The equation p-20=-30 can be rewritten as an addition problem by adding 20 to both sides of the equation. The resulting addition equation is p = -30 + 20.
Explanation:The equation p-20=-30 can be written as an addition problem by moving -20 to the other side of the equation. This is done by adding 20 to both sides of the equation, which results in the equation p = -30 + 20. In other words, you can add -30 and 20 together to find the value of p. This is how you write the equation p-20=-30 as an addition problem. So the equation becomes p= -30+20
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A study of bulimia among college women studied the connection between childhood sexual abuse and a measure of family cohesion (the higher the score, the greater the cohesion). The sample mean on the family cohesion scale was 1.9 for 13 sexually abused students (sequals2.1) and 5.2 for 17 nonabused students (sequals3.5). a. Find the standard error for comparing the means. b. Construct a 95% confidence interval for the difference between the mean family cohesion for sexually abused students and non-abused students. Interpret.
Answer: a) 1.029, b) (-5.318, -1.282).
Step-by-step explanation:
Since we have given that
[tex]n_1=13\\\\n_2=17\\\\\bar{x_1}=1.9\\\\\bar{x_2}=5.2[/tex]
and
[tex]s_1=2.1\\\\s_2=3.5[/tex]
So, the standard error for comparing the means :
[tex]SE=\sqrt{\dfrac{s^2_1}{n_1}+\dfrac{s^2_2}{n_2}}\\\\SE=\sqrt{\dfrac{2.1^2}{13}+\dfrac{3.5^2}{17}}\\\\SE=\sqrt{1.0598}\\\\SE=1.029[/tex]
At 95% confidence interval, z = 1.96
So, Confidence interval would be
[tex]\bar{x_1}-\bar{x_2}\pm z\times SE\\\\=(1.9-5.2)\pm 1.96\times 1.0294\\\\=-3.3\pm 2.017624\\\\=(-3.3-2.018,-3.3+2.018)\\\\=(-5.318,-1.282)[/tex]
Hence, a) 1.029, b) (-5.318, -1.282).
According to a survey, 62% of murders committed last year were cleared by arrest or exceptional means. Fifty murde committed last year are randomly selected, and the number cleared by arrest or exceptional means is recorded. When technology is used, use the Tech Help button for further assistance.
a. Find the probability that exactly 41 of the murders were cleared.
b. Find the probability that between 36 and 38 of the murders, inclusive, were cleared.
c. Would it be unusual if fewer than 19 of the murders were cleared? Why or why not?
a.The probability that exactly 41 of the murders were cleared is ____. (Round to four decimal places as needed.)
Answer:
a) The probability that exactly 41 of the murders were cleared is 0.0013
b) The probability that between 36 and 38 of the murders, inclusive, were cleared is 0.0809.
c) Yes, it would be unusual
Step-by-step explanation:
Let p=62% considered as the probability of having a commited that is cleared by arres or exceptional means. We assume that choosing each of the 50 commited is independent of each choose. Then, let X be the number of cleared. In this case, X is distributed as a binomial random variable. Recall that, in this case,
[tex] P(X=k) = \binom{50}{k} p^{k}(1-p)^{50-k}[/tex] for[tex]0\leq k \leq 50[/tex], with p=0.62
a) We have that
[tex] P(X=40) = \binom{50}{40} p^{40}(1-p)^{50-40} =0.001273487 [/tex]
b) We are asked for the following
[tex]P(36\leq X \leq 38) = P(X=36)+P(X=37)+P(X=38) = 0.080888936
[/tex] (The specific calculation is omitted.
c) We will check for the following probability [tex]P(X\leq 19)[/tex]
[tex]P(X\leq 19 ) = \sum_{k=0}^{19} P(X=k) = 0.000499222 [/tex]
Given that the probability of this event is really close to 0, it would be unusual if less than 19 murders are cleared.
The question deals with the application of binomial probability distribution in a real life situation involving crime investigation. The probability values for a certain range or exact number of cleared murders can be calculated by using the binomial probability formula. It would be statistically unusual for fewer than 19 murders to be cleared given a 62% clearance rate.
