Answer:
[tex]\frac{2}{5}[/tex] • [tex]\frac{3}{7}[/tex] = [tex]\frac{6}{35}[/tex]
Answer: 0.171
Step-by-step explanation:
First, do 2/5 which would equal 0.4
Second, so 3/7 which would equal 0.428571428571429
Lastly multiply the two answers together to get 0.171428571428571
1/3 of a shipment of books weights 28 pounds
Answer:
84 pounds
Step-by-step explanation:
If 1/3 of a book is equal to 28 pounds then 28*3 will give you your answer
Which expression is equal to (1+6i)−(7+3i) ?
Answer:
- 6+3iStep-by-step explanation:
[tex](1+6i)-(7+3i) ?\\Group\:the\:real\:part\:and\:the\:imaginary\\\:part\:of\:the\:complex\:number\\\left(a+bi\right)\pm \left(c+di\right)=\left(a\:\pm \:c\right)+\left(b\:\pm \:d\right)i\\=\left(1-7\right)+\left(6-3\right)i\\1-7=-6\\6-3=3\\=-6+3i[/tex]
A study of 25 graduates of four-year public colleges revealed the mean amount owed by a student in student loans was $55,051. The standard deviation of the sample was $7,568.
Required:
a. Construct a 90% confidence interval for the population mean.
b. Confidence interval for the population men between _______ up to_______________
Answer:
a
The 90% confidence interval is [tex]52561.13 < \mu < 57540.8[/tex]
b
Confidence interval for the population men between $52561.13 up to $57540.8
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 25[/tex]
The sample mean is [tex]\= x = \$ 55,051[/tex]
The standard deviation is [tex]\sigma = \$ 7,568[/tex]
Given that the confidence level is 90% then the level of confidence is mathematically represented as
[tex]\alpha = 100 -90[/tex]
[tex]\alpha = 10\%[/tex]
[tex]\alpha = 0.10[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table the values is
[tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{ \alpha }{2} } * \frac{ \sigma }{\sqrt{n} }[/tex]
substituting values
[tex]E = 1.645 * \frac{ 7568}{ \sqrt{ 25} }[/tex]
[tex]E = 2489.9[/tex]
The 90% confidence interval is mathematically evaluated as
[tex]\= x -E < \mu < \= x +E[/tex]
substituting values
[tex]55051 - 2489.8 < \mu < 55051 + 2489.8[/tex]
[tex]52561.13 < \mu < 57540.8[/tex]
According to the Census Bureau, 3.34 people reside in the typical American household. A sample of 26 households in Arizona retirement communities showed the mean number of residents per household was 2.70 residents. The standard deviation of this sample was 1.17 residents. At the .10 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.34 persons?
(a) State the null hypothesis and the alternate hypothesis. (Round your answer to 2 decimal places.)
H0: ? ?
H1: ? <
(b)
State the decision rule for .10 significance level. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
Reject H0 if t <
(c)
Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
Value of the test statistic
(d)
Is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.34 persons?
H0. Mean number of residents less than 3.34 persons.
Answer:
Step-by-step explanation:
Given that:
Mean = 3.34
sample size = 26
sample mean = 2.7
standard deviation = 1.17
level of significance = 0.10
The null hypothesis and the alternative hypothesis can be computed as follows:
[tex]\mathtt{H_o: \mu \geq 3.34} \\ \\ \mathtt{H_1: \mu < 3.34}[/tex]
degree of freedom = n - 1
degree of freedom = 26 -1
degree of freedom = 25
level of significance = 0.10
Since the alternative hypothesis contains <, then the test is left tailed
[tex]\mathtt{t_{\alpha, df} = t_{0.10, 25}}[/tex]
[tex]\mathtt{t_{0.10, 25}}[/tex] = - 1.316
The rejection region therefore consist of all values smaller than - 1.316, therefore ; reject [tex]H_o[/tex] if t < -1.316
The test statistics can be computed as follows:
[tex]t = \dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]t = \dfrac{2.7 - 3.34}{\dfrac{1.17}{\sqrt{26}}}[/tex]
[tex]t = \dfrac{-0.64}{\dfrac{1.17}{5.099}}[/tex]
t = - 2.789
Decision Rule: To reject the null hypothesis if the t test lies in the rejection region or less than the rejection region.
