Answer:
0
Step-by-step explanation:
Let's first divide the two roots of [tex]\sqrt{6}[/tex] and [tex]\sqrt{3}[/tex].
[tex]\sqrt{\frac{6}{3} }[/tex]
[tex]\sqrt{2}[/tex]
Now, [tex]\sqrt{2}[/tex] - [tex]\sqrt{2}[/tex] = 0.
So, this equation comes out to be 0.
Hope this helped!
A researcher tests five individuals who have seen paid political ads about a particular issue. These individuals take a multiple-choice test about the issue in which people in general (who know nothing about the issue) usually get 40 questions correct. The number correct for these five individuals was 48, 41, 40, 51, and 50. Using the .05 level of significance, two-tailed, do people who see the ads score differently on this test
Use steps of hypothesis testing sketch distribution involved
Answer:
[tex]t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606[/tex]
The degrees of freedom are given by:
[tex]df=n-1=5-1=4[/tex]
The p value wuld be given by:
[tex]p_v =2*P(t_{(4)}>2.606)=0.060[/tex]
For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40
The distribution with the critical values are in the figure attached
Step-by-step explanation:
Information given
48, 41, 40, 51, and 50
The sample mean and deviation can be calculated with these formulas:
[tex]\bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
[tex]\bar X=46[/tex] represent the mean height for the sample
[tex]s=5.148[/tex] represent the sample standard deviation
[tex]n=5[/tex] sample size
[tex]\mu_o =40[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test
Hypothesis to test
We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 40[/tex]
Alternative hypothesis:[tex]\mu \neq 40[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing we got:
[tex]t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606[/tex]
The degrees of freedom are given by:
[tex]df=n-1=5-1=4[/tex]
The p value wuld be given by:
[tex]p_v =2*P(t_{(4)}>2.606)=0.060[/tex]
For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40
The distribution with the critical values are in the figure attached
pleassseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
Answer:
The answer is 0.4
Step-by-step explanation:
2/5 is equal to 0.4
Answer:
0.4
Step-by-step explanation:
start with 2/5.
multipy the 5 by 20 to get 100, and the 2 by 20 to get 40.
so, now your fraction should look like this:
40/100
then, shift the decimal in 40 over 2 spots to get 0.40, or 0.4
hope this helps :)
In a study of 205 adults, the average heart rate was 75 beats per minute. Assume the population of heart rates is known to be approximately normal, with a standard deviation of 8 beats per minute. What does a margin of error of 1.1 for the 95% confidence interval of the average beats per minute mean? There is a 95% chance that the population mean is between 67 and 83 beats per minute. There is a 95% chance that the population mean is between 73.9 and 76.1 beats per minute. There is a 5% chance that the population mean is less than 75 beats per minute. There is a 5% chance that the population mean is more than 75 beats per minute.
Answer:
There is a 95% chance that the population mean is between 73.9 and 76.1 beats per minute.
Step-by-step explanation:
i have the test
There is a 95% chance that the population mean is between 73.9 and 76.1 beats per minute.
Calculation of margin of error:Since
The average heart rate was 75 beats per minute.
The standard deviation is 8 beats per minute
And, there is the study of 205 adults
Now the following formula is to be used
Since
[tex]x \pm z \frac{\sigma}{\sqrt{n} }[/tex]
Here
z = 1.96 at 95% confidence interval
So,
[tex]= 75 \pm 1.96 \frac{8}{\sqrt{205} } \\\\= 75 - 1.96 \frac{8}{\sqrt{205} } , 75 + 1.96 \frac{8}{\sqrt{205} }[/tex]
= 73.9 ,76.1
Hence, the above statement should be true.
Learn more about standard deviation here: https://brainly.com/question/20529928
what is 3/5 of 1800
Answer:
1080
Step-by-step explanation:
first do 3 times 1800, because they are both the numerators. Then divide that number, which is 5400, by the denominator: 5. You will get 1080.
Please help me I’ll mark brainliest
15 3/4% is equal to which decimal?