Explanation:This question can be approached using the binomial distribution, where the number of successes in a sequence of n independent experiments (in this case, the number of murders being cleared) follows a binomial distribution.
a. The probability that exactly 41 of the murders were cleared can be found by calculating the binomial probability. This can be done by using the formula: P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)). In this case, n=50 (number of trials/murders), k=41 (number of successes/murders cleared), and p=0.62 (probability of success/clearing a murder). You need to substitute these values into the formula and calculate the value.
b. Finding the probability that between 36 and 38 murders were cleared involves the same process, but you need to calculate for k=36, 37, and 38, and then add the results together to get the total probability.
c. If fewer than 19 of the murders were cleared, it would be statistically unusual considering the 62% clearance rate according to the survey. The reasoning being that, given a 62% probability, the expectation would be significantly higher.
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Look at the number line below. The letters f, g, h, and i all represent integers. Write two inequalities to compare f and g.
Without the provided number line, we cannot determine the exact relationship between f and g. Inequalities f < g or f > g represent f being less than or greater than g, respectively. To write the correct inequality, one must refer to the positions of f and g on the number line.
Explanation:Since the number line is not provided, we cannot see the exact positions of f and g. However, we can discuss how to write inequalities to compare two integers based on their positions on a number line. If f is located to the left of g on the number line, it means that f is less than g. The inequality for this scenario would be f < g. On the other hand, if f is located to the right of g, then f is greater than g, and the corresponding inequality would be f > g.
You can use an inequality symbol to show how two metric measurements are related. If two numbers are the same, the inequality symbol would be the equal sign, representing they are equivalent. However, without the number line, we cannot determine the exact relationship between f and g, so one must look at the number line to ascertain the correct inequality to use.
Show that (2, 1) is a solution of the system of equations.
x + 3y = 5,
y = –x + 3
1. Substitute (2, 1) into x + 3y = 5 to get .____
Options: 1 + 32 = 5, 1 + 3(2) = 5, 2 + 31 = 5, 2 + 3(1) = 5
2. Simplify the equation to get ____
Options: 5 = 5 is true, 7 = 7 is true, 7 = 5 is false, 33 = 5 is false
3. Substitute (2, 1) into y = –x + 3 to get ____
Options: 1 = -(2) + 3, 1 = -(2 + 3), 2 = -(1) + 3, 2 = -(1 + 3)
4. Simplify the equation to get .____
Options: 1 = 1 is true, 1 = -1 is false, 2 = 2 is true, 2 = -2 is false
Answer: The answer is 1. Substitute (2,1) into x+3y=5
Step-by-step explanation: Remember, (2,1) 2 is substituted for x and 1 is used as the y substitute
Answer:
1. 2+3(1)=5
2. 5=5 is true
3. 1=-(2)+3
4. 1=1 is true
D. Now calculate resting Vm given the following relative permeability ratios. Assume that there is no meaningful permeability to Ca2+. (1pt) PK : PNa : PCl = 1.0 : 0.04 : 0.45 E. Let’s say that the K+ permeability increases, for example because of expression of more K+ channels. Calculate Vm with new the permeability values and discuss how the membrane potential has changed in relation to the potassium equilibrium potential EK. (roughly 1 sentence) (1pt)+
Answer:EX = RT ln [X]o
.........zF.....[X]i
EX = (1.987 cal/deg.mol)(293 deg) ln [X]o
.........z(23,062 cal/volt.mol)................[X]i
OR
EX = (8.315 joules/deg.mol)(293 deg) ln [X]o
.........z(96,485 joules/volt.mol)................[X]i
EK+ = 0.025 ln(12/400) = -0.088 V = -88 mV
ENa+ = 0.025 ln(450/55) = 0.053 V = 53 mV
ECa+2 = 0.0126 ln(10/0.0001) = 0.145 V = 145 mV
ECl- = -0.025 ln(550/56) = -0.058 V = -58 mV
Step-by-step explanation:
The Goldman equation is used to calculate resting membrane potential (Vm) considering the relative permeabilities of different ions. Given the permeabilities, the resting Vm can be estimated. If the K+ permeability increases, Vm will move closer to the Potassium equilibrium potential (EK).
Explanation:The resting membrane potential (Vm) can be calculated using the Goldman equation, which considers the relative permeabilities and concentrations of different ions. The equation is: Vm = 61.5 log ((PK[K+]out + PNa[Na+]out + PCl[Cl-]in) / (PK[K+]in + PNa[Na+]in + PCl[Cl-]out)), where Px indicates the relative permeability of each ion & the square brackets contain the ion concentrations inside (in) or outside (out) the cell.
Given the permeabilities PK : PNa : PCl = 1.0 : 0.04 : 0.45, assuming concentrations inside and outside the cell in a balanced condition with no net movement of any ion, you might estimate the resting Vm.