Conclusion: We reject the null hypothesis since t = (- 2.789) < -1.316. Then we conclude that the mean number of residents in the retirement community household is less than 3.34 persons.
Compute (3/4)*(8/9)*(15/16)*(24/25)*(35/36)*(48/49)*(63/64)*(80/81)*(99/100) Express your answer in the simplest way possible. (Suggestion: First, try computing 3/4*8/9 then 3/4*8/9*15/16 and so on. Look for patterns.
Answer:
[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]
Step-by-step explanation:
Given
[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100})[/tex]
Required
Simplify
For clarity, group the expression in threes
[tex]((\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
Evaluate the first group [Divide 8 by 4]
[tex]((\frac{3}{1})*(\frac{2}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[Divide 9 by 3]
[tex]((\frac{1}{1})*(\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[tex]((\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[Divide 15 by 3]
[tex]((\frac{2}{1})*(\frac{5}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[Divide 16 by 2]
[tex]((\frac{1}{1})*(\frac{5}{8}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[tex](\frac{5}{8})*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
Evaluate the second group [Divide 35 and 25 by 5]
[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{7}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[Divide 49 by 7]
[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{1}{3})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[Divide 24 by 3]
[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{1}{1})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
Merge the first and second group
[tex]((\frac{5}{8})*(\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[tex](1*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]
Evaluate the last group [Divide 99 by 9]
[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{9})*(\frac{11}{100}))[/tex]
[Divide 63 by 9]
[tex](\frac{4}{7})*((\frac{7}{64})*(\frac{80}{1})*(\frac{11}{100}))[/tex]
[Divide 64 and 80 by 8]
[tex](\frac{4}{7})*((\frac{7}{8})*(\frac{10}{1})*(\frac{11}{100}))[/tex]
[Divide 10 and 4 by 2]
[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{5}{1})*(\frac{11}{100}))[/tex]
[Divide 100 by 5]
[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{1}{1})*(\frac{11}{20}))[/tex]
[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{11}{20}))[/tex]
[tex](\frac{4}{7})*(\frac{7}{4})*(\frac{11}{20})[/tex]
[tex]1*(\frac{11}{20})[/tex]
[tex]\frac{11}{20}[/tex]
Hence;
[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]
A population of bacteria P is changing at a rate of dP/dt = 3000/1+0.25t where t is the time in days. The initial population (when t=0) is 1000. Write an equation that gives the population at any time t. Then find the population when t = 3 days.
Answer:
- At any time t, the population is:
P = 375t² + 3000t + 1000
- At time t = 3 days, the population is:
P = 13,375
Step-by-step explanation:
Given the rate of change of the population of bacteria as:
dP/dt = 3000/(1 + 0.25t)
we need to rewrite the given differential equation, and solve.
Rewriting, we have:
dP/3000 = (1 + 0.25t)dt
Integrating both sides, we have
P/3000 = t + (0.25/2)t² + C
P/3000 = t + 0.125t² + C
When t = 0, P = 1000
So,
1000/3000 = C
C = 1/3
Therefore, at any time t, the population is:
P/3000 = 0.125t² + t + 1/3
P = 375t² + 3000t + 1000
At time t = 3 days, the population is :
P = 375(3²) + 3000(3) + 1000
= 3375 + 9000 + 1000
P = 13,375
Please answer asap this person made a mistake what is the error and correct solution to this problem
Answer:
6
Step-by-step explanation:
Hello, please consider the following.