Answer:
2/4%
Step-by-step explanation:
The dimensions of a closed rectangular box are measured as 96 cm, 58 cm, and 48 cm, respectively, with a possible error of 0.2 cm in each dimension. Use differentials to estimate the maximum error in calculating the surface area of the box.
Answer:
161.6 cm²Step-by-step explanation:
Surface Area of the rectangular box = 2(LW+LH+WH)
L is the length of the box
W is the width of the box
H is the height of the box
let dL, dW and dH be the possible error in the dimensions L, W and H respectively.
Since there is a possible error of 0.2cm in each dimension, then dL = dW = dH = 0.2cm
The surface Area of the rectangular box using the differentials is expressed as shown;
S = 2{(LdW+WdL)+(LdH+HdL)+(WdH+HdW)]
Also given L = 96cm W = 58cm and H = 48cm, on substituting this given values and the differential error, we will have;
S = 2{(96*0.2+58*0.2) + (96*0.2+48*0.2)+(58*0.2+48*0.2)}
S = 2{19.2+11.6+19.2+9.6+11.6+9.6}
S = 2(80.8)
S = 161.6 cm²
Hence, the surface area of the box is 161.6 cm²
For 9$ a shoekeeper buys 13 dozen pencils.However 3 dozen broke in transit. At what price per dozen must the shoekeeper sell the remaining pencils to make back 1/3 of the whole cost
Answer: $0.30 per dozen.
Step-by-step explanation:
Given: The whole cost of 13 dozen pencils = $9
If 3 dozen broke in transit, remaining dozens of pencils = 13-3 = 10 dozens
Also, Selling price of theses 10 dozens = [tex]\dfrac{1}{3}\times\text{whole cost}[/tex] [given]
[tex]=\dfrac{1}{3}\times9=\$3[/tex]
Then, the selling price of each dozen = [tex]\dfrac{\$3}{10}=\$0.30[/tex]
Hence, the shopkeeper sells the remaining pencils at $0.30 per dozen.
Which expressions are equivalent to -3(2w+6)-4
Answer:
B is the answer
Step-by-step explanation:
-3(2w+6)-4
-6w-18-4
-6w-22
Answer:
B = 2(−3w + (−11)) is the answer.Step-by-step explanation:
-3(2w + 6) - 4
1. Distribute
= -3*2w = -6w
= -3 * 6 = -18
= -6w -18
2. Simplify like terms
= -18 - 4
= -22
3. Place variables and numbers together
= -6w - 22
-6w -22 is the answer.So, B is the answer.Explanation:
2 * -3w = -6w
2*-11 = -22
Place them together and you get the answer!
A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 418 gram setting. It is believed that the machine is underfilling the bags. A 9 bag sample had a mean of 413 grams with a standard deviation of 20. A level of significance of 0.1 will be used. Assume the population distribution is approximately normal. Is there sufficient evidence to support the claim that the bags are underfilled?
Answer:
No. At a significance level of 0.1, there is not enough evidence to support the claim that the bags are underfilled (population mean significantly less than 418 g.)
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the bags are underfilled (population mean significantly less than 418 g.)
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=418\\\\H_a:\mu< 418[/tex]
The significance level is 0.1.
The sample has a size n=9.
The sample mean is M=413.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=20.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{20}{\sqrt{9}}=6.6667[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{413-418}{6.6667}=\dfrac{-5}{6.6667}=-0.75[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=9-1=8[/tex]
This test is a left-tailed test, with 8 degrees of freedom and t=-0.75, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=P(t<-0.75)=0.237[/tex]
As the P-value (0.237) is bigger than the significance level (0.1), the effect is not significant.
The null hypothesis failed to be rejected.
At a significance level of 0.1, there is not enough evidence to support the claim that the bags are underfilled (population mean significantly less than 418 g.)
Proofs are used to show that a mathematical statement is true. The most common form of mathematical statements are if-then statements. Give an example of a true mathematical statement and a false mathematical statement in if-then form. For the false statement, include a counterexample showing that the statement isn’t true.
Answer:
True mathematical statement.
"If x = 0, then for any real number y, we have: y*x = 0."
