Answer:
to get x you need to divide
210/42= 5
x=5
210 = 42x
210/42=5
42*5=210
so remove the x it'd be 210 = 42(5)
x = 5
A researcher has obtained the number of hours worked per week during the summer for a sample of 15 students. 40 25 35 30 20 40 30 20 40 10 30 20 10 5 20 Using this data set, compute the following: a. Median b. Mean c. Mode d. 40th percentile e. Range f. Sample variance g. Standard deviation
Answer:
Mean = 25
Median = 25
Mode = 6
Step-by-step explanation:
Given the data :
Using calculator :
Mode = highest occurring data point
Range = max - min = 40 - 5 = 35
Sample variance = 128.57 (calculator)
Sample standard deviation = sqrt(variance) = 11.34
17.
What is the value of the expression
2a + 5b + 3c for a = 12, b = 6, and
C=3?
A 10
B 21
C49
D 63
D. 63
2a+5b+3c
2(12)+5(6)+3(3)
24+30+9=63
Hope this helps! :)
What is the value of x for x +20 x=22
Answer:
x(1+20)=22
21x=22
x= 22/21
x= 1.05
Step-by-step explanation:
Answer:
2!!!
Step-by-step explanation:
the person is wrong.. 2+20=22
Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.
x + y - z = -2
2x - y + 3z = 9
x - 4y - 2z = 1
Answer:
x = 1 , y = - 1 , z = 2
Step-by-step explanation:
[tex]\begin{pmatrix}1&1&-1&-2\\ 2&-1&3&9\\ 1&-1&-2&1\end{pmatrix}\\\\\\= \begin{pmatrix}2&-1&3&9\\ 1&1&-1&-2\\ 1&-1&-2&1\end{pmatrix}[/tex] [tex][ \ swap \ R_1 \ and \ R_2 \ ][/tex]
[tex]=\begin{pmatrix}2&-1&3&9\\ 0&\frac{3}{2}&-\frac{5}{2}&-\frac{13}{2}\\ 1&-4&-2&1\end{pmatrix}[/tex] [tex][ \ R_2 = R_2 - \frac{1}{2} R_1 \ ][/tex]
[tex]=\begin{pmatrix}2&-1&3&9\\ 0&\frac{3}{2}&-\frac{5}{2}&-\frac{13}{2}\\ 0&-\frac{7}{2}&-\frac{7}{2}&-\frac{7}{2}\end{pmatrix}[/tex] [tex][ \ R_3 = R_3 - \frac{1}{2} R_1 \ ][/tex]
[tex]=\begin{pmatrix}2&-1&3&9\\ 0&-\frac{7}{2}&-\frac{7}{2}&-\frac{7}{2}\\ 0&\frac{3}{2}&-\frac{5}{2}&-\frac{13}{2}\end{pmatrix}[/tex] [tex][ \ swap \ R_2 \ and \ R_3 \ ][/tex]
[tex]=\begin{pmatrix}2&-1&3&9\\ 0&-\frac{7}{2}&-\frac{7}{2}&-\frac{7}{2}\\ 0&0&-4&-8\end{pmatrix}[/tex] [tex][ \ R_3 = R_3 + \frac{3}{7}R_2 \ ][/tex]
Therefore,
[tex]-4z = - 8\\\\z = 2[/tex]
[tex]-\frac{7}{2} y - \frac{7}{2}z = -\frac{7}{2}\\\\y + z = 1\\\\y + 2 = 1 \\\\y = 1 - 2 = - 1[/tex]
[tex]2x - 1y + 3z = 9\\\\2x -1(-1) + 3(2) = 9\\\\2x + 1 + 6 = 9\\\\2x = 9 - 7\\\\2x = 2 \\\\x = 1[/tex]
7. Find the length of X (in the picture) plssss I need help. (GIVING PPINTS AND BRAINLY)
Answer:
[tex] \frac{8}{5} = \frac{4}{x} \\ 8x = 20 \\ x = 2.5[/tex]
Answer:
here the resize factor is 5/8
so we need to calculate
x = 4 *5/8
= 20/8
= 2.5
I assume you get the basic idea from the other question you posted.
Use the substitution method to solve the system of equations. Choose the correct ordered pair. 4(x + 4) = 8(y + 2); 18y - 22 = 3x + 2
x = 30
y = 2
Get the explanation from the image I have shared.