[tex](4+x)^2=4^2+2\cdot 4\cdot x+x^2=16+\boxed{8}x+x^2\\\\\text{ ... and not ...}\\\\16+\boxed{4}x+x^2[/tex]
So the correct equation becomes.
[tex]x^2+64=16+8x+x^2\\\\8x=64-16=48\\\\\text{ we divide by 8 both sides of the equation.}\\\\x=\dfrac{45}{8}=6[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
Answer:
Error : The expression ( 4 + x )² was expanded incorrectly.
Correct Solution : x = 6
Step-by-step explanation:
The planning of the solution is correct, by Pythagorean Theorem you can say that PQ² + QO² = PO², and hence through substitution x² + 8² = ( 4 + x )². Let's look into the calculations.
PQ² + QO² = PO²,
x² + 8² = ( 4 + x )²,
x² + 8² = 16 + 8x + x²,
64 = 16 + 8x,
48 = 8x,
x = 48 / 8 = 6, x = 6
As you can see, the only error in the calculations was expanding the expression ( 4 + x )². ( 4 + x )² = 4² + 2 [tex]*[/tex] 4 [tex]*[/tex] x + x² = 4² + 8x + x² = 16 + 8x + x², not 16 + 4x + x².
Given the graph, find an equation for the parabola.
Answer:
[tex]\Large \boxed{\sf \bf \ \ y=\dfrac{1}{16}(a-3)^2-2 \ \ }[/tex]
Step-by-step explanation:
Hello, please consider the following.
When the parabola equation is like
[tex]y=a(x-h)^2+k[/tex]
The vertex is the point (h,k) and the focus is the point (h, k+1/(4a))
As the vertex is (3,-2) we can say that h = 3 and k = -2.
We need to find a.
The focus is (3,2) so we can say.
[tex]2=-2+\dfrac{1}{4a}\\\\\text{*** We add 2. ***}\\\\\dfrac{1}{4a}=2+2=4\\\\\text{*** We multiply by 4a. ***}\\\\16a=1\\\\\text{*** We divide by 16. ***}\\\\a=\dfrac{1}{16}[/tex]
So an equation for the parabola is.
[tex]\large \boxed{\sf y=\dfrac{1}{16}(a-3)^2-2 }[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
A type of related samples design in which participants are observed more than once is called a
A. repeated measures design
B. matched pairs design
C. matched samples design
D. both matched pairs design and matched samples design
Answer:
Option A (repeated measures design) is the correct option.
Step-by-step explanation:
Researchers as well as statisticians vary in terms of methods used mostly for repetitive measurements. Besides illustration, repeated models of measurements are however recognized as repeated analyzes of variance measurements, standardized considerations of measurements, or layouts of objects throughout them.The other three options are not related to the given instance. So that alternative A would be the correct choice.
Find the work W done by a force of 7pounds acting in the direction 30 degreesto the horizontal in moving an object 7feet from (0 comma 0 )to (7 comma 0 ).
Answer:
The work done by the force is 42.4 Joules
Step-by-step explanation:
The force F = 7 pounds
angle to the horizontal that the force acts ∅ = 30°
The object is moved a distance d = 7 feet
The coordinate (0 comma 0 )to (7 comma 0 ), indicates that the movement started from the origin, and is along the x-axis.
The work done by this force = F cos ∅ x d
==> 7 cos 30° x 7
==> 7 x 0.866 x 7 = 42.4 Joules
Find the intervals on which f is increasing and the intervals on which it is decreasing. f(x)=-2cos^(2)x
Answer:
Increasing
0°≤x≤180°
Decreasing
180°≤x≤360°
Given these four points: A(3, 3), B(−5, 7), C(2, 11), and D(9, −2), find the coordinates of the midpoint of line segments AB and CD.
