Answer: 8
Hope this help :)
Solve x∕3 < 5 Question 5 options: A) x ≥ 15 B) x > 15 C) x < 15 D) x ≤ 15
Answer:
C
Step-by-step explanation:
Given
[tex]\frac{x}{3}[/tex] < 5 ( multiply both sides by 3 to clear the fraction )
x < 15 → C
the number of multiples of a given number is infinite ( )
Answer:
make an 8 horizontal
oooookkkk
Answer:
TRUE
The number of multiples of a given number is finite is a false statement. The number of multiples of a given number is infinite.
Examples:
Multiples of 2 = 2,4,6,8,10,…..
Multiples of 3 = 3,6,9,12,15,18,…
Multiples of 4 = 4, 8, 12, 16, 120, 24….
∴ The number of multiples of a given number is infinite .
Answer From Gauth Math
Use the commutative law of multiplication to rewrite 67 x 13.
A. 3 X 671
B. 13 x 67
C.6 X 7 X1 X3
D.80
Answer:
A. 671*3
B. 67*13
C. 3*1*7*6
D. 1*80
QUESTION 3.1 POINT
An investment pays 25% interest compounded monthly. What percent, as a decimal, is the effective annual yied? Enter your
answer as a decimal rounded to four decimal places.
9514 1404 393
Answer:
0.2807
Step-by-step explanation:
The relationship between the effective annual yield (e) and the nominal annual interest rate (r) compounded n times per year is ...
e = (1 +r/n)^n -1
e = (1 +0.25/12)^12 -1 = 0.2807 . . . . . . about 28.07%
190 of 7
6 7 8 9 10
-3
4
5
6
The slope of the line shown in the graph is
and the intercept of the line is
Answer:slope 2/3
Y-int 6
Step-by-step explanation:
PLEASEE HELP ME ASAPPP (geometry)
Answer:AE=EC và BF=FC => EF là đường trung bình của tam giác ABC
=> EF // và bằng 1/2 AB
=> AB = 16
Step-by-step explanation:
Answer:
AB=16
Step-by-step explanation:
Midsegment Theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long.
The mid-segment of a triangle, which joins the midpoints of two sides of a triangle, is parallel to the third side of the triangle and half the length of that third side of the triangle.
AD=DB
AD+DB=AB=2EF
AB=2×8=16
pls help me asap !!!
Answer:
11
Step-by-step explanation:
Hopefully you can see that this is an isosceles triangle and remembering the inequality theorem of a triangle (4,4,11 triangle cannot exist). Iso triangle has two side the same length - as well as two angles the same.
The distance between Ali's house and 1 point
college is exactly 135 miles. If she
drove 2/3 of the distance in 135
minutes. What was her average speed
in miles per hour?
Ali's average speed was 40 miles per hour.
What is an average speed?
The total distance traveled is to be divided by the total time consumed brings us the average speed.
How to calculate the average speed of Ali?
The total distance between the college from Ali's house is 135 miles.
She drove 2/3rd of the total distance in 135 minutes.
She drove =135*2/3miles
=90miles.
Ali can drive 90miles in 135 mins.
Therefore, her average speed is: 90*60/135 miles per hour.
=40 miles per hour.
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a farmer needs 5 men to clear his farm in 10 days. How many men will he need if he must finish clearing the farm in two days if they work at the same rate?
Answer:
25 workers
Step-by-step explanation:
If you like my answer than please mark me brainliest thanks
,
Evaluate:
[tex]{ \int \limits^\pi_{ \frac{1}{4}\pi}{ {e {}^{2 \sigma} (\sqrt{1 - { \sigma}^{2} } ) d \sigma}}}[/tex]
Answer:
hope this answer helps.
The PTA sells 100 tickets for a raffle and puts them in a bowl. They will randomly pull out a ticket for the first prize and then another ticket for the second prize. You have 10 tickets and your friend has 10 tickets. What is the probability that your friend wins the first prize and you win the second prize?
Doyle Company issued $500,000 of 10-year, 7 percent bonds on January 1, 2018. The bonds were issued at face value. Interest is payable in cash on December 31 of each year. Doyle immediately invested the proceeds from the bond issue in land. The land was leased for an annual $125,000 of cash revenue, which was collected on December 31 of each year, beginning December 31, 2018
Answer:
f
Step-by-step explanation:
4. Manuel swims at a speed of 1 yard per second. How many feet per minute does he swim?
Answer:
180 Feet Per Minute
Step-by-step explanation:
Please mark me as brainliest and rate 5 stars!
