Design and complete a frequency table for Belinda.
Belinda ask 20 people, how many hours of TV did you watch last week?
Here is the results
3,17,4,4,6,11,14,14,1,20,9,8,9,6,12,7,8,13,13,9.
Belinda wants to show these result in a frequency table.
She will use 4 equal groups.
The first group will start with 1 hour and the last group will end with 20 hours.
Answer:
Step-by-step explanation:
Since she will use 4 groups or class intervals, the the class width would be 20/4 = 5 hours
The class groups would be
1 to 5
5 to 10
10 to 15
15 to 20
The class mark for each class is the average of the minimum and maximum value of each class. Therefore, the class marks are
(1 + 5)/2 = 3
(5 + 10)/2 = 7.5
(10 + 15)/2 = 12.5
(15 + 20)/2 = 17.5
The frequency table would be
Class group Frequency
1 - 5 4
5 - 10 8
10 - 15 6
15 - 20 2
The total frequency is 4 + 8 + 6 + 2 = 20
1. In an arithmetic sequence, the first term is -2, the fourth term is 16, and the n-th term is 11,998
(a) Find the common difference d
(b) Find the value of n.
pls help...
Answer:
see explanation
Step-by-step explanation:
The n th term of an arithmetic sequence is
[tex]a_{n}[/tex] = a₁ + (n - 1)d
(a)
Given a₁ = - 2 and a₄ = 16, then
a₁ + 3d = 16 , that is
- 2 + 3d = 16 ( add 2 to both sides )
3d = 18 ( divide both sides by 3 )
d = 6
--------------
(b)
Given
[tex]a_{n}[/tex] = 11998 , then
a₁ + (n - 1)d = 11998 , that is
- 2 + 6(n - 1) = 11998 ( add 2 to both sides )
6(n - 1) = 12000 ( divide both sides by 6 )
n - 1 = 2000 ( add 1 to both sides )
n = 2001
------------------
Find the constant of variation for the relation and use it to write and solve the equation.
if y varies directly as x and as the square of z, and y=25/9 when x=5 and z=1, find y when x=1 and z=4
Answer:
When x = 1 and z = 4, [tex]y=\frac{80}{9}[/tex]
Step-by-step explanation:
The variation described in the problem can be written using a constant of proportionality "b" as:
[tex]y=b\,\,x\,\,z^2[/tex]
The other piece of information is that when x = 5 and z = 1, then y gives 25/9. So we use this info to find the constant "b":
[tex]y=b\,\,x\,\,z^2\\\frac{25}{9} =b\,\,(5)\,\,(1)^2\\\frac{25}{9} =b\,\,(5)\\b=\frac{5}{9}[/tex]
Knowing this constant, we can find the value of y when x=1 and z=4 as:
[tex]y=b\,\,x\,\,z^2\\y=\frac{5}{9} \,\,x\,\,z^2\\y=\frac{5}{9} \,\,(1)\,\,(4)^2\\y=\frac{5*16}{9}\\y=\frac{80}{9}[/tex]
a football team had 50 players at the start of the season, but then some players left the team. After that, the team had 42 players
Answer:
50 = p + 42
Step-by-step explanation:
The unknown part of this equation is the variable p, the number of people that left. So you want to add p to 42 and that will give you the total number of football players, which is 50. In order to get p, you need to get it by itself and make it equal something. Subtract 42 from both sides and you are stuck with 50-42 = p
p = 8
Answer:
50-p=42
Step-by-step explanation:
A commuter uses a bus and a train to get to work. The train is more than 5 minutes late 1/6 of the times they use it The bus is more than 5 minutes late 3/5 of the times they use it. What is the probability that both the bus and train will be more than 5 minutes late?
Answer:
10% probability that both the bus and train will be more than 5 minutes late
Step-by-step explanation:
Independent events:
If two events, A and B, are independent, we have that:
[tex]P(A \cap B) = P(A)*P(B)[/tex]
What is the probability that both the bus and train will be more than 5 minutes late?