Midpoint formula: (x1 + x2)/2 , (y1 + y2)/2
Midpoint AB = (3 +-5)/2, (3 + 7)/2 = -2/2 , 10/2 = (-1,5)
Midpoint CD = (2 +9)/2, (11 + -2)/2 = (11/2,9/2)
Many elementary school students in a school district currently have ear infections. A random sample of children in two different schools found that 16 of 42 at one school and 18 of 34 at the other have ear infections. At the 0.05 level of significance, is there sufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools? Group of answer choices
Answer:
Step-by-step explanation:
The summary of the given data includes;
sample size for the first school [tex]n_1[/tex] = 42
sample size for the second school [tex]n_2[/tex] = 34
so 16 out of 42 i.e [tex]x_1[/tex] = 16 and 18 out of 34 i.e [tex]x_2[/tex] = 18 have ear infection.
the proportion of students with ear infection Is as follows:
[tex]\hat p_1 = \dfrac{16}{42}[/tex] = 0.38095
[tex]\hat p_2 = \dfrac{18}{34}[/tex] = 0.5294
Since this is a two tailed test , the null and the alternative hypothesis can be computed as :
[tex]H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0[/tex]
level of significance ∝ = 0.05,
Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.
The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.
[tex]\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}[/tex]
[tex]\bar p = \dfrac{16 +18}{42 +34}[/tex]
[tex]\bar p = \dfrac{34}{76}[/tex]
[tex]\bar p = 0.447368[/tex]
[tex]\bar p + \bar q = 1 \\ \\ \bar q = 1 -\bar p \\ \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632[/tex]
The pooled standard error can be computed by using the formula:
[tex]S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} + \dfrac{\bar p \bar p}{n_2} }[/tex]
[tex]S.E = \sqrt{ \dfrac{ 0.447368 * 0.552632}{ 42} + \dfrac{ 0.447368 * 0.447368}{34} }[/tex]
[tex]S.E = \sqrt{ \dfrac{ 0.2472298726}{ 42} + \dfrac{ 0.2001381274}{34} }[/tex]
[tex]S.E = \sqrt{ 0.01177284105}[/tex]
[tex]S.E = 0.1085[/tex]
The test statistics is ;
[tex]z = \dfrac{\hat p_1 - \hat p_2}{S.E}[/tex]
[tex]z = \dfrac{0.38095- 0.5294}{0.1085}[/tex]
[tex]z = \dfrac{-0.14845}{0.1085}[/tex]
z = - 1.368
Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.
Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools
The base of a triangle is 4 cm greater than the
height. The area is 30 cm. Find the height and
the length of the base
h
The height of the triangle is
The base of the triangle is
Answer:
Step-by-step explanation:
Formula for area of a triangle:
Height x Base /2
Base (b) = h +4
Height = h
h + 4 x h /2 = 30cm
=> h +4 x h = 60
=> h+4h =60
=> 5h = 60
=> h = 12
Height = 12
Base = 12 +4 = 16
Evaluate the expression: -(31 + 2) +7² - (-5²)
A) -9
B) -5
C) 41
OD -40
Answer: C. 41
Step-by-step explanation:
[tex]-\left(31+2\right)+7^2-\left(-5^2\right)[/tex]
[tex]=-33+7^2-\left(-5^2\right)[/tex]
[tex]\left(-5^2\right)=-25[/tex]
[tex]=-33+7^2-\left(-25\right)[/tex]
[tex]7^2=49[/tex]
[tex]=-33+49-\left(-25\right)[/tex]
[tex]-33+49=16[/tex]
[tex]=16-\left(-25\right)[/tex]
[tex]\mathrm{Apply\:rule\:}-\left(-a\right)\:=\:+a[/tex]
[tex]16+25=41[/tex]
A sample of a radioactive substance decayed 11% over the course of 3 weeks. How many grams were in the sample originally if 30.26 grams of the substance were remaining after the 3 weeks?
Answer:
34 grams
Step-by-step explanation:
If the remaining sample has 30.26 grams of radioactive substance, and 11% of it decayed, that means that 30.26 grams is 89% of the original. Let the original be x.
30.26=0.89x
Multiply both by one hundred
3026=89x
Divide both by 89
34=x
x=original, so the original was 34 grams.
Sarah had a balance of $155 in her bank account at the start of the week. She withdrew $65.50 on Monday, $23.25 on Wednesday, and $26.45 on Thursday. On Friday she deposited $165.30. Write an expression that represents Sarah's spending.
Answer:
155 + 165.3 - 65.5 - 23.25 - 26.45
Step-by-step explanation:
She had $155 dollars in the starting = +155
She withdrew $65.5 = -65.5
She withdrew another $23.25 = -23.25
She withdrew another $26.45 = -26.45
The deposited $165.3 = +165.3
The expression looks like:
155 + 165.3 - 65.5 - 23.25 - 26.45
We could simplify the expression:
155 + 165.3 - 65.5 - 23.25 - 26.45
=> 320.3 - 88.75 - 26.45
=> 320.3 -115.2
=> 205.1
At the end of the week, she had a total of $205.10.
Lydia drives from city a to city b to transport goods. her return speed is 3 times her departure speed and she takes 40 minutes less on her return trip. how long did her departure trip take?
Answer:
1 hour
Step-by-step explanation:
Hello, let's say that her departure trip takes t in minutes, as her return speed is 3 times her departure speed, she took t/3 for the return and we know that this 40 minutes less, so we can write.
t/3=t-40
We can multiply by 3
t = 3t -40*3 = 3t - 120
This is equivalent to
3t -120 = t
We subtract t
2t-120 = 0
2t = 120
We divide by 2
t = 120/2 = 60
So this is 60 minutes = 1 hour.
Thank you.
What are the solutions of the equation x4 + 6x2 + 5 = 0? Use u substitution to solve.
x = i and x = i5
x=+ i and x
x= +115
O x=V-1 and x = = -5
x=+ -1 and x = = -5
Answer:
A; The first choice.
Step-by-step explanation:
We have the equation [tex]x^4+6x^2+5=0[/tex] and we want to solve using u-substitution.
When solving by u-substitution, we essentially want to turn our equation into quadratic form.
So, let [tex]u=x^2[/tex]. We can rewrite our equation as:
[tex](x^2)^2+6(x^2)+5=0[/tex]
Substitute:
[tex]u^2+6u+5=0[/tex]
Solve. We can factor:
[tex](u+5)(u+1)=0[/tex]
Zero Product Property:
[tex]u+5=0\text{ and } u+1=0[/tex]
Solve for each case:
[tex]u=-5\text{ and } u=-1[/tex]
Substitute back u:
[tex]x^2=-5\text{ and } x^2=-1[/tex]
Take the square root of both sides for each case. Since we are taking an even root, we need plus-minus. Thus:
[tex]x=\pm\sqrt{-5}\text{ and } x=\pm\sqrt{-1}[/tex]
Simplify:
[tex]x=\pm i\sqrt{5}\text{ and } x=\pm i[/tex]
Our answer is A.
Describe how the graph of y= x2 can be transformed to the graph of the given equation. y = (x - 6)2 Shift the graph of y = x2 down 6 units. Shift the graph of y = x2 right 6 units. Shift the graph of y = x2 up 6 units. Shift the graph of y = x2 left 6 units.
Answer:
Shift the graph of y = x2 right 6 units.
Hi I cannot seem to get this question correct
Answer:
150000 doctoral degrees
Step-by-step explanation:
From the graph attached,
At t = 0 represents the year 2005 and each unit of y-axis represents 10000 degrees.
Number of approximate number of degrees awarded to women in 2008,
f'(3) = 2.25
Similarly, number of doctoral degrees awarded from 2009 to 2013 are,
f'(4) = 2.5
f'(5) = 2.5
f'(6) = 2.5
f'(7) = 2.5
f'(8) = 2.75
Total number of degrees awarded to women from the start of 2008 to the start of 2014 = (2.25 + 2.5 + 2.5 + 2.5 + 2.5 + 2.75 ) × 10000
= 15 × 10000
= 150000
Step 1: Subtract 3 from both sides of the inequality
Step 2
Step 3: Divide both sides of the inequality by the
coefficient of x.
What is the missing step in solving the inequality 5 -
8x < 2x + 3?
O Add 2x to both sides of the inequality
O Subtract 8x from both sides of the inequality
O Subtract 2x from both sides of the inequality
Add 8x to both sides of the inequality.
Mark this and return
Save and Exit
Intext
Submit
Answer:
add 8x to both sides
Step-by-step explanation:
5-8x<2x+3
first step, subtract 3 from both sides:
2-8x<2x
second step,?
2<?x
so you need to add 8x first
A schoolteacher would like to know whether or not toothpaste brands are used differentially in her classroom (in other words, is one brand preferred over the others?). She asks her students to report which of three brands they use: Crest, Colgate, or Aquafresh. Below are the numbers of students who use each type of toothpaste (notice there are 45 students total in her class). Test her hypothesis using an alpha level of .05.
Crest Colgate Aquafresh
24 13 8
a. What test is appropriate for this analysis?
b. State the null hypothesis:
c. State the alternative hypothesis:
d. Find the critical value:
e. Calculate the test statistic:
f. Make a decision:
Answer:
Step-by-step explanation:
a. What test is appropriate for this analysis?
A Chi-square test of independence is appropriate for this analysis because it is used to compare two variables or testing relationship on categorical variables.
b. State the null hypothesis:
The null hypothesis is the default hypothesis
[tex]\mathtt{H_o:}[/tex] There is no particular preference for any brand of toothpaste among students.
c. State the alternative hypothesis:
The alternative hypothesis is the research hypothesis which comes in place to challenge the validity of the null hypothesis.
[tex]\mathtt{H_a:}[/tex] There is particular preference for brands of toothpaste among students.
d. Find the critical value:
degree of freedom = n-1
degree of freedom = 3 - 1
degree of freedom = 2
At the level of significance ∝ = 0.05
The confidence interval = 0.95 and degree of freedom = 2, the critical value from the chi-square distribution table = 5.991
e. Calculate the test statistic:
Using the chi square test statistics; we have the following:
Crest Colgate Aquafresh Total
24 13 8 45
Since we have three brands. Then, for each brand, the expected value
= Total /3
= 45/3
=15
Thus:
Chi -square [tex]\mathtt{X^2 = \dfrac{(observed \ value - expected \ value)^2}{expected \ value}}[/tex]
[tex]\mathtt{X^2 = \dfrac{(24 - 15)^2}{15} + \dfrac{(13 - 15)^2}{15} + \dfrac{(8 - 15)^2}{15} }[/tex]
[tex]\mathtt{X^2 = \dfrac{81}{15} + \dfrac{4}{15} + \dfrac{49}{15} }[/tex]
[tex]\mathtt{X^2 = \dfrac{81+4+49}{15}}[/tex]
[tex]\mathtt{X^2 = \dfrac{134}{15}}[/tex]
[tex]\mathtt{X^2 =8.93 }[/tex]
f. Make a decision:
Since the chi-square value is greater than the critical value , we reject the null hypothesis and conclude that the students have particular preference for brands of toothpaste.
The image of (-2, 7) reflected across the x-axis is
2
(-2,-7)
b)
(2,7)
(2, -7)
d)
(-2, 7)
Answer:
(-2,-7)
Step-by-step explanation:
because it's reflected across the x-axis, only the y-intercept will change
Answer:
(-2,-7)
Step-by-step explanation:
All you have to do is draw a graph and draw the point across the x axis in the same row and same distance from the x axis.The distance is 7 so you just change it to -7.
the production of a printer consists of the cost of raw material at 100 dollars the cost of overheads at 80$ and wages at 120$ if the cost of raw materials and overheads are increased by 11% and 20% respectively while wages are decreased by 15% find the percentage increase or decrease in the production cost of the printer
Answer:
The percentage increase in the production cost of the printer is 3%.