If lan does a job in 132 hours and with the help of Danielle they can do it together in 44 hours, how long
would it take Danielle to do it alone
Answer:
88 hours
Step-by-step explanation:
Ian = 132
Daniel + Ian = 44
Daniel = Ian - 44
= 132-44 = 88 hours
If a normally distributed population has a mean (mu) that equals 100 with a standard deviation (sigma) of 18, what will be the computed z-score with a sample mean (x-bar) of 106 from a sample size of 9?
Answer:
Z = 1
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes 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.
Mean (mu) that equals 100 with a standard deviation (sigma) of 18
[tex]\mu = 100, \sigma = 18[/tex]
Sample of 9:
This means that [tex]n = 9, s = \frac{18}{\sqrt{9}} = 6[/tex]
What will be the computed z-score with a sample mean (x-bar) of 106?
This is Z when X = 106. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{106 - 100}{6}[/tex]
[tex]Z = 1[/tex]
So Z = 1 is the answer.
Add :-
a+2b-3c, -3a+b+2cand 2a -3b+c
Answer:
[tex]a + 2b -3 c + - 3a + b + 2c + 2a - 3b + c \\ = a - 3a + 2a + 2b + b - 3b - 3c + 2c + c \\ 0a + 0b + 0c \\ thank \: you[/tex]
a+2b-3c+(-3a+b+2c) +(2a-3b+c)
=a+2b-3c-3a+b+2c+2a-3b+c
=a-3a+2b+2b+b-3b-3c++2b+c
=0a+0b+0c
=0
Therefore, the addition of the expressions, a+2b-3c+(-3a+b+2c) +(2a-3b+c) is zero or 0.
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Use the tables to answer the question.
How can you determine which store offers a better price on guitar lessons?
A. You cannot determine this from the information given.
B. Compare the most expensive price from each store’s table.
C. Compare the least expensive price from each store’s table.
D. Find the cost of 1 week of guitar lessons for each store and compare.
Answer:
D. Find the cost of 1 week of guitar lessons for each store and compare.
Step-by-step explanation:
Find the cost for 1 week of guitar lessons by finding the slope of the data.
[tex]\frac{y_2-y_1}{x_2-x_1}\\\\ A.)\frac{132-66}{6-3}=\frac{66}{3}=22\\\\ B.)\frac{168-84}{8-4}=\frac{84}{4}=21[/tex]
Therefore, Store B would be the best choice.
Solve. x+y+z=6 3x−2y+2z=2−2x−y+3z=−4
Answer:
-4?
hope dis helps ^-^
There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.10 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation 0.04 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm. What is the probability that the first machine produces an acceptable cork
Answer:
0.6827
Step-by-step explanation:
Given that :
Mean, μ = 3
Standard deviation, σ = 0.1
To produce an acceptable cork. :
P(2.9 < X < 3.1)
Recall :
Z = (x - μ) / σ
P(2.9 < X < 3.1) = P[((2.9 - 3) / 0.1) < Z < ((3.1 - 3) / 0.1)]
P(2.9 < X < 3.1) = P(-1 < Z < 1)
Using a normal distribution calculator, we obtain the probability to the right of the distribution :
P(2.9 < X < 3.1) = P(1 < Z < - 1) = 0.8413 - 0.1587 = 0.6827
Hence, the probability that the first machine produces an acceptable cork is 0.6827
The ratio of frogs to toads was 3 to 7. If there were 1280 frogs and toads in all, how many were frogs?
Answer:
348 frogs
Step-by-step explanation:
ratio = 3:7
total of ratio = 10
frogs = 3/10 × 1280 = 348 frogs
Answer:
let the ratio be 3x and 7x.
3x+7x=1280
10x=1280
x=128
Now
frogs =3x=3*128=384
toads =7x=7*128=896
hlo anyone free .... im bo r ed
d
Step-by-step explanation:
Excuse me! Who r u? where r u frm? tell me tht frst.
Answer:
Oop
Step-by-step explanation:
I’m bored
The resistors produced by a manufacturer are required to have an average resistance of 0.150 ohms. Statistical analysis of the output suggests that the resistances can be approximated by a normal distribution with known standard deviation of 0.005 ohms. We are interested in testing the hypothesis that the resistors conform to the specifications.
Requied:
a. Determine whether a random sample of 10 resistors yielding a sample mean of 0.152 ohms indicates that the resistors are conforming. Use alpha = 0.05.
b. Calculate a 95% confidence interval for the average resistance. How does this interval relate to your solution of part (a)?
Answer:
a) The p-value of the test is 0.2076 > 0.05, which means that the sample indicates that the resistors are conforming.
b) The 95% confidence interval for the average resistance is (0.147, 0.153). 0.152 is part of the confidence interval, which means that as the test statistic in item a), it indicates that the resistors are conforming.
Step-by-step explanation:
Question a:
The resistors produced by a manufacturer are required to have an average resistance of 0.150 ohms.
At the null hypothesis, we test if this is the average resistance, that is:
[tex]H_0: \mu = 0.15[/tex]
We are interested in testing the hypothesis that the resistors conform to the specifications.
At the alternative hypothesis, we test if it is not conforming, that is, the mean is different of 0.15, so:
[tex]H_1: \mu \neq 0.15[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.15 is tested at the null hypothesis:
This means that [tex]\mu = 0.15[/tex]
Sample mean of 0.152, sample of 10, population standard deviation of 0.005.
This means that [tex]X = 0.152, n = 10, \sigma = 0.005[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.152 - 0.15}{\frac{0.005}{\sqrt{10}}}[/tex]
[tex]z = 1.26[/tex]
P-value of the test and decision:
The p-value of the test is the probability of the sample mean differing from 0.15 by at least 0.152 - 0.15 = 0.002, which is P(|z| > 1.26), given by two multiplied by the p-value of z = -1.26.
Looking at the z-table, z = -1.26 has a p-value of 0.1038.
2*0.1038 = 0.2076
The p-value of the test is 0.2076 > 0.05, which means that the sample indicates that the resistors are conforming.
Question b:
We have to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]
Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a p-value of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.96\frac{0.005}{\sqrt{10}} = 0.003[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 0.15 - 0.003 = 0.147.
The upper end of the interval is the sample mean added to M. So it is 0.15 + 0.003 = 0.153.
The 95% confidence interval for the average resistance is (0.147, 0.153). 0.152 is part of the confidence interval, which means that as the test statistic in item a), it indicates that the resistors are conforming.
Please help!!
Find BD
Answer: [tex]8\sqrt{2}[/tex]
==========================================================
Work Shown:
Focus entirely on triangle ABD (or on triangle BCD; both are identical)
The two legs of this triangle are AB = 8 and AD = 8. The hypotenuse is unknown, so we'll say BD = x.
Apply the pythagorean theorem.
[tex]a^2 + b^2 = c^2\\\\c = \sqrt{a^2 + b^2}\\\\x = \sqrt{8^2 + 8^2}\\\\x = \sqrt{2*8^2}\\\\x = \sqrt{8^2*2}\\\\x = \sqrt{8^2}*\sqrt{2}\\\\x = 8\sqrt{2}\\\\[/tex]
So that's why the diagonal BD is exactly [tex]8\sqrt{2}\\\\[/tex] units long
Side note: [tex]8\sqrt{2} \approx 11.3137[/tex]
5x-22 3x +105 x minus 22 3 X + 10
-291x+10
:)))))) Have fun
-5 + 3 and also what is 1/4 of 24
What is the answer i am struggling
Answer:
-5+3=-2
1/4 of 24 = 6
Step-by-step explanation:
A rectangle has a length of 7 in. and a width of 2 in. if the rectangle is enlarged using a scale factor of 1.5, what will be the perimeter of the new rectangle
Answer:
27 inch
Step-by-step explanation:
Current perimeter=18
New perimeter=18*1.5=27 in
A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with the mean 3, find the probability that the shop sells. . (a) At least 3 in a week. (b) At most 7 in a week. (c) More than 20 in a month (4 weeks).
Answer:
a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.
b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.
c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.
Step-by-step explanation:
For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]
In which
x is the number of successes
e = 2.71828 is the Euler number
[tex]\lambda[/tex] is the mean in the given interval.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
The Poisson distribution can be approximated to the normal with [tex]\mu = \lambda, \sigma = \sqrt{\lambda}[/tex], if [tex]\lambda>10[/tex].
Poisson variable with the mean 3
This means that [tex]\lambda= 3[/tex].
(a) At least 3 in a week.
This is [tex]P(X \geq 3)[/tex]. So
[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]
In which:
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
Then
[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498[/tex]
[tex]P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494[/tex]
[tex]P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240[/tex]
So
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768[/tex]
0.5768 = 57.68% probability that the shop sells at least 3 in a week.
(b) At most 7 in a week.
This is:
[tex]P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498[/tex]
[tex]P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494[/tex]
[tex]P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240[/tex]
[tex]P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240[/tex]
[tex]P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680[/tex]
[tex]P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008[/tex]
[tex]P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504[/tex]
[tex]P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216[/tex]
Then
[tex]P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988[/tex]
0.988 = 98.8% probability that the shop sells at most 7 in a week.
(c) More than 20 in a month (4 weeks).
4 weeks, so:
[tex]\mu = \lambda = 4(3) = 12[/tex]
[tex]\sigma = \sqrt{\lambda} = \sqrt{12}[/tex]
The probability, using continuity correction, is P(X > 20 + 0.5) = P(X > 20.5), which is 1 subtracted by the p-value of Z when X = 20.5.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20 - 12}{\sqrt{12}}[/tex]
[tex]Z = 2.31[/tex]
[tex]Z = 2.31[/tex] has a p-value of 0.9896.
1 - 0.9896 = 0.0104
0.0104 = 1.04% probability that the shop sells more than 20 in a month.
The probability of the selling the video recorders for considered cases are:
P(At least 3 in a week) = 0.5768 approximately.P(At most 7 in a week) = 0.9881 approximately.P( more than 20 in a month) = 0.0839 approximately.What are some of the properties of Poisson distribution?Let X ~ Pois(λ)
Then we have:
E(X) = λ = Var(X)
Since standard deviation is square root (positive) of variance,
Thus,
Standard deviation of X = [tex]\sqrt{\lambda}[/tex]
Its probability function is given by
f(k; λ) = Pr(X = k) = [tex]\dfrac{\lambda^{k}e^{-\lambda}}{k!}[/tex]
For this case, let we have:
X = the number of weekly demand of video recorder for the considered shop.
Then, by the given data, we have:
X ~ Pois(λ=3)
Evaluating each event's probability:
Case 1: At least 3 in a week.
[tex]P(X > 3) = 1- P(X \leq 2) = \sum_{i=0}^{2}P(X=i) = \sum_{i=0}^{2} \dfrac{3^ie^{-3}}{i!}\\\\P(X > 3) = 1 - e^{-3} \times \left( 1 + 3 + 9/2\right) \approx 1 - 0.4232 = 0.5768[/tex]
Case 2: At most 7 in a week.
[tex]P(X \leq 7) = \sum_{i=0}^{7}P(X=i) = \sum_{i=0}^{7} \dfrac{3^ie^{-3}}{i!}\\\\P(X \leq 7) = e^{-3} \times \left( 1 + 3 + 9/2 + 27/6 + 81/24 + 243/120 + 729/720 + 2187/5040\right)\\\\P(X \leq 7) \approx 0.9881[/tex]
Case 3: More than 20 in a month(4 weeks)
That means more than 5 in a week on average.
[tex]P(X > 5) = 1- P(X \leq 5) =\sum_{i=0}^{5}P(X=i) = \sum_{i=0}^{5} \dfrac{3^ie^{-3}}{i!}\\\\P(X > 5) = 1- e^{-3}( 1 + 3 + 9/2 + 27/6 + 81/24 + 243/120)\\\\P(X > 5) \approx 1 - 0.9161 \\ P(X > 5) \approx 0.0839[/tex]
Thus, the probability of the selling the video recorders for considered cases are:
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3 write the factor of the following (1) 48 (2) 36 (3) 28 (4) 100 (5) 125
Answer:
FACTORING THE NUMBERS :-
well u didnt say to prime factors so i am writing all factors
1) 48 => 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48 ( all are plus and minus )
2) 36 => 1, 2, 3, 4, 6, 9, 12, 18 and 36 ( all are plus and minus )
3) 28 => 1, 2, 4, 7, 14 and 28 ( all are plus and minus )
4) 100 => 1, 2, 4, 5, 10, 20, 25, 50, and 100 ( all are plus and minus )
5) 125 => 1, 5, 25, 125 ( all are plus and minus )
is it worth brainliest...
yes ofc
Police sometimes measure shoe prints at crime scenes so that they can learn something about criminals. Listed below are shoe print lengths, foot lengths, and heights of males. Construct a scatterplot, find the value of the linear correlation coefficient r, and find the P-value of r. Determine whether there is sufficient evidence to support a claim of linear correlation between the two variables. Based on these results, does it appear that police can use a shoe print length to estimate the height of a male? Use a significance level of α=0.01
It does not appear that police can use a shoe print length to estimate the height of a male.
The given parameters are:
[tex]\begin{array}{cccccc}{Shoe\ Print} & {28.6} & {29.4} & {32.2} & {32.4} & {27.3} \ \\ Height (cm) & {172.5} & {176.7} & {188.4} & {170.1} & {179.2} \ \end{array}[/tex]
Rewrite as:
[tex]\begin{array}{cccccc}{x} & {28.6} & {29.4} & {32.2} & {32.4} & {27.3} \ \\ y & {172.5} & {176.7} & {188.4} & {170.1} & {179.2} \ \end{array}[/tex]
See attachment for scatter plot
To determine the correlation coefficient, we extend the table as follows:
[tex]\begin{array}{cccccc}{x} & {28.6} & {29.4} & {32.2} & {32.4} & {27.3} & y & {172.5} & {176.7} & {188.4} & {170.1} & {179.2} & x^2 & {817.96} & {864.36} & {1036.84} & {1049.76} & {745.29} & y^2 & {29756.25} & {31222.89} & {35494.56} & {28934.01} & {32112.64} & x \times y & {4933.5} & {5194.98} & {6066.48} & {5511.24} & {4892.16} \ \end{array}[/tex]
The correlation coefficient (r) is:
[tex]r = \frac{\sum(x - \bar x)(y - \bar y)}{\sqrt{SS_x * SS_y}}[/tex]
We have:
[tex]n =5[/tex]
[tex]\sum xy =4933.5+5194.98+6066.48+5511.24+4892.16 =26598.36[/tex]
[tex]\sum x =28.6+29.4+32.2+32.4+27.3=149.9[/tex]
[tex]\sum y =172.5+176.7+188.4+170.1+179.2=886.9[/tex]
[tex]\sum x^2 =817.96+864.36+1036.84+1049.76+745.29=4514.21[/tex]
[tex]\sum y^2 =29756.25+31222.89+35494.56+28934.01+32112.64=157520.35[/tex]
Calculate mean of x and y
[tex]\bar x = \frac{\sum x}{n} = \frac{149.9}{5} = 29.98[/tex]
[tex]\bar y = \frac{\sum y}{n} = \frac{886.9}{5} = 177.38[/tex]
Calculate SSx and SSy
[tex]SS_x = \sum (x - \bar x)^2 =(28.6-29.98)^2 + (29.4-29.98)^2 + (32.2-29.98)^2 + (32.4-29.98)^2 + (27.3-29.98)^2 =20.208[/tex]
[tex]SS_y = \sum (y - \bar x)^2 =(172.5-177.38)^2 + (176.7-177.38)^2 + (188.4-177.38)^2 + (170.1-177.38)^2 + (179.2-177.38)^2 =202.028[/tex]
Calculate [tex]\sum(x - \bar x)(y - \bar y)[/tex]
[tex]\sum(x - \bar x)(y - \bar y) = (28.6-29.98)*(172.5-177.38) + (29.4-29.98)*(176.7-177.38) + (32.2-29.98)*(188.4-177.38) + (32.4-29.98)*(170.1-177.38) + (27.3-29.98) *(179.2-177.38) =9.098[/tex]
So:
[tex]r = \frac{\sum(x - \bar x)(y - \bar y)}{\sqrt{SS_x * SS_y}}[/tex]
[tex]r = \frac{9.098}{\sqrt{20.208 * 202.028}}[/tex]
[tex]r = \frac{9.098}{\sqrt{4082.581824}}[/tex]
[tex]r = \frac{9.098}{63.90}[/tex]
[tex]r = 0.142[/tex]
Calculate test statistic:
[tex]t = \frac{r}{\sqrt{\frac{1 - r^2}{n-2}}}[/tex]
[tex]t = \frac{0.142}{\sqrt{\frac{1 - 0.142^2}{5-2}}}[/tex]
[tex]t = \frac{0.142}{\sqrt{\frac{0.979836}{3}}}[/tex]
[tex]t = \frac{0.142}{\sqrt{0.326612}}[/tex]
[tex]t = \frac{0.142}{0.5715}[/tex]
[tex]t = 0.248[/tex]
Calculate the degrees of freedom
[tex]df = n - 2 = 5 - 2 = 3[/tex]
The [tex]t_{\alpha/2}[/tex] value at:
[tex]df =3[/tex]
[tex]t = 0.248[/tex]
[tex]\alpha = 0.01[/tex]
The value is:
[tex]t_{0.01/2} = \±5.841[/tex]
This means that we reject the null hypothesis if the t value is not between -5.841 and 5.841
We calculate the t value as:
[tex]t = 0.248[/tex]
[tex]-5.841 < 0.248 < 5.841[/tex]
Hence, we do not reject the null hypothesis because they do not appear to have any correlation.
Read more about regression at:
https://brainly.com/question/18405415
[tex]\sqrt{25}[/tex]
Answer:
5
Step-by-step explanation:
Calculate the square root of 25 and get 5.