The bus being more than 5 minutes late is independent of the train, and vice versa. So
Event A: Bus more than 5 minutes late
Event B: Train more than 5 minutes late
The train is more than 5 minutes late 1/6 of the times they use it
This means that [tex]P(B) = \frac{1}{6}[/tex]
The bus is more than 5 minutes late 3/5 of the times they use it.
This means that [tex]P(A) = \frac{3}{5}[/tex]
Then
[tex]P(A \cap B) = \frac{3}{5}*\frac{1}{6} = \frac{3}{30} = 0.1[/tex]
10% probability that both the bus and train will be more than 5 minutes late
What’s the correct answer for this?
Answer:
C.
Step-by-step explanation:
Density = Mass / Volume
2.7 = 54 / V
V = 54 / 2.7
V = 20 cubic cm
Isabella averages 152 points per bowling game with a standard deviation of 14.5 points. Suppose Isabella's points per bowling game are normally distributed. Let X= the number of points per bowling game. Then X∼N(152,14.5)______.
If necessary, round to three decimal places.
Suppose Isabella scores 187 points in the game on Sunday. The z-score when x=187 is ___ The mean is _________
This z-score tells you that x = 187 is _________ standard deviations.
Answer:
The z-score when x=187 is 2.41. The mean is 187. This z-score tells you that x = 187 is 2.41 standard deviations above the mean.
Step-by-step explanation:
Problems of normally distributed samples are 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 question:
[tex]\mu = 152, \sigma = 14.5[/tex]
The z-score when x=187 is ...
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{187 - 152}{14.5}[/tex]
[tex]Z = 2.41[/tex]
The z-score when x=187 is 2.41. The mean is 187. This z-score tells you that x = 187 is 2.41 standard deviations above the mean.
Which expression is equivalent to 5^10 times 5^5. 5^2 5^5 5^15 5^50
Answer:
5^15
Step-by-step explanation:
(5^10)(5^5)= 5^10+5= 5^15
Find an Equation of a line with the given slope that passes through the point. Write the equation in the form Ax + By=C
M=3/2, (7,-2) -problem
Bridge math sails
Module 4B2
Answer:
c = 24 can i get brainliest
Step-by-step explanation:
The sum of two fractions can always be written as a
Answer: decimal
Step-by-step explanation:
because i did this quiz
Find the general solution to y′′+6y′+13y=0. Give your answer as y=.... In your answer, use c1 and c2 to denote arbitrary constants and x the independent variable. Enter c1 as c1 and c2 as c2.
Answer:
[tex]y(x)=c_1e^{-3x} cos(2x)+c_2e^{-3x} sin(2x)[/tex]
Step-by-step explanation:
In order to find the general solution of a homogeneous second order differential equation, we need to solve the characteristic equation. This is basically as easy as solving a quadratic.
For a second order differential equation of type:
[tex]ay''+by'+cy=0[/tex]
Has characteristic equation:
[tex]a r^{2} +br+c=0[/tex]
Whose solutions [tex]r_1 , r_2 ,.., r_n[/tex] are the roots from which the general solution can be formed. There are three cases:
Real roots:
[tex]y(x)=c_1e^{r_1 x} +c_2e^{r_2 x}[/tex]
Repeated roots:
[tex]y(x)=c_1e^{r x} +c_1 xe^{r x}[/tex]
Complex roots:
[tex]y(x)=c_1e^{\lambda x}cos(\mu x) +c_2e^{\lambda x}sin(\mu x)\\\\Where:\\\\r_1_,_2=\lambda \pm \mu i[/tex]
Therefore:
The characteristic equation for:
[tex]y''+6y'+13y=0[/tex]
Is:
[tex]r^{2} +6r+13=0[/tex]
Solving for [tex]r[/tex] :
[tex]r_1_,_2= -3 \pm 2i[/tex]
So:
[tex]\mu = 2\\\\and\\\\\lambda=-3[/tex]
Hence, the general solution of the differential equation will be given by:
[tex]y(x)=c_1e^{-3x} cos(2x)+c_2e^{-3x} sin(2x)[/tex]
HELP! Let f(x) = x + 1 and g(x)=1/x The graph of (fg)(x) is shown below.
