500. According to the cellular device contract, you will need to be
$0.15 per minute for the first 6oo minutes.
a. Write the function that models the total cost of the cell phone bill for the first 600 minutes
b. Write the domain of the function interval notation).
a. As described in the problem you will be paying in the bill $0.15 per minute which means you have a linear relationship where both the total cost of the bill and the minutes will grow at the same rate (the price of the minute times the minutes) like this
[tex]C(t)=0.15t[/tex]
where C(t) is the total cost of the cell phone bill and t will be the time in minutes
b. The domain of the function will be the values that we can enter to the function. It is defined in the problem that this cost of the minutes its only up to 600 minutes so there is our limitation for the values that will enter the function. The domain will be between 0 and 600 both included because if he calls 0 minutes the bill will be 0 and if he calls 600 he would pay 0.15 for this last minute as well.
[tex][0,600][/tex]
Bob, the proprietor of Midland Lumber, believes that the odds in favor of a business deal going through are 9 to 5. What is the (subjective) probability that this deal will not materialize? (Round your answer to three decimal places.)
Answer:
There is a 35.7% probability that this deal will not materialize.
Step-by-step explanation:
This problem can be solved by a simple system of equations.
-I am going to say that x is the probability that this deal materializes and y is the probability that this deal does not materialize.
The sum of all probabilities is always 100%. So
[tex]1) x + y = 100[/tex].
Bob, the proprietor of Midland Lumber, believes that the odds in favor of a business deal going through are 9 to 5.
Mathematically, this means that:
[tex]2) \frac{x}{y} = \frac{9}{5}[/tex]
We want to find the value of y. So, we can write x as a function of y in equation 2), and replace it in equation 1).
Solution:
[tex]\frac{x}{y} = \frac{9}{5}[/tex]
[tex]x = \frac{9y}{5}[/tex]
[tex]x + y = 100[/tex]
[tex]\frac{9y}{5} + y = 100[/tex]
[tex]\frac{14y}{5} = 100[/tex]
[tex]14y = 500[/tex]
[tex]y = \frac{500}{14}[/tex]
[tex]y = 35.7[/tex]
There is a 35.7% probability that this deal will not materialize.
In the following ordinary annuity, the interest is compounded with each payment, and the payment is made at the end of the compounding period.
An individual retirement account, or IRA, earns tax-deferred interest and allows the owner to invest up to $5000 each year. Joe and Jill both will make IRA deposits for 30 years (from age 35 to 65) into stock mutual funds yielding 9.3%. Joe deposits $5000 once each year, while Jill has $96.15 (which is 5000/52) withheld from her weekly paycheck and deposited automatically. How much will each have at age 65? (Round your answer to the nearest cent.
Joe $
Jill $
Answer:
Ans. Joe will have $720,862.28 and Jill will have $819,348.90 after 30 years.
Step-by-step explanation:
Hi, since the interest is compounded with each payment, the effective rate of Joe is exactly equal to its compounded rate, that is 9.3%, but in the case of Jill, this rate is compounded weekly, this means that we have to divide 9.3% by 52 (which are the weeks in a year) in order to obtain an effective rate, in our case, effective weekly.
On the other hand, the time for Joe is pretty straight forward, he saves for 30 years at an effective annual interest rate of 9.3%, but Jill saves for 30*52=1560 weeks, at a rate of 0.1788% effective weekly.
They both have to use the following formula in order to find how much money will they have after 30 years of savings.
[tex]FutureValue=\frac{A((1+r)^{n}-1) }{r}[/tex]
In the case of Joe, this should look like this
[tex]FutureValue=\frac{5,000((1+0.093)^{30}-1) }{0.093} =720,862.28[/tex]
In the case of Jill, this is how this should look like.
[tex]FutureValue=\frac{96.15((1+0.001788)^{1560}-1) }{0.001788} =819,348.90[/tex]
Best of luck.
For what values of q are the two vectors A = i + j + kq each other and B-iq-23 + 2kg perpendicular to
Answer:
The value of q are 0.781,-1.281.
Step-by-step explanation:
Given : Two vectors [tex]A=i+j+kq[/tex] and [tex]B=iq-2j+2kq[/tex] are perpendicular to each other.
To find : The value of q ?