If the K+ permeability increases, Vm would reportedly move closer towards the Potassium equilibrium potential (EK). This is because the membrane is becoming more permeable to K+ and less responsive to the influences of other ions.
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If the drawing has dimensions 7.5 cm × 5 cm, what is the size of the soccer field?
Answer: 35.5cm^2
Step-by-step explanation:
I believe we're talking about a rectangular. So it would be length * width or 7.5cm * 5cm = 35.5cm^2
Two dice: Find the probability of rolling an odd number on the first die and an even number on the second die.
Answer:
1/4; 25%
Step-by-step explanation:
Both events happen with probability 1/2: there are 3 even numbers and 3 odd numbers in a die.
Since the two events are also independent (the outcome of the first die doesn't affect the outcome of the second), we have to multiply those probability.
So, you roll an odd number on the first die and an even number on the second die with probability
[tex]\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}[/tex]
x g(x)
−2 1/4
-1 1/2
0 1
1 2
3 8
Consider that f(x) = x + 2, while the table represents y = g(x). Which statement is true when comparing the rate of change for the functions?
A) The rate of increase for the functions is the same.
B) f(x) has a greater rate of increase than function g(x).
C) g(x) has a greater rate of increase than function f(x).
D) g(x) has a greater rate of decrease than function f(x).
Answer:
A
Step-by-step explanation:
Cause I know
Answer: C!!
Step-by-step explanation:
USA Test Prep told me! :)
What is the slope of this line?
(1,4) (6,-1)
Answer:
The slope is -1
Step-by-step explanation:
Let's find the slope between your two points.
(1,4);(6,−1)
(x1,y1)=(1,4)
(x2,y2)=(6,−1)
Use the slope formula:
m= y2−y1/x2−x1 = −1−4/6−1
= −5/5
= −1
Hope this is a better explanation :)
Hope it helps!!!!!!!!!
A study is conducted to investigate whether customer satisfaction is greater among computer companies that offer tech support versus those that do not offer tech support. A random sample of 50 customers are selected from among those that purchased computers that offer tech support. A separate random sample of 40 customers are selected from among those that purchased computers that do not offer tech support.
The study found that the mean satisfaction rating was significantly greater among customers that purchased computers that offer tech support.
Which of the following is the best description of this study?
(A) An experiment using a completely randomized design.
(B) An experiment using a randomized block design.
(C) An experiment using a matched pairs design
(D) An observational study using a simple random sample.
Answer:
The correct answer is (E).
This is not an experiment because no treatment is being imposed upon the customers. Additionally, this study used a stratified sample because independent random samples were selected from two distinct populations of customers.
Step-by-step explanation:
The correct answer is (E) An observational study using a stratified sample.
What is stratified sample?Stratified sampling is also known as stratified random sampling. The stratified sampling process starts with researchers dividing a diverse population into relatively homogeneous groups called strata, the plural of stratum. Then, they draw a random sample from each group (stratum) and combine them to form their complete representative sample.
Given that a data of a survey, the study found that the mean satisfaction rating was significantly greater among customers that purchased computers that offer tech support.
We need to find which is the best description of this study,
This study used a stratified sample because independent random samples were selected from two distinct populations of customers.
This is not an experiment because no treatment is being imposed upon the customers.
Hence, the best description is an observational study using a stratified sample.
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The complete question is attached.
If n - 2 = 6 -2(7 + n), then n = ?
Use the given information to bound the p-value of the F statistic for a one-tailed test with the indicated degrees of freedom. F = 4.23, df1 = 4, df2 = 5 p-value < 0.005 0.005 < p-value < 0.010 0.010 < p-value < 0.025 0.025 < p-value < 0.050 0.050 < p-value < 0.100 p-value > 0.100
Answer:
The range of the p-value is: 0.050 < p-value < 0.100.
Step-by-step explanation:
For checking the equivalence of two population variances of independent samples, we use the f-test.
The test statistic is given by:
[tex]F=\frac{S_{1}^{2}}{S_{2}^{2}}\sim F_{\alpha, (n_{1}-1)(n_{2}-1)}[/tex]
It is provided that the hypothesis test is one-tailed.
The computed value of the test statistic is:
F = 4.23.
The degrees of freedom of the numerator and denominator are:
[tex]df_{1}=4\\df_{2}=5[/tex]
Use MS-Excel to compute the p-value as follows:
Step 1: Select function fX → F.DIST.RT.
Step 2: A dialog box will open. Enter the values of f-statistic and the two degrees of freedom.