Hope it helps you
How do you solve x[tex]x^{2} +4x+3=0[/tex]?
Answer:
[tex]{ \tt{ {x}^{2} + 4x + 3 = 0}} \\ { \tt{(x + 1)(x + 3) = 0}} \\ \\ { \tt{x = - 1 \: \: and \: \: - 3}}[/tex]
3. Betty rode her bike down a block that was 50 meters long. How many millimeters did she ride?
Answer:
50,000 millimeters
Step-by-step explanation:
Hope this helps : ) I have no explanation but it's correct
Find the sum of: 7a² - 9a + 5 and 11a + a² + 8
Answer:
Add them:
Squares to squares, a to and number to number:
8a^2+3a+13
The simplified sum of the given expressions 7a² - 9a + 5 and 11a + a² + 8 is 8a² + 2a + 13.
To find the sum of the expressions 7a² - 9a + 5 and 11a + a² + 8, we simply combine like terms.
The given expressions contain terms with different powers of "a."
Combining the terms with the same powers, we have:
7a² + a² = 8a² (the coefficient for the "a²" term is 7 + 1 = 8)
-9a + 11a = 2a (the coefficient for the "a" term is -9 + 11 = 2)
Finally, we combine the constant terms:
5 + 8 = 13
Putting it all together, the sum of the expressions 7a² - 9a + 5 and 11a + a² + 8 is:
8a² + 2a + 13
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Instructions: Find the value of x
I’ll mark brainliest please help
The little lines theu each section are telling you all those sections are identical, which mean they are the same length.
You are told one section is 10, which means x is also 10
X = 10
Find Length of x line OR
Both figures r similar (given )
so :-[tex] \frac{5}{2.5} = \frac{3}{1.5} = \frac{2}{1} = \frac{4}{x} \\ \frac{2}{1} = \frac{4}{x} \\ \frac{ \cancel{2}}{1} = \frac{ \cancel{4}^ { \tiny{2}}}{x} \\ x = 2 \: \: ans[/tex]
What is the probability of rolling a number less than or equal to 8 with the
sum of two dice, given that at least one of the dice must show a 6?
Answer:
I hope this helps
the outcomes are the compulsory 6, and 1 or 2
Step-by-step explanation:
[tex] \frac{3}{6} \\ \frac{1}{2} or \: 0.5[/tex]
If the set of values in the table represents points on a line, what is the value of a?
Answer:
3
Step-by-step explanation:
Answer:
7
Step-by-step explanation:
just did it
ANSWER ASAP IM TIMED 30 POINTS
Which shape below will form a tessellation?
A. regular heptagons
B. regular pentagons
C. regular octagons
D. regular hexagons
Answer:
Step-by-step explanation:
D ANSWER D
Answer:
d
Step-by-step explanation:
In a regression analysis involving 30 observations, the following estimated regressionequation was obtained.y^ =17.6+3.8x 1 −2.3x 2 +7.6x 3 +2.7x 4For this estimated regression equation SST = 1805 and SSR = 1760. a. At \alpha =α= .05, test the significance of the relationship among the variables.Suppose variables x 1 and x 4 are dropped from the model and the following estimatedregression equation is obtained.y^ =11.1−3.6x 2 +8.1x 3For this model SST = 1805 and SSR = 1705.b. Compute SSE(x 1 ,x 2 ,x 3 ,x 4 )c. Compute SSE (x2 ,x3 ) d. Use an F test and a .05 level of significance to determine whether x1 and x4 contribute significantly to the model.