Step-by-step explanation:
We are given that the production of a printer consists of the cost of raw material at 100 dollars the cost of overheads at 80$ and wages at 120$.
Also, the cost of raw materials and overheads are increased by 11% and 20% respectively while wages are decreased by 15%.
Cost of raw material = $100
Cost of overheads = $80
Cost of wages = $120
So, the total cost of the printer = $100 + $80 + $120
= $300
Now, the increase in the cost of raw material = $100 + 11% of $100
= [tex]\$100 + (\frac{11}{100} \times \$100)[/tex]
= $100 + $11 = $111
The increase in the cost of overheads = $80 + 20% of $80
= [tex]\$80 + (\frac{20}{100} \times \$80)[/tex]
= $80 + $16 = $96
The decrease in the cost of wages = $120 - 15% of $120
= [tex]\$120 - (\frac{15}{100} \times \$120)[/tex]
= $120 - $18 = $102
So, the new cost of a printer = $111 + $96 + $102 = $309
Now, the percentage increase in the production cost of the printer is given by;
% increase = [tex]\frac{\text{Net increase in the cost of printer}}{\text{Original cost of printer}} \times 100[/tex]
= [tex]\frac{\$309- \$300}{\$300} \times 100[/tex]
= 3%
Hence, the percentage increase in the production cost of the printer is 3%.
x power 8 + x power 4 + 1
factorize
Answer:
[tex]1(x {}^{8} + x {}^{4} + 1)[/tex]
Step-by-step explanation:
[tex]x {}^{8} + {x}^{4} + 1 =1( x {}^{8} + x {}^{2} + 1)[/tex]
Hope this helps ;) ❤❤❤
Let me know if there is an error in my answer.
) A random sample of size 36 is selected from a normally distributed population with a mean of 16 and a standard deviation of 3. What is the probability that the sample mean is somewhere between 15.8 and 16.2
Answer:
The probability is 0.31084
Step-by-step explanation:
We can calculate this probability using the z-score route.
Mathematically;
z = (x-mean)/SD/√n
Where the mean = 16, SD = 3 and n = 36
For 15.8, we have;
z = (15.8-16)/3/√36 = -0.2/3/6 = -0.2/0.5 = -0.4
For 16.2, we have
z = (16.2-16)/3/√36 = 0.2/3/6 = 0.2/0.5 = 0.4
So the probability we want to calculate is;
P(-0.4<z<0.4)
We can get this using the standard normal distribution table;
So we have;
P(-0.4 <z<0.4) = P(z<-0.4) - P(z<0.4)
= 0.31084
2. A 10 Mg truck hauls a 20 Mg trailer. If the unit starts from rest on a level road with a
tractive force of 20 kN between the driving wheels of the truck and the road, calculate the
acceleration of the unit and the tension in the horizontal draw-bar.
Drawbar
20 Mg Trailer
10 Mg Truck
a=0.667 m/s2
T= 13.3 KN
Oro
W
Answer:
The acceleration on the unit is 0.667 m/s^2
The tension on the draw-bar is 13.34 kN
Step-by-step explanation:
The mass of the truck = 10 Mg = 10 x 10^3 kg
The mass of the trailer = 20 Mg = 20 x 10^3 kg
Tractive force from the truck = 20 kN = 20 x 10^3 N
The total mass of the unit = 10 Mg + 20 Mg = 30 Mg = 30 x 10^3 kg
The tractive force on the unit will produce an acceleration that is given as
F = ma
where
F is the tractive = 20 x 10^3 N
m is the mass of the unit = 30 x 10^3 kg
a is the acceleration of the unit = ?