This is true, and we can prove it with the axioms of the real set.
A false mathematical statement can be:
"if n and x are integer numbers, then n/x is also an integer number."
And a counterexample of this is if we took n = 1 and x = 2, both are integer numbers, so the first part is true, but:
n/x = 1/2 = 0.5 is not an integer number, then the statement is false,
Answer:
True mathematical statement.
"If x = 0, then for any real number y, we have: y*x = 0."
This is true, and we can prove it with the axioms of the real set.
A false mathematical statement can be:
"if n and x are integer numbers, then n/x is also an integer number."
And a counterexample of this is if we took n = 1 and x = 2, both are integer numbers, so the first part is true, but:
n/x = 1/2 = 0.5 is not an integer number, then the statement is false,
Step-by-step explanation:
A rectangular box has a base that is 4 times as long as it is wide. The sum of the height and the girth of the box is 200 feet. (a) Express the volume V of the box as a function of its width w. Determine the domain of V (w).
Answer:
(a) [tex]V = (-8W^3 + 800W^2)/3[/tex]
(b) [tex]W > 100[/tex]
Step-by-step explanation:
Let's call the length of the box L, the width W and the height H. Then, we can write the following equations:
"A rectangular box has a base that is 4 times as long as it is wide"
[tex]L = 4W[/tex]
"The sum of the height and the girth of the box is 200 feet"
[tex]H + (2W + 2H) = 200[/tex]
[tex]2W + 3H = 200 \rightarrow H = (200 - 2W)/3[/tex]
The volume of the box is given by:
[tex]V = L * W * H[/tex]
Using the L and H values from the equations above, we have:
[tex]V = 4W * W * (200 - 2W)/3[/tex]
[tex]V = (-8W^3 + 800W^2)/3[/tex]
The domain of V(W) is all positive values of W that gives a positive value for the volume (because a negative value for the volume or for the width doesn't make sense).
So to find where V(W) > 0, let's find first when V(W) = 0:
[tex](-8W^3 + 800W^2)/3 = 0[/tex]
[tex]-8W^3 +800W^2 = 0[/tex]
[tex]W^3 -100W^2 = 0[/tex]
[tex]W^2(W -100) = 0[/tex]
The volume is zero when W = 0 or W = 100.
For positive values of W ≤ 100, the term W^2 is positive, but the term (W - 100) is negative, then we would have a negative volume.
For positive values of W > 100, both terms W^2 and (W - 100) would be positive, giving a positive volume.
So the domain of V(W) is W > 100.
Current and Quick Ratios The Nelson Company has $1,250,000 in current assets and $500,000 in current liabilities. Its initial inventory level is $335,000, and it will raise funds as additional notes payable and use them to increase inventory. How much can Nelson's short-term debt (notes payable) increase without pushing its current ratio below 2.2? Do not round intermediate calculations. Round your answer to the nearest dollar.
Answer:
Step-by-step explanation:
$1,250,000 in current assets (cr) and $500,000 in current liabilities(cl)
current ratio (cr)=1250000/500000=2.5
Δ note payable ( change in note payable NP)
minimum current=2.2
2.2=(1250000+ΔNP)/(500000+ΔNP)
2.2(500000+ΔNP)=1250000+ΔNP
1100000+2.2ΔNP=1250000+ΔNP
2,2ΔNP-ΔNP=1250000-1100000
1.2ΔNP=150000
ΔNP=150000/1.2=125000
Nelson's short-term debt (notes payable) increase without pushing its current ratio below 2.2=125000
assuming this amount used to increase the inventory
new inventory=335000+125000=360000
current asset= 1250000+125000=1375000
the new ratio=(1375000-360000)/500000+125000
new ratio = 1.624
How do you pronounce B"?
Answer:
bee
Step-by-step explanation:
If C(x) = 16000 + 600x − 1.8x2 + 0.004x3 is the cost function and p(x) = 4200 − 6x is the demand function, find the production level that will maximize profit. (Hint: If the profit is maximized, then the marginal revenue equals the marginal cost.)
Answer:
Quantity that will maximize profit=1000
Step-by-step explanation:
Assume quantity=x
Revenue=price*quantity
=(4200-6x)x
=4200x-6x^2
Marginal revenue(MR) =4200-12x
Cost(x)= 16000 + 600x − 1.8x2 + 0.004x3
Marginal cost(MC) =600-3.6x+0.012x^2
Marginal cost=Marginal revenue
600-3.6x+0.012x^2=4200-12x
600-3.6x+0.012x^2-4200+12x=0
0.012x^2-8.4x-3600=0
Solve the quadratic equation using
x= -b +or- √b^2-4ac/2a
a=0.012
b=-8.4
c=-3600
x=-(-8.4) +or- √(-8.4)^2- (4)(0.012)(-3600) / (2)(0.012)
= 8.4 +or- √70.56-(-172.8) / 0.024
= 8.4 +or- √70.56+172.8 / 0.024
= 8.4 +or- √243.36 / 0.024
= 8.4 +or- 15.6/0.024
= 8.4/0.024 +15.6/0.024
= 350+650
x=1000
OR
= 8.4/0.024 -15.6/0.024
= 350 - 650
= -300
x=1000 or -300
Quantity that maximises profits can not be negative
So, quantity that maximises profits=1000
Use the Laplace transform to solve the given initial-value problem.
y' + 3y = f(t), y(0) = 0
where f(t) = t, 0 ≤ t < 1 0, t ≥ 1
Answer:
The solution to the given Initial - Value - Problem is [tex]y(t) = \frac{-1}{9} + \frac{1}{3}t + \frac{1}{9}e^{-3t} - [\frac{-1}{9} + \frac{1}{3}t - \frac{2}{9}e^{-3(t-1)}]u(t-1)[/tex]
Step-by-step explanation:
y' + 3y = f(t).................(1)
f(t) = t when 0 ≤ t < 1
f(t) = 0 when t ≥ 1
Step 1: Take the Laplace transform of the LHS of equation (1)
That is L(y' + 3y) = sY(s) + 3Y(s) = Y(s)[s + 3]..............(*)
Step 2: Get an expression for f(t)
For f(t) = t when 0 ≤ t < 1
f₁(t) = t (1 - u(t - 1)) ( there is a time shift of the unit step)
For f(t) = 0 when t ≥ 1
f₂(t) = 0(u(t-1))
f(t) = f₁(t) + f₂(t)
f(t) = t - t u(t-1)................(2)
Step 3: Taking the Laplace transform of equation (2)
[tex]F(s) = \frac{1}{s^2} - e^{-s} ( \frac{1}{s^2} + \frac{1}{s})[/tex]...............(**)
Step 4: Equating * and **
[tex]Y(s) [s + 3]=\frac{1}{s^2} - e^{-s} ( \frac{1}{s^2} + \frac{1}{s}) \\Y(s) = \frac{1}{s^2(s+3)} - e^{-s} ( \frac{1}{s^2(s+3)} + \frac{1}{s(s+3)})[/tex].......................(3)
Since y(t) is the solution we are looking for we need to find the Inverse Laplace Transform of equation (3) by first breaking every fraction into partial fraction:
[tex]\frac{1}{s^2 (s+3)} = \frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)}[/tex]
[tex]\frac{1}{s (s+3)} = \frac{1}{3s} + \frac{1}{3(s+3)}[/tex]
We can rewrite equation (3) by representing the fractions by their partial fractions.
[tex]Y(s) = \frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)} - e^{-s} [\frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)} + \frac{1}{3s} + \frac{1}{3(s+3)}]\\Y(s) = \frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)} - e^{-s}[\frac{2}{9s} + \frac{1}{3s^2} - \frac{2}{9(s+3)}][/tex]................(4)
step 5: Take the inverse Laplace transform of equation (4)
[tex]y(t) = \frac{-1}{9} + \frac{1}{3}t + \frac{1}{9}e^{-3t} - u(t-1)[\frac{2}{9} + \frac{1}{3}(t-1) - \frac{2}{9}e^{-3(t-1)}][/tex]
Simplifying the above equation:
[tex]y(t) = \frac{-1}{9} + \frac{1}{3}t + \frac{1}{9}e^{-3t} - [\frac{-1}{9} + \frac{1}{3}t - \frac{2}{9}e^{-3(t-1)}]u(t-1)[/tex]
The Laplace transform is use to solve the differential equation problem.
The solution for the given initial-value problem is,
[tex]y(t)=\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{1}{9}e^-3t-\left[\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{2}{9}e^{-3(t-1)}\right]u(t-1)[/tex]
Given:
The given initial value problem is [tex]y' + 3y = f(t)[/tex].
Consider the left hand side of the given equation.
[tex]y'+3y[/tex]
Take the Laplace transform.
[tex]L(y' + 3y) = sY(s) + 3Y(s) \\L(y' + 3y) = Y(s)[s + 3][/tex]
Consider the right hand side and get the expression for [tex]f(t)[/tex].
[tex]f(t) = t[/tex] when 0 ≤ t < 1
From time shift of the unit step
[tex]f_1(t) = t (1 - u(t - 1))[/tex]
For f(t) = 0 when t ≥ 1
Now,
[tex]f_2(t) = 0(u(t-1))f(t) = f_1(t) + f_2(t)f(t) = t - t u(t-1)[/tex]
Take the Laplace for above expression.
[tex]F(s)=\dfrac{1}{s^2}-e^{-s}\left(\dfrac{1}{s^2}+\dfrac{1}{s}\right)[/tex]
Now, the equate the above two equation.
[tex]Y(s)\left[s+3\right ]=\dfrac{1}{s^2}-e^{-s}\left(\dfrac{1}{s^2}+\dfrac{1}{s}\right)\\Y(s)=\dfrac {1}{(s^2(s+3))}-e^{-s}\left(\dfrac{1}{(s^2(s+3))}+\dfrac{1}{s(s+3)\right)}[/tex]
Find the inverse Laplace for the above equation.
[tex]\dfrac{1}{(s^2(s+3))}=\dfrac{-1}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}\\\dfrac{1}{(s(s+3))}=\dfrac{1}{3s}+\dfrac{1}{3(s+3)}[/tex]
Calculate the partial fraction of above equation.
[tex]Y(s)=\dfrac{-1}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}-e^{-s}\left[\dfrac{-1}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}+\dfrac{1}{3s}+\dfrac{1}{3(s+3)}\right]\\Y(s)=\dfrac{2}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}-e^{-s}\left[\dfrac{2}{9s}+\dfrac{1}{3s^2}-\dfrac{2}{9(s+3)}\right][/tex]
Take the inverse Laplace of the above equation.
[tex]y(t)=\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{1}{9}e^-3t-\left[\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{2}{9}e^{-3(t-1)}\right]u(t-1)[/tex]
Thus, the solution for the given initial-value problem is,
[tex]y(t)=\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{1}{9}e^-3t-\left[\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{2}{9}e^{-3(t-1)}\right]u(t-1)[/tex]
Learn more about what Laplace transformation is here:
https://brainly.com/question/14487937
Find sets of parametric equations and symmetric equations of the line that passes through the two points (if possible). (For each line, write the direction numbers as integers.) (0, 0, 25), (10, 10, 0)
Answer:
a)Parametric equations are
X= -10t
Y= -10t and
z= 25+25t
b) Symmetric equations are
(x/-10) = (y/-10) = (z- 25)/25
Step-by-step explanation:
We were told to fin two things here which are ; a) the parametric equations and b) the symmetric equations
The given two points are (0, 0, 25)and (10, 10, 0)
The direction vector from the points (0, 0, 25) and (10, 10, 0)
(a,b,c) =( 0 -10 , 0-10 ,25-0)
= < -10 , -10 ,25>
The direction vector is
(a,b,c) = < -10 , -10 ,25>
The parametric equations passing through the point (X₁,Y₁,Z₁)and parallel to the direction vector (a,b,c) are X= x₁+ at ,y=y₁+by ,z=z₁+ct
Substitute (X₁ ,Y₁ ,Z₁)= (0, 0, 25), and (a,b,c) = < -10 , -10 ,25>
and in parametric equations.
Parametric equations are X= 0-10t
Y= 0-10t and z= 25+25t
Therefore, the Parametric equations are
X= -10t
Y= -10t and
z= 25+25t
b) Symmetric equations:
If the direction numbers image and image are all non zero, then eliminate the parameter image to obtain symmetric equations of the line.
(x-x₁)/a = (y-y₁)/b = (z-z₁)/c
CHECK THE ATTACHMENT FOR DETAILED EXPLANATION
The following chart represents the record low temperatures recorded in Phoenix for April-November. Select the answer below that best describes the mean and the median of the data set (round answers to the nearest tenth). A graph titled Phoenix Low Temperatures has month on the x-axis and temperature (degrees Fahrenheit) on the y-axis. April, 32; May, 40; June, 50; July, 61; August, 60; September, 47; October, 34; November, 25. a. The mean is 43.5°F, and the median is 43.6°F. b. The mean is 60.5°F, and the median is 60.5°F. c. The mean is 60°F, and the median is 61°F. d. The mean is 43.6°F, and the median is 43.5°F.
Answer:
d. The mean is 43.6°F, and the median is 43.5°F.
Step-by-step explanation:
Hello!
The data corresponds to the low temperatures in Phoenix recorded for April to November.
April: 32ºF
May: 40ºF
June: 50ºF
July: 61ºF
August: 60ºF
September: 47ºF
October: 34ºF
November: 25ºF
Sample size: n= 8 months
The mean or average temperature of the low temperatures in Phoenix can be calculated as:
[tex]\frac{}{X}[/tex]= ∑X/n= (32+40+50+61+60+47+34+25)/8= 43.625ºF (≅ 43.6ºF)
The Median (Me) is the value that separates the data set in two halves, first you have to calculate its position:
PosMe= (n+1)/2= (8+1)/2= 4.5
The value that separates the sample in halves is between the 4th and the 5th observations, so first you have to order the data from least to greatest:
25; 32; 34; 40; 47; 50; 60; 61
The Median is between 40 and 47 ºF, so you have to calculate the average between these two values:
[tex]Me= \frac{(40+47)}{2} = 43.5[/tex] ºF
The correct option is D.
I hope this helps!
Answer:
it is d
Step-by-step explanation:
Given the following functions, evaluate each of the following
Answer:
[tex](f+g)(5) = 40\\(f-g)(5) = 22\\(f*g)(5) = 279[/tex]
[tex](f/g)(5) = 31/9[/tex]
Step-by-step explanation:
[tex]f(5) = (5)^2+2(5)-4\\f(5) = 25+10-4\\f(5) = 31[/tex]
[tex]g(5) = 5+4\\g(5) = 9[/tex]
[tex](f+g)(5) = 31+9\\(f+g)(5) = 40[/tex]
[tex](f-g)(5) = 31-9\\(f-g)(5) = 22[/tex]
[tex](f*g)(5) = 31*9\\(f*g)(5) = 279\\[/tex]
[tex](f/g)(5) = 31/9[/tex]
The length of a rectangle is 5M more than twice the width and the area of the rectangle is 63M to find the dimension of the rectangle
Answer:
width = 4.5 m
length = 14 m
Step-by-step explanation:
okay so first you right down that L = 5 + 2w
then as you know that Area = length * width so you replace the length with 5 + 2w
so it's A = (5 +2w) * w = 63
then 2 w^2 + 5w - 63 =0
so we solve for w which equals 4.5 after that you solve for length : 5+ 2*4.5 = 14
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▹ Answer
0.25 = 1/4 because 25/100 = 1/4
▹ Step-by-Step Explanation
0.25 to a fraction → 25/100
25/100 = 1/4
Therefore, this statement is true. (0.25 = 1/4 because 25/100 = 1/4)
Hope this helps!
- CloutAnswers ❁
Brainliest is greatly appreciated!
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Find the remainder when f(x)=2x3−x2+x+1 is divided by 2x+1.
Step-by-step explanation:
it can be simply done by using remainder theorem.
Jina wants to measure the width of a river. She marks off two right triangles, as shown in the figure. The base of the larger triangle has a length of 56m, and the base of the smaller triangle has a length of 26m. The height of the smaller triangle is 20.9m. How wide is the river? Round your answer to the nearest meter.
Answer:
width of a river = 45m
Step-by-step explanation:
ration and proportion
let x = width of a river
x 20.9 m
------ = --------
56 m 26 m
x = (20.9 * 56) / 26
x = 45 m
therefore the width of a river is 45 m
Write an equation in slope-intercept form of the line that passes through the point (-6,-5) with slope 6.
Answer:
y=6x+31
Step-by-step explanation:
Since we are given a point and a slope, we can use the slope-intercept formula.
[tex]y-y_{1} =m(x-x_{1})[/tex]
where (x1,y1) is a point on the line and m is the slope.
The point given is (-6,-5) and the slope is 6.
x1= -6
y1= -5
m=6
[tex]y--5=6(x--6)[/tex]
A negative number subtracted from another number, or two negative signs, becomes a positive.
[tex]y+5=6(x+6)[/tex]
We want to find the equation of the line, which is y=mx+b (m is the slope and b is the y-intercept). Therefore, we must get y by itself on one side of the equation.
First, distribute the 6. Multiply each term inside the parentheses by 6.
[tex]y+5=(6*x)+(6*6)[/tex]
[tex]y+5=6x+36[/tex]
Subtract 5 from both sides, because it is being added on to y.
[tex]y+5-5=6x+36-5[/tex]
[tex]y=6x+36-5[/tex]
[tex]y=6x+31[/tex]
The equation of the line is y=6x+31
How do I solve this problem
Answer:
It would take 1 more mile if he took route Street A and then Street B rather than just Street C.
Step-by-step explanation:
Pythagorean Theorem: a² + b² = c²
We use the Pythagorean Theorem to find the length of Street C:
2² + 1.5² = c²
c = √6.25
c = 2.5
Now we find how much longer route A and B is compared to C:
3.5 - (2 + 1.5) = 3.5 - 2.5 = 1
You observe the following pattern: J, K, N, B. What is the next letter in the sequence?
a. X
b. B
c. G
d. U
e. E
f. S
Answer:
C. G
Step-by-step explanation:
The National Safety Council (NSC) estimates that off-the-job accidents cost U.S. businesses almost $200 billion annually in lost productivity (National Safety Council, March 2006). Based on NSC estimates, companies with 50 employees are expected to average three employee off-the-job accidents per year. Answer the following questions for companies with 50 employees.a. What is the probability of no off-the-job accidents during a one-year period (to 4 decimals)?b. What is the probability of at least two off-the-job accidents during a one-year period (to 4 decimals)?c. What is the expected number of off-the-job accidents during six months (to 1 decimal)?d. What is the probability of no off-the-job accidents during the next six months (to 4 decimals)?
Answer:
a. 0.0498 = 4.98% probability of no off-the-job accidents during a one-year period
b. 0.8008 = 80.08% probability of at least two off-the-job accidents during a one-year period.
c. The expected number of off-the-job accidents during six months is 1.5.
d. 0.2231 = 22.31% probability of no off-the-job accidents during the next six months.
Step-by-step explanation:
We have the mean during a period, so we use the Poisson distribution.
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^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
Companies with 50 employees are expected to average three employee off-the-job accidents per year.
This means that [tex]\mu = 3n[/tex], in which n is the number of years.
a. What is the probability of no off-the-job accidents during a one-year period (to 4 decimals)?
This is [tex]P(X = 0)[/tex] when [tex]\mu = 3*1 = 3[/tex]. So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498[/tex]
0.0498 = 4.98% probability of no off-the-job accidents during a one-year period.
b. What is the probability of at least two off-the-job accidents during a one-year period (to 4 decimals)?