Solution :
When two vectors are perpendicular to each other then their dot product is zero.
i.e. [tex]\vec{A}\cdot \vec{B}=0[/tex]
Two vectors [tex]A=i+j+kq[/tex] and [tex]B=iq-2j+2kq[/tex]
[tex](i+j+kq)\cdot (iq-2j+2kq)=0[/tex]
[tex](1)(q)+(1)(-2)+(q)(2q)=0[/tex]
[tex]q-2+2q^2=0[/tex]
[tex]2q^2+q-2=0[/tex]
[tex]2q^2+q-2=0[/tex]
Using quadratic formula,
[tex]q=\frac{-1\pm\sqrt{1^2-4(2)(-2)}}{2(2)}[/tex]
[tex]q=\frac{-1\pm\sqrt{17}}{4}[/tex]
[tex]q=\frac{-1+\sqrt{17}}{4},\frac{-1-\sqrt{17}}{4}[/tex]
[tex]q=0.781,-1.281[/tex]
Therefore, The value of q are 0.781,-1.281.
If an intravenous solution containing 123 mg of a drug substance in each 250-mL bottle is to be administered at the rate of 200 μg of drug per minute, how many milliliters of the solution would be given per hour?
Answer:
24.39mL of the solution would be given per hour.
Step-by-step explanation:
This problem can be solved by direct rule of three, in which there are a direct relationship between the measures, which means that the rule of three is a cross multiplication.
The first step to solve this problem is to see how many mg of the solution is administered per hour.
Each minute, 200 ug are administered. 1mg has 1000ug, so
1mg - 1000 ug
xmg - 200 ug
[tex]1000x = 200[/tex]
[tex]x = \frac{200}{1000}[/tex]
[tex]x = 0.2mg[/tex]
In each minute, 0.2 mg are administered. Each hour has 60 minutes. How many mg are administered in 60 minutes?
1 minute - 0.2 mg
60 minutes - x mg
[tex]x = 60*0.2[/tex]
[tex]x = 12mg[/tex]
In an hour, 12 mg of the drug is administered. In 250 mL, there is 123 mg of the drug. How many ml are there in 12 mg of the drug.
123mg - 250mL
12 mg - xmL
[tex]123x = 250*12[/tex]
[tex]x = \frac{250*12}{123}[/tex]
[tex]x = 24.39[/tex]mL
24.39mL of the solution would be given per hour.
Final answer:
Approximately 24.39 milliliters of the intravenous solution would be administered per hour to deliver an hourly drug rate of 12 milligrams based on the given concentration.
Explanation:
To calculate how many milliliters of the intravenous solution would be given per hour, first convert the rate of drug administered from micrograms to milligrams: 200 \5g is equal to 0.2 mg. Since the drug administration rate is 0.2 mg per minute, we need to multiply this by 60 minutes to get the hourly rate:
0.2 mg/minute x 60 minutes/hour = 12 mg/hour.
Next, we need to find out how many milliliters of the solution contain 12 mg of the drug. Since we have 123 mg in 250 mL, we can set up a proportion to solve for the volume needed:
(123 mg/250 mL) = (12 mg/V mL)
V = (12 mg x 250 mL) / 123 mg = 24.39 mL.
Therefore, approximately 24.39 mL of the solution would be administered per hour.
Before the industrial revolution in 1800 the concentration of carbon in Earth’s atmo- sphere was 280 ppm. The concentration in 2015 was 399 ppm. What is the percent increase in the amount of carbon in the atmosphere?
Answer: There is increase of 4.255 in the amount of carbon in the atmosphere.
Step-by-step explanation:
Since we have given that
Concentration of carbon in Earth's atmosphere in 1800 = 280 ppm
Concentration of carbon in Earth's atmosphere in 2015 = 399 ppm
We need to find the percentage increase in the amount of carbon in the atmosphere.
So, Difference = 399-280 = 119 ppm
so, percentage increase in the amount of carbon is given by
[tex]\dfrac{Difference}{Original}\times 100\\\\=\dfrac{119}{280}\times 100\\\\=\dfrac{11900}{280}\\\\=42.5\%[/tex]
Hence, there is increase of 4.255 in the amount of carbon in the atmosphere.
In studying different societies, an archeologist measures head circumferences of skulls Choose the correct answer below O A. The data are qualitative because they don't measure or count anything O B. The data are qualitative because they consist of counts or measurements. O c. The data are quantitative because they don't measure or count anything. O D. The data are quantitative because they consist of counts or measurements. Click to select your answer Reflect in ePortfolio Download Print
Answer: The data are quantitative because they consist of counts or measurements.
Step-by-step explanation:
The definition of quantitative data says that if we can count or measure some thing in our data such as number of apples on each bag , length, width, etc then the data is said to be quantitative.
On the other hand in qualitative data we can obverse characteristics and features but can't be counted or measured such as honesty , color, tastes etc.
Given : In studying different societies, an archaeologist measures head circumferences of skulls.
Since here we are measuring circumferences of skulls, therefor it comes under quantitative data.
Hence, the correct answer is : The data are quantitative because they consist of counts or measurements.
The correct answer is D. The data are quantitative because they consist of counts or measurements. Quantitative data is numerical and can be measured and analyzed statistically.
Step by Step Solution:
When an archaeologist measures head circumferences of skulls, they are collecting quantitative data. Quantitative data consists of counts or measurements that are numerical in nature and can be subjected to statistical analysis. Examples of quantitative data in archaeology include measuring the length of projectile points, counting pollen grains, or recording quantities of animal bones at a site.
Therefore, the correct answer is:
D. The data are quantitative because they consist of counts or measurements.
Suppose that for some [tex]a,b,c[/tex] we have [tex]a+b+c = 1[/tex], [tex]ab+ac+bc = abc = -4[/tex]. What is [tex] a^3+b^3+c^3?[/tex]
Consider the cubic polynomial,
[tex](x+a)(x+b)(x+c)[/tex]
Expanding this gives
[tex]x^3+(a+b+c)x^2+(ab+ac+bc)x+abc=x^3+x^2-4x-4[/tex]
We can factor this by grouping,
[tex]x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x^2-4)(x+1)=(x-2)(x+2)(x+1)[/tex]
Then letting [tex]a=-2[/tex], [tex]b=2[/tex], and [tex]c=1[/tex] gives [tex]a^3+b^3+c^3=-8+8+1=\boxed1[/tex]
The physician orders an IV infusion of D5W 1000 ml to infuse over the next eight hours. The IV tubing that you are using delivers 10 gtt/ml. What is the correct rate of flow (drops per minute)? _gtt/min (rounded to the nearest drop)
Answer: 10gtt/ml means that in 10 drops there is a ml of the solution.
Now, you need 1000ml in 8 hours, and want to know the correct rate of flow in drops per minute.
first, 8 hours are 8*60 = 480 minutes.
then you need to infuse 1000ml in 480 minutes, so if you infuse at a constant rate, you need to infuse 1000/480 = 2.083 ml/min.
And we know that 10 drops are equivalent to 1 ml, then 2.083*10= 20.8 drops are equivalent a 2.083 ml, rounding it up, you get 21 drops for the dose.
So the correct rate of flow will be 21 drops per minute.
To find the correct rate of flow for an IV infusion, convert the time to minutes, divide the total volume by the total time to find the rate in ml/min, then multiply by the drip factor to convert to drops/min. Rounding to the nearest drop, we get 21 gtt/min.
Explanation:To calculate the correct rate of flow for an IV infusion, we need to use the given information: the volume of the IV infusion (D5W 1000 ml), the time over which it must infuse (8 hours), and the IV tubing drip factor (10 gtt/ml).
First, convert the time from hours to minutes as we're interested in drops per minute: 8 hours * 60 minutes/hour = 480 minutes.
Next, we divide the total volume by the total time: 1000 ml / 480 minutes = ~2.08 ml/min. This is the rate in ml/min.
Finally, we multiply by the drip factor to get the rate in drops per minute: 2.08 ml/min * 10 gtt/ml = 20.8 gtt/min.
Rounding to the nearest drop gives us a rate of 21 gtt/min.
Learn more about IV Flow Rate here:https://brainly.com/question/34306099
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A local food mart donates 20% of it's friday's sales to charity.
This friday the food mart had sales totaling 320.00 dollars. how
much of fridays sales will be donated to the charity?
Answer:
$64
Step-by-step explanation:
We have been given that a local food mart donates 20% of it's Friday's sales to charity. This Friday the food mart had sales totaling 320.00 dollars.