*See the attachment below.
Step 3: Press OK.
The p-value is, 0.0728.
The range of the p-value is:
0.050 < p-value < 0.100
Add together 8.03 m 1.26 m 0.5 m 4.09 m 3.5 m
Answer:
17.38m
Step-by-step explanation:
Answer:
17.38m
Step-by-step explanation:
What do common multiples of 4 and 6 mean
Answer:
so the common multiple of 4 and 6 is 2 i hope this helps! :)
Step-by-step explanation:
well the common multiples of 4 and 6 is 2
2 * 2 = 4
2 * 3 = 6
so there has to be a certain number that goes into both of the numbers you are working with
Answer:
Common multiples of 4 and 6 are 12, 24, 36, 48, 60, 72, 84, 96...
Jay was reaching into her purse and accidentally spilled her coin purse. 10 pennies fell on the floor. Jay noticed that only 2 of the pennies landed on heads. What is the theoretical probability of this happening?
Answer:
The theoretical probability of landing on 2 heads, when 10 coins are tossed is 0.0439 or 4.39%.
Step-by-step explanation:
Number of coins that fell on the floor = 10
Number of coins that landed on heads = 2
We have to find the theoretical probability of getting 2 coins landing of heads when 10 coins are tossed.
Notice that there are only 2 possible outcomes: Either that coin will land on head or it won't. Landing of each coin is independent of the others coins. Probability of each coin landing on head is constant i.e. 0.5 or 1/2. Number of trials, i.e. the number of times the experiment will be done is fixed, which is 10.
All the 4 conditions for an experiment to be considered a Binomial Experiment are satisfied. So we will use Binomial Probability to solve this problem.
Probability of success = Probability of coin landing on head = 0.5
Number of trials = n = 10
Number of success = r = 2
The formula for Binomial Probability is:
[tex]P(X = x) =^{n}C_{r}(p)^{r}(1-p)^{n-r}[/tex]
Using the values, we get:
[tex]P(X=2)=^{10}C_{2}(0.5)^2(0.5)^8=0.0439[/tex]
Thus, the theoretical probability of landing on 2 heads, when 10 coins are tossed is 0.0439 or 4.39%.
I don’t know how to do this. Care to explain?
Answer:
y=6x+7
Step-by-step explanation:
−6x+y=7
Step 1: Add 6x to both sides
y=6x+7
Answer:
y=6x+7
Step-by-step explanation:
whenever you solve for a letter you have to get it by itself. That means it has to be alone on the opposite side of the equation
move the -6x to the other side of the equation
that -6 changes to a positive 6 because, your moving it to the other side of the equation
your left with y=6x+7
pls mark me brainliest
In a one-way ANOVA, if the computed F statistic is greater than the critical F value you may Question 1 options: reject H0 since there is evidence that not all the means are different. not reject H0 since there is no evidence of a difference in the means. not reject H0 because a mistake has been made. reject H0 since there is evidence all the means differ.
Answer:
And the F statistic calculated from the mean squares if [tex]F_{calc}[/tex]. And for this case we know that [tex] F_{calc}>F_{critical}[/tex]. So then we can reject the null hypothesis that all the means are equal at a significance level given [tex]\alpha[/tex]. And the best conclusion would be:
reject H0 since there is evidence all the means differ.
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
Solution to the problem
The hypothesis for this case are:
Null hypothesis: [tex]\mu_{A}=\mu_{B}=....=\mu_{k}[/tex]
Alternative hypothesis: Not all the means are equal [tex]\mu_{i}\neq \mu_{j}, i,j=A,B,...,k[/tex]
If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:
[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]
[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]
And we have this property
[tex]SST=SS_{between}+SS_{within}[/tex]
And the F statistic calculated from the mean squares if [tex]F_{calc}[/tex]. And for this case we know that [tex] F_{calc}>F_{critical}[/tex]. So then we can reject the null hypothesis that all the means are equal at a significance level given [tex]\alpha[/tex]. And the best conclusion would be:
reject H0 since there is evidence all the means differ.
In a one-way ANOVA, if the computed F statistic is greater than the critical F value you may eject H0 since there is evidence all the means differ.
What is F-test?F test is a method used in statistics to determine which models best fits the population from which the sample is derived.
The formula for calculating the F-test statistic = explained variance / unexplained variance
The null hypothesis usually states that the means are the means are the same while the alternative hypothesis states that the means are not the same.
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An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm exam. After all the students had turned in their exam pa- pers, the instructor randomly ordered them before grading. Consider the first 15 graded exam papers. (a) Find the probability that exactly 10 of these are from the second section. Find the probability that at least 10 of these are from the second section. Find the probability that at least 10 of these are from the same section?
Answer:
The answers are for option (a) 0.2070 (b)0.3798 (c) 0.3938
Step-by-step explanation:
Given:
Here Section 1 students = 20
Section 2 students = 30
Here there are 15 graded exam papers.
(a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070
(b) Here if x is the number of students copies of section 2 out of 15 exam papers.
here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15
Then,
Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798
(c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1)
so,
Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938
Note : Here the given distribution is Hyper-geometric distribution
where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.
Use this information to answers Questions 1 through 7. The U.S. government provides money to each state to maintain the interstate highway system in the state. The U.S. can revoke or reduce the money if the states do not safely maintain the highways. The U.S. government is particularly concerned about the speed of traffic on Kansas highways. If there is convincing evidence that the average speed of all interstate highway vehicles in Kansas exceeds the posted speed limit of 70 mph, the federal government will reduce the amount of funding it provides. If there is not convincing evidence, then the government will not reduce the funding.
Kansas Highway Patrol recorded the speed of 450 interstate vehicles and found the mean speed of 70.2 mph with standard deviation 1.6 mph. The U.S. government will use this information to conduct a hypothesis test at significance level 0.01 to decide whether or not to reduce to money sent to Kansas.
Suppose that mu is the true mean speed of all vehicles on the Kansas interstate highway system.
1. What are the null and alternative hypotheses that the U.S. government should test?
2. What is the value of the test statistic?
Answer:
1. Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 70 mph
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 70 mph
2. Value of test statistics is 2.652.
Step-by-step explanation:
We are given that Kansas Highway Patrol recorded the speed of 450 interstate vehicles and found the mean speed of 70.2 mph with standard deviation 1.6 mph.
We have to conduct a hypothesis test at significance level 0.01 to decide whether or not to reduce the money sent to Kansas.
Let [tex]\mu[/tex] = true mean speed of all vehicles on the Kansas interstate highway system.
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 70 mph {means that the federal government will not reduce the amount of funding it provides as the speed limit is less than or equal to 70 mph}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 70 mph {means that the federal government will reduce the amount of funding it provides as the speed limit exceed 70 mph}
The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean speed limit of 450 interstate vehicles = 70.2 mph
s = sample standard deviation = 1.6 mph
n = sample of vehicles = 450
So, test statistics = [tex]\frac{70.2-70}{\frac{1.6}{\sqrt{450} } }[/tex] ~ [tex]t_4_4_9[/tex]
= 2.652
Hence, the value of test statistics is 2.652.
A study by a federal agency concludes that polygraph tests given to truthful persons have probability 0.2 of suggesting that the person is deceptive. A firm asks 12 job applicants about thefts from previous employers, using a polygraph to assess their truthfulness. Suppose that all 12 answer truthfully. Let X be the number of applicants who are classified deceptive. a) Describe the probability mass function of X. b) What is the probability that the polygraph says at least 1 is deceptive? c) What is the mean number among 12 truthful persons who will be classified as deceptive? What is the standard deviation of this number? d) What is the probability that the number classified deceptive is less than the mean?
Answer:
b) 0.9313
c) mean = 2.4
standard deviation= 1.3856
d) 0.5583
Step-by-step explanation:
Given:
p = 0.2
n = 12
a) X= number of applicants classified as deceptive.
Probability mass function of X will be:
[tex] P(X=x) = \left(\begin{array}{c}12\\x\end{array}\left) (0.2)^x(1-0.2)^1^2^-^x,x=0, 1, 2, .....,12 [/tex]
b) Probability that the polygraph says at least 1 is deceptive:
[tex] P(X≥1) = 1 -P(X=0) = 1 -\left(\begin{array}{c}12\\0\end{array}\left) (0.2)^0(1-0.2)^1^2^-^0[/tex]
= 1 - 0.0687
= 0.9313
c) The mean number among 12 truthful persons who will be classified as deceptive:
E(X) = n•p
= 12 * 0.2
= 24
Standard deviation:
[tex]s.d = \sqrt{12*0.2*(1-0.2)}[/tex]
= 1.3856
d) Probability that the number classified deceptive is less than the mean:
[tex] P(X<2.4) = P(X≤2) = E^2_x_=_0 \left(\begin{array}{c}12\\0\end{array}\left) (0.2)^0(1-0.2)^1^2^-^0 [/tex]
= 0.5583
When rolling two fair 6 sided dice, what is the probability that the total is at most 10?
Answer:
[tex]\frac{33}{36}[/tex]
Step-by-step explanation:
Combinations greater than a 10
5 - 5
5 - 6
6 - 5
There are total 36 combinations (6 * 6).
3 of these combinations are higher than 10.
So 36 - 3 combinations are less than 10.
Qual o valor que Z assume?
-Cz+ 6z = Tz+ 83
Answer:
[tex]\large \boxed{z = \dfrac{83}{6 - C - T}}[/tex]
Step-by-step explanation:
[tex]\begin{array}{rccl}-Cz + 6z & = & Tz + 83 & \\6z -Cz - Tz &= & 83 & \text{Subtracted Tz from each side}\\z(6 - C - T) & = & 83 & \text{Removed the common factor}\\z& = & \mathbf{\dfrac{83}{6 - C - T}} &\text{Divided each side by 6 - C - T}\\\end{array}\\\\\large \boxed{\mathbf{z = \dfrac{83}{6 - C - T}}}[/tex]
A store sells televisions for $360 and video cassette recorders for $270. At the beginning of the week its entire stock is worth $56,430. During the week it sells three quarters of the televisions and one third of the video cassette recorders for a total of $32,310. How many televisions and video cassette recorders did it have in its stock at the beginning of the week
Answer:
The number of Television at the beginning was 90
The number of video cassette recorders at the beginning was 89
Step-by-step explanation:
Selling Price of 1 Television =$360.
Selling Price of 1 video cassette recorders for $270.
Let the number of Television at the beginning=x
Let the number of video cassette recorders at the beginning =y
Opening Stock =$56,430.
Therefore:
360x+270y=$56,430.It sells three quarters of the televisions and one third of the video cassette recorders for a total of $32,310.
[tex]\frac{3}{4}(360)x+\frac{1}{3}(270)y= \$32,310[/tex]
270x+90y=32310
We then solve the two equations to obtain x and y.
360x+270y=$56,430. (Multiply by 270)270x+90y=32310 (Multiply by 360)97200x+72900y=15236100
97200x+32400y=11631600
Subtract
40500y=3604500
y=89
Substitute y=89 into 270x+90y=32310 to obtain x
270x+90(89)=32310
270x=32310-8010=24300
x=90
Therefore:
The number of Television at the beginning was 90
The number of video cassette recorders at the beginning was 89
Answer:
90 televisions and 89 video cassette recorders
Step-by-step explanation:
The unit cost of television = $360
The unit cost of video cassette recorder = $270
Let "T" represent the number of televisions and "R" represent the number of recorders, so that we can make representations using equations from the statements.
At the beginning of the week, Total Stock is worth $56,430, where
Total Stock = Total cost of televisions + Total cost of recorders
Total Cost = Unit Cost × Number of items
$56,430 = 360T + 270R This is the first equation
Next, During the week, Number of Sales = [tex]\frac{3}{4}[/tex] T + [tex]\frac{1}{3}[/tex] R
Total Sales Price = 360 ([tex]\frac{3}{4}[/tex] T) + 270 ([tex]\frac{1}{3}[/tex] R)
$32,310 = 270T + 90R This is the second equation
Solving both equations simultaneously, let us use elimination method which involves equating one of the two terms in both equations. Let us multiply the second equation by 3. This doesn't affect the equation, since we are doing it to all the terms in it.
56,430 = 360T + 270R
32,310 = 270T + 90R × 3
So, we have;
56,430 = 360T + 270R
96,930 = 810T + 270R
Subtracting both equations, we have;
96,930 - 56,430 = 810T - 360T
40,500 = 450T
T = [tex]\frac{40,500}{450}[/tex] = 90
Since we now have the number of televisions, we can get the number of recorders by putting 90 in any (say, the second) equation.
32,310 = 270 (90) + 90R
32,310 = 24,300 + 90R
32,310 - 24,300 = 90R
8010 = 90R
R = [tex]\frac{8010}{90}[/tex] = 89
At the beginning of the week, the store had 90 televisions and 89 video cassette recorders
In this assignment, your team is managing a software development project with a total project budget of $178,500. Total work effort is 1,536 hours and the timeline for completion 24 weeks. At the end of week 12, the plan was to have completed 55% of the project scope. However, actual progress was calculated at 650 worth of hours completed on project activities and actual cost (in hours) for these activities is 780. With your team, please answer the following questions. For each response, show the pertinent formula(s) as it applies and "show your work" in addition to providing a narrative statement.
a. How much money was supposed to have been spent at the end of week 12?b. Will the project finish on time and within the given budget?