Answer:
(a) There is a significant relationship between y and [tex]x_1, x_2, x_3, x_4[/tex]
(b) [tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45[/tex]
(c) [tex]SSE_{(x_2,x_3)} = 100[/tex]
(d) [tex]x_1[/tex] and [tex]x_4[/tex] are significant
Step-by-step explanation:
Given
[tex]y = 17.6+3.8x_1 - 2.3x_2 +7.6x_3 +2.7x_4[/tex] --- estimated regression equation
[tex]n = 30[/tex]
[tex]p = 4[/tex] --- independent variables i.e. x1 to x4
[tex]SSR = 1760[/tex]
[tex]SST = 1805[/tex]
[tex]\alpha = 0.05[/tex]
Solving (a): Test of significance
We have:
[tex]H_o :[/tex] There is no significant relationship between y and [tex]x_1, x_2, x_3, x_4[/tex]
[tex]H_a :[/tex] There is a significant relationship between y and [tex]x_1, x_2, x_3, x_4[/tex]
First, we calculate the t-score using:
[tex]t = \frac{SSR}{p} \div \frac{SST - SSR}{n - p - 1}[/tex]
[tex]t = \frac{1760}{4} \div \frac{1805- 1760}{30 - 4 - 1}[/tex]
[tex]t = 440 \div \frac{45}{25}[/tex]
[tex]t = 440 \div 1.8[/tex]
[tex]t = 244.44[/tex]
Next, we calculate the p value from the t score
Where:
[tex]df = n - p - 1[/tex]
[tex]df = 30 -4 - 1=25[/tex]
The p value when [tex]t = 244.44[/tex] and [tex]df = 25[/tex] is:
[tex]p =0[/tex]
So:
[tex]p < \alpha[/tex] i.e. [tex]0 < 0.05[/tex]
Solving (b): [tex]SSE(x_1 ,x_2 ,x_3 ,x_4)[/tex]
To calculate SSE, we use:
[tex]SSE = SST - SSR[/tex]
Given that:
[tex]SSR = 1760[/tex] ----------- [tex](x_1 ,x_2 ,x_3 ,x_4)[/tex]
[tex]SST = 1805[/tex]
So:
[tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4)} = 1805 - 1760[/tex]
[tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45[/tex]
Solving (c): [tex]SSE(x_2 ,x_3)[/tex]
To calculate SSE, we use:
[tex]SSE = SST - SSR[/tex]
Given that:
[tex]SSR = 1705[/tex] ----------- [tex](x_2 ,x_3)[/tex]
[tex]SST = 1805[/tex]
So:
[tex]SSE_{(x_2,x_3)} = 1805 - 1705[/tex]
[tex]SSE_{(x_2,x_3)} = 100[/tex]
Solving (d): F test of significance
The null and alternate hypothesis are:
We have:
[tex]H_o :[/tex] [tex]x_1[/tex] and [tex]x_4[/tex] are not significant
[tex]H_a :[/tex] [tex]x_1[/tex] and [tex]x_4[/tex] are significant
For this model:
[tex]y =11.1 -3.6x_2+8.1x_3[/tex]
[tex]SSE_{(x_2,x_3)} = 100[/tex]
[tex]SST = 1805[/tex]
[tex]SSR_{(x_2 ,x_3)} = 1705[/tex]
[tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45[/tex]
[tex]p_{(x_2,x_3)} = 2[/tex]
[tex]\alpha = 0.05[/tex]
Calculate the t-score
[tex]t = \frac{SSE_{(x_2,x_3)}-SSE_{(x_1,x_2,x_3,x_4)}}{p_{(x_2,x_3)}} \div \frac{SSE_{(x_1,x_2,x_3,x_4)}}{n - p - 1}[/tex]
[tex]t = \frac{100-45}{2} \div \frac{45}{30 - 4 - 1}[/tex]
[tex]t = \frac{55}{2} \div \frac{45}{25}[/tex]
[tex]t = 27.5 \div 1.8[/tex]
[tex]t = 15.28[/tex]
Next, we calculate the p value from the t score
Where:
[tex]df = n - p - 1[/tex]
[tex]df = 30 -4 - 1=25[/tex]
The p value when [tex]t = 15.28[/tex] and [tex]df = 25[/tex] is:
[tex]p =0[/tex]
So:
[tex]p < \alpha[/tex] i.e. [tex]0 < 0.05[/tex]
Hence, we reject the null hypothesis
Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights of oranges are also normally distributed with a mean of 131 grams and a standard deviation of 20 grams. Amy has an apple that weighs 90 grams and an orange that weighs 155 grams.
Required:
a. Find the probability a randomly chosen apple exceeds 100 g in weight.
b. What weight do 80% of the apples exceed?
Answer:
a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.
b) The weight that 80% of the apples exceed is of 78.28g.
Step-by-step explanation:
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.
Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.
This means that [tex]\mu = 85, \sigma = 8[/tex]
a. Find the probability a randomly chosen apple exceeds 100 g in weight.