To find the the amount donated to the charity, we will find 20% of $320.
[tex]\text{The amount donated to the charity}=\$320\times\frac{20}{100}[/tex]
[tex]\text{The amount donated to the charity}=\$320\times0.20[/tex]
[tex]\text{The amount donated to the charity}=\$64[/tex]
Therefore, $64 were donated to the charity.
Cheese costs $4.40 per pound. Find the cost per kilogram. (1kg = 2.2lb)
Answer:
The cost is $9.70 per kilogram.
Step-by-step explanation:
This can be solved by a rule of three.
In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.
When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too. In this case, the rule of three is a cross multiplication.
When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease. In this case, the rule of three is a line multiplication.
In this problem, the measures are the weight of the cheese and the price. As the weight increases, so does the price. It means that this is a direct rule of three.
Solution:
The problem states that cheese costs $4.40 per pound. Each kg has 2.2 pounds. How many kg are there in 1 pound. So:
1 pound - xkg
2.2 pound - 1 kg
[tex]2.2x = 1[/tex]
[tex]x = \frac{1}{2.2}[/tex]
[tex]x = 0.45[/tex]kg
Since cheese costs $4.40 per pound, and each pound has 0.45kg, cheese costs $4.40 per 0.45kg. How much does is cost for 1kg?
$4.40 - 0.45kg
$x - 1kg
[tex]0.45x = 4.40[/tex]
[tex]x = \frac{4.40}{0.45}[/tex]
[tex]x = 9.70[/tex]
The cost is $9.70 per kilogram.
"The cost per kilogram of cheese is approximately $2.00.
To find the cost per kilogram, we need to convert the cost from dollars per pound to dollars per kilogram using the conversion factor between pounds and kilograms. Given that 1 kilogram is equal to 2.2 pounds, we can set up the following conversion:
Cost per pound of cheese = $4.40
Conversion factor = 2.2 pounds/kilogram
Now, to find the cost per kilogram, we divide the cost per pound by the conversion factor:
Cost per kilogram = Cost per pound / Conversion factor
Cost per kilogram = $4.40 / 2.2 pounds/kilogram
Performing the division, we get:
Cost per kilogram ≈ $2.00
(7)-0, at the points x 71, 72, 73, 74, and 7.5 Use Euler's method with step size 0.1 to approximate the solution to the initial value pro oblemy - 2x+y The approximate solution to y'=2x-y?.y(7)=0, at the point x = 71 is (Round to five decimal places as needed.)
Answer:
2.68
Step-by-step explanation:
We are given that [tex]x_0=7,x_1=7.1,x_2=7.2,x_3=7.3,x_4=7.4,x_5=7.5[/tex]
h=0.1
y'=2x-y
y(7)=0,f(x,y)=2x-y
[tex]x_0=7,y_0=0[/tex]
We have to find the approximate solution to the initial problem at x=7.1
[tex]y_1=y_0+hf(x_0,y_0)[/tex]
Substitute the value then, we get
[tex]y_1=0+(0.1)(2(7)-0)=0+(0.1)(14)=1.4[/tex]
[tex]y_1=1.4[/tex]
[tex]x_1=x_0+h=7+0.1=7.1[/tex]
[tex]y_2=y_1+hf(x_1,y_1)[/tex]
Substitute the values then, we get
[tex]y_2=1.4+(0.1)(2(7.1)-1.4)=1.4+(0.1)(14.2-1.4)=1.4+(0.1)(12.8)=1.4+1.28[/tex]
[tex]y_2=1.4+1.28=2.68[/tex]
Hence, the approximation solution to the initial problem at x=7.1 is =2.68
Round the following number to the indicated place. 66.1086 to hundredths
Answer:
66.11
Step-by-step explanation:
We are given that a number
66.1086
We have to round the number to hundredths
Place of 6=One;s
Place of second 6=Tens
Place of 1=Tenths
Place of 0=Hundredths
Place of 8=Thousandths
Place of 6=Ten thousandths
Thousandths place is 8 which is greater than 5 therefore, one will be added to hundredth place and other number on the left side of hundredth place remain same and the numbers on the right side of hundredth place will be replace by zero.
Therefore, the given number round to hundredths=66.11