substituting into the equation
20 x 10^3 = 30 x 10^3 x a
a = (20 x 10^3)/(30 x 10^3) = 0.667 m/s^2
the tension on the draw-bar T is gotten from considering only the mass that is pulled by the draw-bar which is 20 Mg
The acceleration on the unit = 0.667 m/s^2
The drawn mass = 20 Mg = 20 x 10^3 kg
The tension on the draw bar = ma = 20 x 10^3 x 0.667 = 13340 N
= 13.34 kN
The acceleration is 0.00067m/s^2, while the tension on the horizontal bar is 13.4 N
The given parameters are:
[tex]\mathbf{m = 10Mg}[/tex] -- mass of the truck
[tex]\mathbf{M = 20Mg}[/tex] -- mass of the trailer
[tex]\mathbf{F_T = 20kN}[/tex] --- tractive force
Start by calculating the total mass
[tex]\mathbf{M_T = m + M}[/tex]
So, we have:
[tex]\mathbf{M_T = 10Mg + 20Mg}[/tex]
[tex]\mathbf{M_T = 30Mg}[/tex]
Convert to kilograms
[tex]\mathbf{M_T = 30 \times 10^3kg}[/tex]
[tex]\mathbf{M_T = 30000 kg}[/tex]
Force is calculated as:
[tex]\mathbf{F =ma}[/tex]
So, we have:
[tex]\mathbf{20kN =30000kg \times a}[/tex]
Divide both sides by 30000
[tex]\mathbf{a = 0.00067ms^{-2}}[/tex]
The tension on the horizontal bar (i.e. the 20 Mg trailer) is:
[tex]\mathbf{T=ma}[/tex]
So, we have:
[tex]\mathbf{T=20Mg \times 0.00067ms^{-2}}[/tex]
Rewrite as:
[tex]\mathbf{T=20 \times 10^3 kg \times 0.00067m/s}[/tex]
[tex]\mathbf{T=13.4N}[/tex]
Hence, the acceleration is 0.00067m/s^2, while the tension on the horizontal bar is 13.4 N
Read more about force and acceleration at:
https://brainly.com/question/20511022
200,000=2x10 to the power of 6
False.
2x10^6 you move the decimal point 6 places to the right. ( add 6 zeros after the 2)
2x 10^6 = 2,000,000
A sample of 36 observations is selected from one population with a population standard deviation of 4.2. The sample mean is 101.5. A sample of 50 observations is selected from a second population with a population standard deviation of 5.0. The sample mean is 100.1. Conduct the following test of hypothesis using the 0.10 significance level.H0 : μ1 = μ2H1 : μ1 ≠ μ2a. This is a_________tailed test.b. State the decision rule. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)The decision rule is to reject H0 if z is ___________the interval.c. Compute the value of the test statistic. (Round your answer to 2 decimal places.) Value of the test statistic d. What is your decision regarding H0?e. What is the p-value? (Round your answer to 4 decimal places.)
Answer:
Step-by-step explanation:
Given that :
the null and the alternative hypothesis are computed as :
[tex]H_o: \mu_1 = \mu_2[/tex]
[tex]H_a : \mu _1 \neq \mu_2[/tex]
This is a two tailed test
This is because of the ≠ sign in the alternative hypothesis which signifies that the rejection region in the alternative hypothesis are at the both sides of the hypothesized mean difference .
Decision Rule: at the level of significance ∝ = 0 . 10
The decision rule is to reject the null hypothesis if z < - 1 . 64 and z > 1 . 64
NOTE: DURING THE MOMENT OF TYPING THIS ANSWER THERE IS A TECHNICAL ISSUE WHICH MAKES ME TO BE UNABLE TO SUBMIT THE FULL ANSWER BUT I'VE MADE SCREENSHOTS OF THEM AND THEY CAN BE FOUND IN THE ATTACHED FILE BELOW