Answer:
Step-by-step explanation:
all numbers except y = 1
because (f*g)(x) = 1+1/x
and 1/x cannot be equal to 0
The following histogram shows the exam scores for a Prealgebra class. Use this histogram to answer the questions.Prealgebra Exam ScoresScores 70.5, 75.5, 80.5, 85.5, 90.5, 95.5, 100.5Frequency 0, 4, 8, 12, 16, 20, 24Step 1 of 5:Find the number of the class containing the largest number of exam scores (1, 2, 3, 4, 5, or 6).Step 2 of 5:Find the upper class limit of the third class.Step 3 of 5:Find the class width for this histogram.Step 4 of 5:Find the number of students that took this exam.Step 5 of 5:Find the percentage of students that scored higher than 95.595.5. Round your answer to the nearest percent.
Answer:
The number of the class containing the largest score can be found in frequency 24 and the class is 98 - 103
For the third class 78 - 83 ; the upper limit = 83
The class width for this histogram 5
The number of students that took the exam simply refers to the frequency is 84
The percentage of students that scored higher than 95.5 is 53%
Step-by-step explanation:
The objective of this question is to use the following histogram that shows the exam scores for a Pre-algebra class to answer the question given:
NOW;
The table given in the question can be illustrated as follows:
S/N Class Score Frequency
1 68 - 73 70.5 0
2 73 - 78 75.5 4
3 78 - 83 80.5 8
4 83 - 88 85.5 12
5 88 - 93 90.5 16
6 93 - 98 95.5 20
7 98 - 103 100.5 24
TOTAL: 84
a) The number of the class containing the largest score can be found in frequency 24 and the class is 98 - 103
b) For the third class 78 - 83 ; the upper limit = 83 ( since the upper limit is derived by addition of 5 to the last number showing in the highest value specified by the number in the class interval which is 78 ( i.e 78 + 5 = 83))
c) The class width for this histogram 5 ; since it is the difference between the upper and lower boundaries limit of the given class.
So , from above the difference in any of the class will definitely result into 5
d) The number of students that took the exam simply refers to the frequency ; which is (0+4+8+12+16+20+24) = 84
e) Lastly; the percentage of students that scored higher than 95.5 is ;
⇒[tex]\dfrac{20+24}{84} *100[/tex]
= 0.5238095 × 100
= 52.83
To the nearest percentage ;the percentage of students that scored higher than 95.5 is 53%
Answer:
1. 98-103 (6th class)
2. 88
3. 5
4. 84
5. 52%
Step-by-step explanation:
Find attached the frequency table.
The class of exam scores falls between (1, 2, 3, 4, 5, or 6).
The exam score ranged from 68-103
1) The largest number of exam scores = 24
The largest number of exam scores is in the 6th class = 98 -103
Step 2 of 5:
The upper class limit is the higher number in an interval. Third class interval is 83-88
The upper class limit of the third class 88.
Step 3 of 5:
Class width = upper class limit - lower class limit
We can use any of the class interval to find this as the answer will be the same. Using the interval between 73-78
Class width = 78 - 73
Class width for the histogram = 5
Step 4 of 5:
The total of students that took the test = sum of all the frequency
= 0+4+8+12+16+20+24 = 84
The total of students that took the test = 84
Step 5 of 5:Find the percentage of students that scored higher than 95.5
Number of student that scored higher than 95.5 = 20 + 24 = 44
Percentage of students that scored higher than 95.5 = [(Number of student that scored higher than 95.5)/(total number of students that took the test)] × 100
= (44/84) × 100 = 0.5238 × 100 = 52.38%
Percentage of students that scored higher than 95.5 = 52% (nearest percent)
If you have changed the tires on your car, the original diameter is 24.5 inches. to a new diameter of 26 inches, how fast are you actually going if your speedometer is reading 53 mph? A. 50.5 mph B. 53 mph C. 56.2 mph D. 62.8 mph
Answer: c) 56.2
Step-by-step explanation:
Compare the original rate to the the new rate:
[tex]\dfrac{diameter}{mph}:\dfrac{24.5\ in}{53\ mph}=\dfrac{26\ in}{x}\\\\\\24.5x=53(26)\\\\\\x=\dfrac{53(26)}{24.5}\\\\\\x=\large\boxed{56.2\ mph}[/tex]
which of the following explains expressions are equivalent to - 5/6 /-1/3
Answer:
2.5
Step-by-step explanation:
(-5/6 ) / (-1/3)
multiply the numerator and denominator by the same number -3 gives:
(-5 * -3 /6 ) / (-1* -3/3)
(15/6 ) / (3/3)
(15/6 ) / 1
(15/6 )
12/6 + 3/6
2 3/6
2 1/2
2.5
For what value of the variable will the value of 7y−2 be ten more than the value of 2y?
Answer:
y=2.4
Step-by-step explanation:
7y-2=2y+10
7y-2y=10+2
5y=12
y=12/5=2.4
Please answer this correctly
Answer:
Bailey: 16%
Coco: 28%
Ginger: 32%
Ruby: 24%
I hope this helps!
According to a polling organization, 22% of adults in a large region consider themselves to be liberal. A survey asked 200 respondents to disclose their political philosophy: Conservative, Liberal, Moderate. Treat the results of the survey as a random sample of adults in this region. Do the survey results suggest the proportion is higher than that reported by the polling organization? Use an alphaequals0.01 level of significance.
Answer:
No. There is not enough evidence to support the claim that the proportion of liberals is higher than that reported by the polling organization (P-value = 0.0366).
Step-by-step explanation:
The question is incomplete: there is no information about the results of the survey. We will assume that 55 of the subjects answer "liberal", and test the claim.
This is a hypothesis test for a proportion.
The claim is that the proportion of liberals is higher than that reported by the polling organization.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.22\\\\H_a:\pi>0.22[/tex]
The significance level is 0.01.
The sample has a size n=200.
The sample proportion is p=0.275.
[tex]p=X/n=55/200=0.275[/tex]
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.22*0.78}{200}}\\\\\\ \sigma_p=\sqrt{0.000858}=0.029[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.275-0.22-0.5/200}{0.029}=\dfrac{0.053}{0.029}=1.792[/tex]
This test is a right-tailed test, so the P-value for this test is calculated as:
[tex]\text{P-value}=P(z>1.792)=0.0366[/tex]
As the P-value (0.0366) is greater than the significance level (0.01), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the proportion of liberals is higher than that reported by the polling organization.
A spinner has 10 equally sized sections, 8 of which are gray and 2 of which are blue. The spinner is spun twice. What is the probability that the first spin lands on gray and the second spin lands on blue ? Write your answer as a fraction in simplest form.
Answer:
4/25
Step-by-step explanation:
The probability the first spin lands on gray is 8/10 = 4/5.
The probability the second spin lands on blue is 2/10 = 1/5.
The probability of both events is 4/5 × 1/5 = 4/25.
Find the area of a triangle bounded by the y-axis, the line f(x)=9−2/3x, and the line perpendicular to f(x) that passes through the origin. Area =
Answer:
The area of the triangle is 18.70 sq.units.
Step-by-step explanation:
It is provided that a triangle is bounded by the y-axis, the line [tex]f(x)=y=9-\frac{2}{3}x[/tex].
The slope of the line is: [tex]m_{1}=-\frac{2}{3}[/tex]
A perpendicular line passes through the origin to the line f (x).
The slope of this perpendicular line is:[tex]m_{2}=-\frac{1}{m_{1}}=\frac{3}{2}[/tex]
The equation of perpendicular line passing through origin is:
[tex]y=\frac{3}{2}x[/tex]
Compute the intersecting point between the lines as follows:
[tex]y=9-\frac{2}{3}x\\\\\frac{3}{2}x=9-\frac{2}{3}x\\\\\frac{3}{2}x+\frac{2}{3}x=9\\\\\frac{13}{6}x=9\\\\x=\frac{54}{13}[/tex]
The value of y is:
[tex]y=\frac{3}{2}x=\frac{3}{2}\times\frac{54}{13}=\frac{81}{13}[/tex]
The intersecting point is [tex](\frac{54}{13},\ \frac{81}{13})[/tex].
The y-intercept of the line f (x) is, 9, i.e. the point is (0, 9).
So, the triangle is bounded by the points:
(0, 0), (0, 9) and [tex](\frac{54}{13},\ \frac{81}{13})[/tex]
Consider the diagram attached.
Compute the area of the triangle as follows:
[tex]\text{Area}=\frac{1}{2}\times 9\times \frac{54}{13}=18.69231\approx 18.70[/tex]
Thus, the area of the triangle is 18.70 sq.units.
The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is randomly selected, find the probability that the class length is between 50.1 and 51.1 min. P(50.1 < X < 51.1) =
Answer:
P(50.1 < X < 51.1) = 0.5
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability that we find a value X between c and d is given by the following formula:
[tex]P(c < X < d) = \frac{d - c}{b - a}[/tex]
The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min.
This means that [tex]a = 50, b = 52[/tex]
So
[tex]P(50.1 < X < 51.1) = \frac{51.1 - 50.1}{52 - 50} = 0.5[/tex]
The average math SAT score is 511 with a standard deviation of 119. A particular high school claims that its students have unusually high math SAT scores. A random sample of 55 students from this school was selected, and the mean math SAT score was 528. Is the high school justified in its claim? Explain. ▼ No Yes , because the z-score ( nothing) is ▼ unusual not unusual since it ▼ does not lie lies within the range of a usual event, namely within ▼ 1 standard deviation 2 standard deviations 3 standard deviations of the mean of the sample means. (Round to two decimal places as needed.)
Answer:
No, because the z-score of Z = 1.06 is not unusual since it does not lie within the range of a usual event, namely within 2 standard deviations of the mean of the sample means.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use 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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Unusual
If X is more than two standard deviations from the mean, x is considered unusual.
In this question:
[tex]\mu = 511, \sigma = 119, n = 55, s = \frac{119}{\sqrt{55}} = 16.046[/tex]
A random sample of 55 students from this school was selected, and the mean math SAT score was 528. Is the high school justified in its claim?
If Z is equal or greater than 2, the claim is justified.
Lets find Z.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{528 - 511}{16.046}[/tex]
[tex]Z = 1.06[/tex]
1.06 < 2, so 528 is not unusually high.
The answer is:
No, because the z-score of Z = 1.06 is not unusual since it does not lie within the range of a usual event, namely within 2 standard deviations of the mean of the sample means.
The statement that could be made regarding the high school about the justification of its claim would be:
- No, because the z-score of Z = 1.06 is not unusual since it does not lie within the range of a usual event, namely within 2 standard deviations of the mean of the sample means.
Given that,
μ = 511
σ = 119
Sample(n) = 55
and
s = [tex]119/\sqrt{55}[/tex]
[tex]= 16.046[/tex]
As we know,
The claim of the high school could be valid and justified only when
[tex]Z > 2[/tex]
To find,
The value of Z
So,
[tex]Z = (X -[/tex] μ )/σ
by putting the values using Central Limit Theorem,
[tex]Z = (528 - 511)/16.046[/tex]
∵ [tex]Z = 1.06[/tex]
Since [tex]Z < 2[/tex], the claim is not justified.
Learn more about "Standard Deviation" here:
brainly.com/question/12402189
Determine whether the data described below are qualitative or quantitative and explain why.
The area codes (such as 617 )of the telephones of survey respondents:
a. The data are quantitative because they consist of counts or measurements.
b. The data are quantitative because they don't measure or count anything.
c. The data are qualitative because they don't measure or count anything.
d. The data are qualitative because they consist of counts or measurements.
Answer:
c. The data are qualitative because they don't measure or count anything.
Step-by-step explanation:
In the case of the area codes, the value although is a number and follows some logic, it does not represent a quantity and any mathematical operation on it has no meaning. The number does not measure or count anything.
They have the same meaning as the name of the city or the area.
Simplify the following expression:
-5[(x^3 + 1)(x + 4)]
Answer:
[tex]-5x^{4} -20x^{3} -5x-20[/tex]
Step-by-step explanation:
[tex]-5[(x^{3} +1)(x+4)][/tex]
Use the FOIL method for the last two groups.
[tex]-5(x^{4} +4x^{3} +x+4)[/tex]
Now, distribute the -5 into each term.
[tex]-5x^{4} -20x^{3} -5x-20[/tex]
On Sunday, a local hamburger shop sold a combined of 572 hamburger and cheeseburger. The number of cheeseburgers sold was three times the number of hamburger sold. How many hamburger were sold on Sunday
Given that d is the midpoint of line segment ab and k is the midpoint of line segment bc, which statement must be true? (May give brainliest)
Answer:
B is the midpoint of line segment AC
A random sample of math majors taking an introductory statistics course were surveyed after completing the final exam. They were asked, "How many times did you review your final exam before handing it in to the professor?" The results are displayed in a probability density function for the random variable X, the number of times students reviewed their exam before handing it in. Find the standard deviation of X. Round the final answer to two decimal places. x P(X = x) 1 1/5 2 2/5 7 2/5
Answer:
[tex] E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8[/tex]
Now we can find the second moment with this formula:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]
And replacing we got:
[tex] E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4[/tex]
The variance would be given by:
[tex] Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96[/tex]
And the deviation would be:
[tex] Sd(X) =\sqrt{6.96}= 2.638[/tex]
Step-by-step explanation:
For this case we have the following distribution given:
X 1 2 7
P(X) 1/5 2/5 2/5
We need to begin finding the mean with this formula:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]
And replacing we got:
[tex] E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8[/tex]
Now we can find the second moment with this formula:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]
And replacing we got:
[tex] E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4[/tex]
The variance would be given by:
[tex] Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96[/tex]
And the deviation would be:
[tex] Sd(X) =\sqrt{6.96}= 2.638[/tex]
9. In 2002 the Georgia department of education reported a mean reading test score of 850 from Tattnall County Career Academy with a standard deviation of 50. The sample was taken from 100 11th grade students. Assuming the test scores are normally distributed, what is the standard error
Answer:
The standard error = 5
Step-by-step explanation:
Explanation:-
Given sample size 'n' = 100
Given mean reading test score μ = 850
Given standard deviation of the population 'σ' = 50
The standard error is determined by
Standard error = [tex]\frac{S.D}{\sqrt{n} }[/tex]
S.E = σ/√n
[tex]S.E = \frac{S.D}{\sqrt{n} } = \frac{50}{\sqrt{100} } = 5[/tex]
Final answer:-
The standard error ( S.E) = 5
Which number is irrational
Answer:
Can you give the question. Can you post the picture. I can help solve. I will edit this answer once you have given the question/picture.
What else would need to be congruent to show that ABC=DEF by SAS?
Answer:
A
Step-by-step explanation:
Answer:
The answer here is A.
A) A is congruent to D.
A=
Step-by-step explanation:
AP E