Answer:
im lost good luck
Step-by-step explanation:
The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $490 and standard deviation $10. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1?
Answer:
[tex]z=1.28<\frac{a-490}{10}[/tex]
And if we solve for a we got
[tex]a=490 +1.28*10=502.8[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 502.8.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the weeknly amount of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(490,10)[/tex]
Where [tex]\mu=490[/tex] and [tex]\sigma=10[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.1[/tex] (a)
[tex]P(X<a)=0.9[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.9[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.9[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.28<\frac{a-490}{10}[/tex]
And if we solve for a we got
[tex]a=490 +1.28*10=502.8[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 502.8.
Final answer:
The company should budget approximately $502.80 for weekly repairs and maintenance to ensure that the probability of exceeding this amount is only 0.1.
Explanation:
We want to find how much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1.
Since the weekly amount of money spent is normally distributed with a mean of $490 and a standard deviation of $10, we can find the amount by looking up the z-score that corresponds to the 90th percentile
(since 100% - 10% = 90%) in a standard normal distribution table or using a calculator.
Let the z-score for the 90th percentile be denoted as z.
Looking up the standard normal distribution table or using a calculator, we find that z ≈ 1.28 for 0.9 cumulative probability.
We then use the z-score formula:
z = (X - mean) / standard deviation
Plugging in our z-score and the parameters, we can solve for X:
1.28 = (X - 490) / 10
X - 490 = 12.8
X = $502.80
Therefore, the company should budget approximately $502.80 for weekly repairs and maintenance to ensure that the probability of exceeding this amount is only 0.1.
The following sample data are from a normal population: 10, 8, 12, 15, 13, 11, 6, 5.
a. What is the point estimate of the population mean?10 b. What is the point estimate of the population standard deviation (to 2 decimals)?3.46 c. With confidence, what is the margin of error for the estimation of the population mean (to 1 decimal)?2.9 d. What is the confidence interval for the population mean (to 1 decimal)?
Answer:
10
3.4641
2.8965
12.8965
Step-by-step explanation:
Given: 10, 8, 12, 15, 13, 11, 6, 5
c = 95%
a. The point estimate of the population mean is the sample mean. The mean is the sum of all values divided by the number of values:
x = 10 + 8 + 12 + 15 + 13 + 11 + 6 + 5 /8
= 80/8
= 10
b. The point estimate of the population standard deviation is the sample standard deviation. The variance is the sum of squared deviations from the mean divided by n - 1. The standard deviation is the square root of the variance:
s = /(10 – 10)^2 +.... + (5– 10)^2/8 – 1
s = 3.4641
c. Determine the t-value by looking in the row starting with degrees of freedom df = n-1 = 8 –1 = 7 and in the column with [tex]\alpha[/tex] = (1 – c)/2 = 0.025 in table :
t_[tex]\alpha[/tex]/2 = 2.365
The margin of error is then:
E = t_[tex]\alpha[/tex]/2 * s/√n
= 2.365 x s 3.4641/ √8
= 2.8965
d. The confidence intent)] then becomes:
7.1035 = 10 – 2.8965 = x – E <u<x +E= 10 + 2.8965 = 12.8965
The point estimate of the population mean will be 10.
How to calculate the point estimate?The point estimate of the population mean will be calculated thus:
= (10 + 8 + 12 + 15 + 13 + 11 + 11 + 6 + 5) / 8
= 80/8
= 10
Also, the margin of error will be:
= 2.365 × 3.4641/✓8
= 2.8965
In conclusion, the margin of error is 2.8965.
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Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.542.54 and a standard deviation of 0.420.42. Using the empirical rule, what percentage of the students have grade point averages that are between 1.281.28 and 3.83.8?
Answer:
[tex] P(1.28< X< 3.8) [/tex]
And we can use the z score formula to calculate how many deviations we are within the mean
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And if we use this formula we got:
[tex] z = \frac{1.28-2.54}{0.42}= -3[/tex]
[tex] z = \frac{3.8-2.54}{0.42}= 3[/tex]
And using the empirical rule we know that within 3 deviation from the mean we have 99.7% of the values
Step-by-step explanation:
Previous concepts
The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".
Let X the random variable who represent the grade point averages of undergraduate students.
From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=2.54, Sd(X)=0.42[/tex]
So we can assume [tex]\mu=2.54 , \sigma=0.42[/tex]
On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
For this case we want to find this probability:
[tex] P(1.28< X< 3.8) [/tex]
And we can use the z score formula to calculate how many deviations we are within the mean
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And if we use this formula we got:
[tex] z = \frac{1.28-2.54}{0.42}= -3[/tex]
[tex] z = \frac{3.8-2.54}{0.42}= 3[/tex]
And using the empirical rule we know that within 3 deviation from the mean we have 99.7% of the values
Approximately 99.7 percent of the students have grade point averages between 1.28 and 3.8 according to the empirical rule.
Explanation:The empirical rule states that approximately 68 percent of the data lies within one standard deviation of the mean, 95 percent lies within two standard deviations, and more than 99 percent lies within three standard deviations. In this case, the mean grade point average is 2.54 and the standard deviation is 0.42. To find the percentage of students with grade point averages between 1.28 and 3.8, we need to find the z-scores for both values and calculate the area under the curve between those z-scores.
First, we find the z-score for 1.28 using the formula: z = (x - µ) / σ = (1.28 - 2.54) / 0.42 = -3.0476. Then, we find the z-score for 3.8 using the same formula: z = (3.8 - 2.54) / 0.42 = 3.0476. Using a standard normal distribution table or a calculator, we can find the area between these two z-scores, which is approximately 99.7 percent.
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Penalty Shots in World Cup Soccer A study1 of 138 penalty shots in World Cup Finals games between 1982 and 1994 found that the goalkeeper correctly guessed the direction of the kick only 41% of the time. The article notes that this is ‘‘slightly worse than random chance." We use these data as a sample of all World Cup penalty shots ever. Test at a 5% significance level to see whether there is evidence that the percent guessed correctly is less than 50%. The sample size is large enough to use the normal distribution. The standard error from a randomization distribution under the null hypothesis is SE=0.043. 1St.John, A., ‘‘Physics of a World Cup Penalty-Kick Shootout - 2010 World Cup Penalty Kicks," Popular Mechanics, June 14, 2010.
Final answer:
To test whether the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer, we can use a one-sample proportion test. Using a significance level of 0.05, we find that the test statistic is -2.09. Comparing this to the critical value from the standard normal distribution (-1.645), we reject the null hypothesis and conclude that there is evidence to suggest that the percentage of correctly guessed penalty shots is less than 50%.
Explanation:
To test whether there is evidence that the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer, we can use a one-sample proportion test. We will assume that the null hypothesis is true and that the goalkeeper's correct guesses are no better than random chance. The alternative hypothesis would be that the goalkeeper's correct guesses are significantly less than 50%. Using a significance level of 0.05, we can calculate the test statistic and compare it to the critical value from the standard normal distribution.
Null hypothesis (H0): The percentage of correctly guessed penalty shots is 50%.
Alternative hypothesis (Ha): The percentage of correctly guessed penalty shots is less than 50%.
Test statistic: We can use the z-test statistic since the sample size is large enough. The formula for the z-test statistic is z = (p - P0) / SE, where p is the sample proportion, P0 is the hypothesized proportion, and SE is the standard error. In this case, since the standard error is given as 0.043, we can plug in the values to calculate the test statistic.
Calculate the z-test statistic: z = (0.41 - 0.5) / 0.043 = -2.09
Find the critical value: Since our alternative hypothesis is that the percentage is less than 50%, we will use a one-tailed test. With a significance level of 0.05, the critical value from the standard normal distribution is -1.645.
Compare the test statistic to the critical value: Since the test statistic (-2.09) is less than the critical value (-1.645), we can reject the null hypothesis. There is evidence to suggest that the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer.
Need help with this please thank you!
Answer:
4
Step-by-step explanation:
AC = AE = 4 cm
According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones, etc.) in back-to-college spending per student. Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54. If a family of a returning college student is randomly selected, what is the probability that: (a) They spend less than $160 on back-to-college electronics
Answer:
7.64% probability that they spend less than $160 on back-to-college electronics
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 237, \sigma = 54[/tex]
Probability that they spend less than $160 on back-to-college electronics
This is the pvalue of Z when X = 160. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{160 - 237}{54}[/tex]
[tex]Z = -1.43[/tex]
[tex]Z = -1.43[/tex] has a pvalue of 0.0763
7.64% probability that they spend less than $160 on back-to-college electronics