This is 1 subtracted by the p-value of Z when X = 100. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{100 - 85}{8}[/tex]
[tex]Z = 1.875[/tex]
[tex]Z = 1.875[/tex] has a p-value of 0.9697
1 - 0.9696 = 0.0304
0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.
b. What weight do 80% of the apples exceed?
This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.84 = \frac{X- 85}{8}[/tex]
[tex]X - 85 = -0.84*8[/tex]
[tex]X = 78.28[/tex]
The weight that 80% of the apples exceed is of 78.28g.
HELP PLEASE IM STUCK!
1. Which of these describes the relation for this set of coordinate pairs?
{(-1, 5), (12, 18), (0, 6), (-3, 3), (4, ?), (?, 11)}
a. x - y = 6 b. f(x) = x +6 c. f(x) = 6 d. y = 6x e. None of these
Answer:
b) f(x) = x + 6
Step-by-step explanation:
The coordinate (0, 6) makes the y-intercept = 6. Only one of these functions has that intercept: f(x) = x + 6. If you plug in each coordinate the outputted y-value matches up, making this the right answer.
what percentage of the appies are yellow?
Answer:
20%
Step-by-step explanation:
6 out of 30. = 1/5 = multiply 5*20= 100 and 1*20= 100 so it is 20% of 100.
CAN SOMEONE PLEASE HELP ME GOOD, I need this to graduate ): 5. Given that AABC - ADEC, find the
value of x.
Answer:
ans: 4
Step-by-step explanation:
corresponding sides are proportional since given triangle are similar triangle, I.e
(4/5.5) = { (2x+8)/(6x-2)}
8/11 = ( x+ 4 ) / ( 3x - 1 )
8( 3x - 1 )= 11( x + 4 )
24x - 8 = 11x + 44
13x = 52
x = 4
An expression is shown below:
3(m + 5 + 9m)
Part A: Write two expressions that are equivalent to the given expression. (3 points)
Part B: Show that one of your expressions in Part A is equivalent to the given expression using algebraic properties. Explain which properties you used. (4 points)
Part C: Show that your other expression from Part A is equivalent to the given expression by substituting a number for m. (3 points)
The answers are as follows part A = 30m+15 part B =3m+15+27m
and partC = 30m+15
What is an expression?Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition,substraction, multiplication and division.
Part A:- Two expressions that are equivalent to the given expressions are:-
3m + 15 + 27m
30m + 15
Part B: Show that one of your expressions in Part A is equivalent to the given expression using algebraic properties.
3 ( m + 5 + 9m )
Open the bracket by multiplying 3 by what is in the bracket
3m + 15 + 27m
Part C: Show that your other expression from Part A is equivalent to the given expression by substituting a number for m.
3 ( m + 5 + 9m )
Open the bracket by multiplying 3 by what is in the bracket
3m + 15 + 27m
Collect like terms together
3m + 27m + 15
= 30m + 15
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Which expression are greater than 1/2? Choose all the apply
Answer:
25/30
5/8
Step-by-step explanation:
Which fraction is it out of all of these 6/14,5/8,25/30,or 3/6?
to determine which fractions are greater than 1/2, convert the fractions to decimals
to convert to decimals, divide the numerator by the denominator
1/2 = 0.5 less than half
6/14 = 0.43 less than half
5/8 = 0.625 greater than half
25 / 30 = 0.83 greater than half
3 / 6 = 0.5 equal to half
Simplify the expression.
4(11 + 7) ÷ (7 – 5)
Answer:
36
Step-by-step explanation:
4(11 + 7) ÷ (7 – 5)
4 * 18 ÷ 2
=36
Are these triangles congruent?
Answer:
yes...
Step-by-step explanation:
its a congrate triangle
Oakley babysits on weekends. He charges a flat fee of $12, plus an additional $5 for each hour that he babysits. How much money would Oakely make if he babysits for 4 hours?
Answer:
32
Since theres a flat fee he automatically starts at 12 since he gets 5 every hour for 4 hours (5×4) ending up at 20. You add the total and the flat fee 20+12= 32
Answer:
$32
Step-by-step explanation:
First you must set up the equation.
A flat fee of $12 means it will be added and $5 each hour for 4 hours so 5 will be multiplied by 4.