Either there are less than two accidents, or there are at least two. The sum of the probabilities of these events is 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]. Then
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/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) = P(X = 0) + P(X = 1) = 0.0498 + 0.1494 = 0.1992[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1992 = 0.8008[/tex]
0.8008 = 80.08% probability of at least two off-the-job accidents during a one-year period.
c. What is the expected number of off-the-job accidents during six months (to 1 decimal)?
6 months is half a year, so [tex]n = 0.5[/tex]
[tex]\mu = 3n = 3*0.5 = 1.5[/tex]
The expected number of off-the-job accidents during six months is 1.5.
d. What is the probability of no off-the-job accidents during the next six months (to 4 decimals)?
This is P(X = 0) when [tex]\mu = 1.5[/tex]. So
[tex]P(X = 0) = \frac{e^{-1.5}*1.5^{0}}{(0)!} = 0.2231[/tex]
0.2231 = 22.31% probability of no off-the-job accidents during the next six months.
A class of 30 music students includes 13 who play the piano, 15 who play the guitar, and 9 who play both the piano and the guitar. How many students in the class play neither instrument?
Answer: 2
Step-by-step explanation:
As given, out of 30 students, 15 play guitar and 13 play piano, thats 28.
Among these, 9 play both the guitar and the piano.
That means, only 2 remaining students play neither instrument. (30-15-13)
The two-way table shows the medal count for the top-performing countries in the 2012 Summer Olympics. A 5-column table has 5 rows. The first column has entries United notes, China, Russia, Great Britain, Total. The second column is labeled Gold with entries 46, 38, 24, 29, 137. The third column is labeled Silver with entries 29, 27, 26, 17, 99. The fourth column is labeled Bronze with entries 29, 23, 32, 19, 103. The fifth column is labeled Total with entries 104, 88, 82, 65, 339. Which statement is true?
Which statement is true?
The probability that a randomly selected silver medal was awarded to Great Britain is StartFraction 17 Over 99 EndFraction. The probability that a randomly selected medal won by Russia was a bronze medal is StartFraction 32 Over 103 EndFraction. The probability that a randomly selected gold medal was awarded to China is StartFraction 88 Over 137 EndFraction. The probability that a randomly selected medal won by the United States was a silver medal is StartFraction 104 Over 339 EndFraction.Answer:
(A)The probability that a randomly selected silver medal was awarded to Great Britain is 17/99.
Step-by-step explanation:
The table is given below:
[tex]\left|\begin{array}{l|c|c|c|c|c} &Gold&Silver & Bronze &Total\\United States &46 & 29 & 29 & 104\\China & 38 & 27 & 23 & 88\\Russia & 24 & 26 & 32 &82\\Great Britain & 29 & 17 & 19 & 65\\&&&&&\\Total &137 & 99 & 103 & 339\end{array}\right[/tex]
We calculate the probabilities given in the statements.
(A) The probability that a randomly selected silver medal was awarded to Great Britain
= 17/99
(B)The probability that a randomly selected medal won by Russia was a bronze medal
=32/82
(C)The probability that a randomly selected gold medal was awarded to China
=38/137
(D)The probability that a randomly selected medal won by the United States was a silver medal
=29/104
We can see that only the first statement is true.
Answer: A. The probability that a randomly selected silver medal was awarded to Great Britain is 17/99.
Step-by-step explanation:
I got it right on edge
Type the correct answer in each box. Use numerals instead of words. Please Help!
Answer:
x = -3h
x = -12
Step-by-step explanation:
Given expression is,
[tex]\frac{x}{h}+1=-2[/tex]
By adding 2 on both the sides of the equation,
[tex]\frac{x}{h}+1+2=-2+2[/tex]
[tex]\frac{x}{h}+3=0[/tex]
Now subtract 3 form both the sides,
[tex]\frac{x}{h}+3-3=0-3[/tex]
[tex]\frac{x}{h}=-3[/tex]
Multiply the equation by 'h'
x = -3h
If h = 4,
By substituting h = 4 in the equation,
x = -3(4)
x = -12