$12 + $5 * 4
12 + 5 * 4 (or 12 + (5 * 4), whatever makes you remember it better)
Remember PEMDAS (Parentheses, Exponents, Multiplication, Divisiom, Addition, Subtraction)
Multiplication comes before Addition so you get 12 + 20
12 + 20 = 32
Remember to put units, so you would get $32.
Find the value of x in the given figure using properties of parallel lines.
Answer:
x =42
Step-by-step explanation:
x and 138 are same side interior angles. Same side interior angles are supplementary when the lines are parallel
x+138 = 180
x = 180-138
x =42
Answer:
42°
Step-by-step explanation:
138° + x = 180°
180° - 138° = 42°
Which relationship is always true for the angles x,y and z of triangle ABC
Answer:
B. y + z = x
Step-by-step explanation:
x is an exterior angle of the triangle.
y and z are the opposite angles opposite the exterior angle.
The exterior angle theorem of a triangle states that the measure of an exterior angle equals the measure of the sum of the two angles opposite the exterior angle.
Thus:
y + z = x
Part A: Factor x2b2 − xb2 − 6b2. Show your work.
Part B: Factor x2 + 4x + 4. Show your work.
Part C: Factor x2 − 4. Show your work
Answer:
A.2b2(x2-x-3)
B.x2+2x+2x+4
=x(x+2)+2(x+2)
=(x+2)(x+2)
C.x2-2^2
=(x+2)(x+2)
Which of the following sets of data does not contain an outlier?
A.16, 17, 20, 19.48
B.59. 60. 61, 67.65
C.95.99.97.94.60
D.-1.2.1.0.5.16
Answer:
it is a letter b
Step-by-step explanation:
that does not contain an outlet
Segment [tex]$s_1$[/tex] has endpoints at [tex]$(3+\sqrt{2},5)$[/tex] and[tex]$(4,7)$[/tex]. Segment [tex]$s_2$[/tex] has endpoints at [tex]$(6-\sqrt{2},3)$[/tex] and[tex]$(3,5)$[/tex]. Find the midpoint of the segment with endpoints at the midpoints of [tex]$s_1$[/tex] and [tex]$s_2$[/tex]. Express your answer as [tex]$(a,b)$[/tex].
Answer:
The midpoint of the segment with endpoints at the midpoints of s1 and s2 is (4,5).
Step-by-step explanation:
Midpoint of a segment:
The coordinates of the midpoint of a segment are the mean of the coordinates of the endpoints of the segment.
Midpoint of s1:
Using the endpoints given in the exercise.
[tex]x = \frac{3 + \sqrt{2} + 4}{2} = \frac{7 + \sqrt{2}}{2}[/tex]
[tex]y = \frac{5 + 7}{2} = \frac{12}{2} = 6[/tex]
Thus:
[tex]M_{s1} = (\frac{7 + \sqrt{2}}{2},6)[/tex]
Midpoint of s2:
[tex]x = \frac{6 - \sqrt{2} + 3}{2} = \frac{9 - \sqrt{2}}{2}[/tex]
[tex]y = \frac{3 + 5}{2} = \frac{8}{2} = 4[/tex]
Thus:
[tex]M_{s2} = (\frac{9 - \sqrt{2}}{2}, 4)[/tex]
Find the midpoint of the segment with endpoints at the midpoints of s1 and s2.
Now the midpoint of the segment with endpoints [tex]M_{s1}[/tex] and [tex]M_{s2}[/tex]. So
[tex]x = \frac{\frac{7 + \sqrt{2}}{2} + \frac{9 - \sqrt{2}}{2}}{2} = \frac{16}{4} = 4[/tex]
[tex]y = \frac{6 + 4}{2} = \frac{10}{2} = 5[/tex]
The midpoint of the segment with endpoints at the midpoints of s1 and s2 is (4,5).
Which statements are true of the given function?
Check all that apply.
9514 1404 393
Answer:
B, E
Step-by-step explanation:
The table tells you that f(0) = 3/2. (0 is found in the x-column; 3/2 is found in the f(x) column.)
__
You can find f(4) by evaluating the formula.
f(4) = 1/2·4 +3/2 = 4/2 +3/2
f(4) = 7/2 . . . . agrees with last answer choice
Answer:
A ,B and E ;)
Step-by-step explanation:
