Answer:
-8/5Step-by-step explanation:
Given two lines y=(3a+2)x-2 and 2y=(a-4)x+2, Since both lines are parallel to each other, this means that the slope of both lines are the same
Let's get the slope of both equation. For the first equation;
y=(3a+2)x-2
We can see that the equation is written in this form y = mx+c where m is the slope of the line. On comparison, the slope of the given line is 3a+2
Similarly for the second line;
2y=(a-4)x+2
Re-writing in the standard format we will have;
y = (a-4)x/2+2/2
y = (a-4)x/2 + 1
The slope of the second line is (a-4)/2
On equating the slope of both lines to get the value of 'a' we will have;
3a+2 = (a-4)/2
Cross multiplying
2(3a+2) = a-4
6a+4 = a-4
Collecting like terms;
6a-a = -4-4
5a = -8
a = -8/5
Hence the value of a is -8/5
write the equation for taking away 5 from x gives 10
Answer:
[tex]\boxed{\sf x - 5 = 10}[/tex]
Step-by-step explanation:
Taking away 5 from x ⇒ subtracting 5 from x
[tex]\large {\sf x - 5[/tex]
Gives 10 ⇒ result is 10
[tex]\large {\sf x - 5 = 10[/tex]
Searches related to Searches related to A motorboat travels 135 kilometers in 3 hours going upstream. It travels 183 kilometers going downstream in the same amount of time. What is the rate of the boat in still water? what is the rate of the current?
Answer:
[tex]\large \boxed{\sf \text{The rate of the boat is } 53 \ km/h \text{, the rate of the current is }8\ km/h \ \ }[/tex]
Step-by-step explanation:
Hello, let's note v the rate of the boat and r the rate of the current. We can write the following
[tex]\dfrac{135}{v-r}=3=\dfrac{183}{v+r}[/tex]
It means that
[tex]135(v+r)=183(v-r)\\\\135 v + 135r=183v-183r\\\\\text{ *** We regroup the terms in v on the right and the ones in r to the left***}\\\\(135+183)r=(183-135)v\\\\318r=48v\\\\\text{ *** We divide by 48 both sides ***}\\\\\boxed{v = \dfrac{318}{48} \cdot r= \dfrac{159}{24} \cdot r}[/tex]
But we can as well use the second equation:
[tex]3(v+r)=183\\\\v+r=\dfrac{183}{3}=61\\\\\dfrac{159}{24}r+r=61\\\\\dfrac{159+24}{24}r=61\\\\\boxed{r = \dfrac{61*24}{183}=8}[/tex]
and then
[tex]\boxed{v=\dfrac{159*8}{24}=53}[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
assume that when adults with smartphones are randomly selected 15 use them in meetings or classes if 15 adult smartphones are randomly selected, find the probability that at least 4 of them use their smartphones
Answer:
The probability that at least 4 of them use their smartphones is 0.1773.
Step-by-step explanation:
We are given that when adults with smartphones are randomly selected 15% use them in meetings or classes.
Also, 15 adult smartphones are randomly selected.
Let X = Number of adults who use their smartphones
The above situation can be represented through the binomial distribution;
[tex]P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; n = 0,1,2,3,.......[/tex]
where, n = number of trials (samples) taken = 15 adult smartphones
r = number of success = at least 4
p = probability of success which in our question is the % of adults
who use them in meetings or classes, i.e. 15%.
So, X ~ Binom(n = 15, p = 0.15)
Now, the probability that at least 4 of them use their smartphones is given by = P(X [tex]\geq[/tex] 4)
P(X [tex]\geq[/tex] 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)
= [tex]1- \binom{15}{0}\times 0.15^{0} \times (1-0.15)^{15-0}-\binom{15}{1}\times 0.15^{1} \times (1-0.15)^{15-1}-\binom{15}{2}\times 0.15^{2} \times (1-0.15)^{15-2}-\binom{15}{3}\times 0.15^{3} \times (1-0.15)^{15-3}[/tex]
= [tex]1- (1\times 1\times 0.85^{15})-(15\times 0.15^{1} \times 0.85^{14})-(105 \times 0.15^{2} \times 0.85^{13})-(455 \times 0.15^{3} \times 0.85^{12})[/tex]
= 0.1773
please answer this correctly
How far apart are the gift shop and the science lab
Please answer this correctly without making mistakes
The answer is 86.4 km
Explanation:
The graph shows the gift shop is to the east of the science lab, and, between the gift shop and the science lab it is the art supply. Besides this, the description of the graph provides the distance between the art supply and the science lab, which is 40.0, as well as, the distance between the art supply and the gift shop, which is 46.4 kilometers.
In this context, it is possible to calculate the distance from the science lab to the gift shop by adding the partial distances, considering the art supply as a middle point in the map. This means the distance from the lab to the gift shop = 40.0 km (distance from the lab to the art supply) + 46.4 km (distance from the art supply to the gift shop) = 86.4 km.
Find the volume of the solid shown or described. If necessary, round to the nearest tenth.
Answer:
37.7
Step-by-step explanation:
Well use the formula for the volume of a cylinder which is,
,[tex]\pi r^2 h[/tex]
So the radius is 2 and the height is 3, so we plug those numbers into the formula,
(pi)(2)^2(3)
2^2 is 4 4*3 is 12
12*pi is about 37.7 rounded to the nearest tenth.
If you would like to check look at the image below.
A circle has a center at (4, -7) and a radius of 4 units. Write an equation of this circle.
Answer:
(x – 4)^2 + (y + 7)^2 = 16
Step-by-step explanation:
The formula of a circle is:
(x – h)^2 + (y – k)^2 = r^2
(h, k) represents the coordinates of the center of the circle
r represents the radius of the circle
If you plug in the given information, you get:
(x – 4)^2 + (y – (-7))^2 = 4^2
which simplifies into:
(x – 4)^2 + (y + 7)^2 = 16
When testing the claim that p 1p1equals=p 2p2, a test statistic of zequals=2.04 is obtained. Find the p-value obtained from this test statistic.
Answer:
0.0414 with an upper tailed test
Step-by-step explanation:
Claim: P1P1 = P2P2
The above is a null hypothesis
The alternative hypothesis for a two-tailed test would be:
P1P1 \=/ P2P2
Where "\=/" represents "not equal to".
Using a z-table or z-calculator, we derive the p-value (probability value) for the z-score 2.04
With an upper tailed test, the
2 × [probability that z>2.04] = 2[0.0207] = 0.0414
This is the p-value for the test statistic.
Focus is on the alternative hypothesis.
Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) x4 18x2 4 x5 30x3 20x dx
Your integrand is missing some symbols. My best interpretation is the following integral:
[tex]I=\displaystyle\int\frac{x^4+18x^2+4}{x^5+30x^3+20x}\,\mathrm dx[/tex]
Decompose into partial fractions; we're looking for an expansion of the form
[tex]\dfrac{x^4+18x^2+4}{x^5+30x^3+20x}=\dfrac ax+\dfrac{bx^3+cx^2+dx+e}{x^4+30x^2+20}[/tex]
Now:
[tex]x^4+18x^2+4=a(x^4+30x^2+20)+(bx^3+cx^2+dx+e)x[/tex]
[tex]=(a+b)x^4+cx^3+(30a+d)x^2+ex+20a[/tex]
Matching up coefficients tells us that
[tex]\begin{cases}a+b=1\\c=0\\30a+d=18\\e=0\\20a=4\end{cases}\implies a=\dfrac15,b=\dfrac45,d=12[/tex]
so that
[tex]I=\displaystyle\frac15\int\frac{\mathrm dx}x+\frac45\int\frac{x^3+15x}{x^4+30x^2+20}\,\mathrm dx[/tex]
The integral is trivial:
[tex]\displaystyle\frac15\int\frac{\mathrm dx}x=\frac15\ln|x|+C[/tex]
For the second integral, notice that
[tex]\mathrm d(x^4+30x^2+20)=(4x^3+60x)\,\mathrm dx[/tex]
Distribute the 4 over the numerator, then substitute [tex]u=x^4+30x^2+20[/tex] and [tex]\mathrm du=(4x^3+60x)\,\mathrm dx[/tex]:
[tex]\displaystyle\frac15\int\frac{4x^3+60x}{x^4+30x^2+20}\,\mathrm dx=\frac15\int\frac{\mathrm du}u=\frac15\ln|u|+C=\frac15\ln(x^4+30x^2+20)+C[/tex]
So we have
[tex]I=\dfrac15\ln|x|+\dfrac15\ln(x^4+30x^2+20)+C[/tex]
and with some simplification,
[tex]I=\boxed{\ln\sqrt[5]{|x^5+30x^3+20x|}+C}[/tex]
The Box-and-Whisker plot shows the average temperatures in, atlanta, georgia, in march. which statement about the temperatures in atlanta must be true? A. about half the days in march had average temperatures above 60 degrees. B. about half the days in march had average temperatures either below 60 or above 73 degrees C. the coldest day in march was 51 D. the hottest day in march was 84
Answer:
"B. about half the days in march had average temperatures either below 60 or above 73 degrees"
Step-by-step explanation:
To answer this question, note that a box plot is usually divided into quartiles, each representing approximately 25% each.
In the box plot above,
*about 25% (Q1) represents days with temperature of 60° and below. This is about ¼ of the days in March.
*About 25% (Q2) represents days with temperature between 61° and 68°. That's about ¼ of the days in March
*About 25% (Q3) represents days with temperature between 70° and 73°. That's about ¼ of the days in March
*About 25% (Q4) represents days with temperature between 74° and 82°. That's about ¼ of the days in March
*Coldest day in March has a temperature of 54°
*Hottest day in March is 82°
From the options given, the only statement that is true is "B. about half the days in march had average temperatures either below 60 or above 73 degrees"
¼ of the Days in March has temperatures below 60° (Q1), while ¼ of the days in March has temperatures above 73° (Q4). Therefore, ¼+¼ = ½ of the days in March having average temperatures either below 60 or above 73 degrees.
Answer:
b
Step-by-step explanation:
About half of the days in March had average temperatures either below 60 or above 73 degrees.
CAN SOMEONE PLEASE HELP ME. ITS HARD AND I CANT SOLVE IT. ILL MARK BRAINLIEST
Answer:
Step-by-step explanation:
Hello,
First of all, we need to find the equation of the line.
It will be something like y = ax + b (we need to find a and b).
Then the graph is the part of the plan which is above this line, so the inequality will be
[tex]\large \boxed{\sf \ \ y\geq ax+b \ \ }[/tex]
There is only one line passing by two different points, right?
We need to find two points of this line, the x-axis gives the x of the point and the y-axis give the y of the point.
We can see on the graph that (-1,-1) and (6,13) are two points of this line.
I attached a graph with the two points A (-1,-1) and B (6,13).
We need to solve y = ax + b to find a and b which means:
(-1) = a(-1 )+ b <=> -1 = -a + b <=> -a + b = -1
(13) = a(6) + b <=> 6a + b = 13
To eliminate b, we can subtract the first equation from the second one.
6a + b -(-a) - b = 13 - (-1) = 13 + 1 = 14
<=> 6a + a = 7a = 14 we can divide by 7 both parts of the equation.
[tex]\large \boxed{\sf \ \ a=\dfrac{14}{7}=2 \ \ }[/tex]
And then, b = -1 + a = -1 + 2 = 1
[tex]\large \boxed{\sf \ \ b=1 \ \ }[/tex]
The equation of the line is then [tex]y=2x+1[/tex]
So, the inequality is:
[tex]\Large \boxed{\sf \ \ y\geq 2x+1 \ \ }[/tex]
And this is the answer of the question 1.
2. First, we find the equation of the line by finding two points on this line. Then, we deduce the inequality from the equality as the graph is all the points above the line (as explained above).
3. A real-life example is the following.
My parents gave $1 to my sister and they will give her $2 every day.
So today at x = 0 she has $1 and her pocket money will be represented by the line. And I told my parents that there were no way I can get less than my little sister. To help them understand what can be an acceptable deal I provided this graph, and I told them to only look for t positive of course.
The label for the x-axis is the time.
The label for the y-axis is the pocket money in $
After one day x = 1 my sister has $3 (point on the line), a possible solution for my pocket money is $6, $14 (two points from the graph) is also a solution of course :-)
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
How do I use intercepts to graph 3y= - 5x - 30
Answer:
y-intercept is (0,-10) and x-intercept is (-6,0). Connect them by a straight line to graph the given equation.
Step-by-step explanation:
The given equation of line is
[tex]3y=-5x-30[/tex]
For x=0,
[tex]3y=-5(0)-30[/tex]
[tex]3y=-30[/tex]
[tex]y=-10[/tex]
So, y-intercept is at point (0,-10).
For y=0,
[tex]3(0)=-5x-30[/tex]
[tex]0=-5x-30[/tex]
[tex]5x=-30[/tex]
[tex]x=-6[/tex]
So, x-intercept is at point (-6,0).
Now, plot the point (0,-10) and (-6,0) on a coordinate plane and connect them by a straight line to graph the given line as shown below.
As the Type II error, β,of a statistical test increases, the power of the test _____________.
Answer:
decreases.
Step-by-step explanation:
Type II error is one in which we fail to reject the null hypothesis that is actually false. Null hypothesis is a statement that is to be tested against the alternative hypothesis and then decision is taken whether to accept or reject the null hypothesis. The power of Type II error is 1 - [tex]\beta[/tex]. As the power increases the probability of Type II error decreases.
A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = 0.05. Suppose a random sample of 100 male students is selected and the GPA for each student is calculated. What is the probability that the random sample of 100 male students has a mean GPA greater than 3.42?
Answer: 0.0548
Step-by-step explanation:
Given, A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = 0.05.
Let [tex]\overline{X}[/tex] represents the sample mean GPA for each student.
Then, the probability that the random sample of 100 male students has a mean GPA greater than 3.42:
[tex]P(\overline{X}>3.42)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{3.42-3.5}{\dfrac{0.5}{\sqrt{100}}})\\\\=P(Z>\dfrac{-0.08}{\dfrac{0.5}{10}})\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(Z>1.6)\\\\=1-P(Z<1.6)\\\\=1-0.9452=0.0548[/tex]
hence, the required probability is 0.0548.
What is the five number summary for this data set?
3, 8, 14, 19, 22, 29, 33, 37, 43, 49
Assume the numbers in each answer choice are listed in this order: min, Q1,
median, Q3, max
Answer:
see explanation
Step-by-step explanation:
The median is the middle value of the data set in ascending order. If there is no exact middle then the median is the average of the values either side of the middle.
Given
3 8 14 19 22 29 33 37 43 49
↑ middle is between 22 and 29
median = [tex]\frac{22+29}{2}[/tex] = [tex]\frac{51}{2}[/tex] = 25.5
The upper quartile [tex]Q_{3}[/tex] is the middle value of the data to the right of the median.
29 33 37 43 49
↑
[tex]Q_{3}[/tex] = 37
The lower quartile [tex]Q_{1}[/tex] is the middle value of the data to the left of the median.
3 8 14 19 22
↑
[tex]Q_{1}[/tex] = 14
The min is the smallest value in the data set, that is 3
The max is the largest value in the data set, that is 49
The 5 number summary is
3, 14, 25.5, 37, 49
differentiate with respect to X
[tex] \sqrt{ \frac{cos2x}{1 +sin2x } } [/tex]
Power and chain rule (where the power rule kicks in because [tex]\sqrt x=x^{1/2}[/tex]):
[tex]\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'[/tex]
Simplify the leading term as
[tex]\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}[/tex]
Quotient rule:
[tex]\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}[/tex]
Chain rule:
[tex](\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)[/tex]
[tex](1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)[/tex]
Put everything together and simplify:
[tex]\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}[/tex]
[tex]=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}[/tex]
[tex]=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}[/tex]
[tex]=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}[/tex]
[tex]=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}[/tex]
[tex]=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}[/tex]
[tex]=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}[/tex]
Vector has x and y components of -8.80 cm and 18.0 cm, respectively; vector has x and y components of 12.2 cm and -6.80 cm, respectively. If - + 3 = 0, what are the components of ? x = cm y = cm
Question:
Vector A has x and y components of −8.80 cm and 18.0 cm , respectively; vector B has x and y components of 12.2 cm and −6.80 cm , respectively. If A − B +3 C = 0, what are the components of C?
Answer:
x = ___ cm
y = ___ cm
Answer:
x = 7.0cm
y = -8.27cm
Step-by-step explanation:
For a vector F, with x and y components of a and b respectively, its unit vector representation is as follows;
F = ai + bj [Where i and j are unit vectors in the x and y directions respectively]
Using this analogy, let's represent vectors A and B from the question in their unit vector notation.
A has an x-component of -8.80cm and y-component of 18.0cm
B has an x-component of 12.2cm and y-component of -6.80cm,
In unit vector notation, these become;
A = -8.80i + 18.0j
B = 12.2 i + (-6.80)j = 12.2i - 6.80j
Also, there is a third vector C. Let the x and y components of C be a and b respectively. Therefore,
C = ai + bj
Now,
A - B + 3C = 0 [substitute the vectors]
=> [-8.80i + 18.0j] - [12.2 i -6.80j] + [3(ai + bj)] = 0 [open brackets]
=> -8.80i + 18.0j - 12.2 i + 6.80j + 3(ai + bj) = 0
=> -8.80i + 18.0j - 12.2 i + 6.80j + 3ai + 3bj = 0
=> -8.80i + 18.0j - 12.2 i + 6.80j + 3ai + 3bj = 0 [collect like terms and solve]
=> -8.80i - 12.2 i + 3ai + 6.80j + 18.0j + 3bj = 0
=> -21.0 i + 3ai + 24.8j + 3bj = 0 [re-arrange]
=> 3ai + 3bj = 21.0i - 24.8j
Comparing both sides shows that;
3a = 21.0 -------------(i)
3b = -24.8 -----------(ii)
From equation (i)
3a = 21.0
a = 21.0 / 3 = 7.0
From equation (ii)
3b = -24.8
b = -24.8 / 3
b = -8.27
Therefore, the x-component and y-component of vector B which are a and b, are 7.0cm and -8.27cm respectively.
There are 42 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of 5 min and a standard deviation of 6 min. (Give answers accurate to 3 decimal places.)
(a) If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins?
1
(b) If the sports report begins at 11:10, what is the probability that he misses part of the report if he waits until grading is done before turning on the TV?
2
Answer:
A) 0.99413
B) 0.00022
Step-by-step explanation:
A) First of all let's find the total grading time from 6:50 P.M to 11:00 P.M.:
Total grading time; X = 11:00 - 6:50 = 4hours 10minutes = 250 minutes
Now since we are given an expected value of 5 minutes, the mean grading time for the whole population would be:
μ = n*μ_s ample = 42 × 5 = 210 minutes
While the standard deviation for the population would be:
σ = √nσ_sample = √(42 × 6) = 15.8745 minutes
To find the z-score, we will use the formula;
z = (x - μ)/σ
Thus;
z = (250 - 210)/15.8745
z = 2.52
From the z-distribution table attached, we have;
P(Z < 2.52) ≈ 0.99413
B) solving this is almost the same as in A above, the only difference is an additional 10 minutes to the time.
Thus, total time is now 250 + 10 = 260 minutes
Similar to the z-formula in A above, we have;
z = (260 - 210)/15.8745
z = 3.15
P(Z > 3.15) = 0.00022
A manufacturer makes plastic wrap used in food packaging and aims to have a minimum breaking strength of 0.5 kg. If the mean breaking strength of a sample drops below a critical value, the production process is halted and the machinery is inspected. Which of the following is a Type 1 error in context?
A) Halting the production process when too many rubber bands break.
B) Halting the production process when the true breaking strength is below the desired level.
C) Halting the production process when the true breaking strength is within specifications.
D) Allowing the production process to continue when the true breaking strength is below specifications.
E) Allowing the production process to continue when the true breaking strength is within specifications
Answer:
Option D
Step-by-step explanation:
A type I error occurs when you reject the null hypothesis when it is actually true.
The null hypothesis in this case is minimum breaking strength is less than or equal to 0.5.
A type one error would be allowing the production process to continue when the true breaking strength is below specifications.
In recent survey, a school district randomly dialed 300 parent phone numbers and asked whether the family read to their toddlers at least three nights per week. 271 of the parents said that they did.
(a) Nationally , 72% of families read to their toddlers at least three nights per week. Does this data provide evidence that more parents at this district read to their children than the national average ? Show all steps in your process.
(b) Name two likely sources of bias in this survey. Write one sentence to explain each.
Answer:
a) We reject H₀ we have enough evidence for that
b) 1.-The survey was made over a district, results will surely be different if the survey is carried out over a whole state ( considering urban and rural areas)
2.-In a district we find an equalized level of salaries, which could be associated with a similar level of habits
Step-by-step explanation:
We have to develop a proportion test. One tail-test
Population proportion mean (national proportion) p₀ = 72 %
Sample information
sample proportion 271/300 p = 0,90 p = 90 %
We assume Confidence Interval 90 % then α = 0,1
Test Hypothesis
Null Hypothesis H₀ p = p₀
Alternative Hypothesis Hₐ p > p₀
As α = 0,1 and
We look at z-table for z(c) and find z(c) = 1,28
And we compute z(s) as
z(s) = ( p - p₀ ) √ (p₀q₀/n
z(s) = ( 0,9 - 0,72 ) /√(0,72)*(0,28)/300
z(s) = 0,18 / √0,1296/300
z(s) = 0,18/ 0,026
z(s) = 6,92
z(s) > z(c) 6,92 > 1,28
As z(s) is bigger than z(c), z(s) is in the rejection region, so we reject the null hypothesis. We have enough evidence to claim that the proportion in the district is bigger than the national one.
b)1. The survey was made over a district, results will surely be different if the survey is carried out over a whole state ( considering urban and rural areas)
2.-In a district we find an equalized level of salaries, which could be associated with a similar level of habits
Find the area of the shaded regions.
Answer:
[tex]A = A_c-A_t=4\pi -8=4.5664cm^2[/tex]
Step-by-step explanation:
The area of the shaded region can be calculated as the area of the semicircle less the area of the right triangle.
The area of the right triangle can be calculated as:
[tex]A_t=\frac{b*h}{2} =\frac{LM*MN}{2}[/tex]
Where LM and MN have the same length because the internal angles are L=45°, M=90°, and N=45°. So the area is:
[tex]A_t=\frac{4*4}{2}=8[/tex]
The diameter of the circle can be calculated using the Pythagorean theorem as:
[tex]D=\sqrt{(LM)^2+(MN)^2} =\sqrt{4^2+4^4}=4\sqrt{2}[/tex]
So, the radius is [tex]r=2\sqrt{2}[/tex]
Finally, the area of the semicircle is:
[tex]A_c=\frac{\pi*r^2 }{2}=\frac{\pi*(2\sqrt{2})^2 }{2}=4\pi[/tex]
So, the area of the shaded region is:
[tex]A = A_c-A_t=4\pi -8=4.5664cm^2[/tex]
The total cost for my brother's bowling party was $140. It cost $50to reserve a bowling lane plus the cost of renting shoes for the 9 people attending.
Answer:
$10 to rent shoes for 9 people
Step-by-step explanation:
Total amount of the party = $140
A bowling lane = $50
$140 - $50 = $90
$90 divided by 9 = 10
$10 to rent shoes for 9 people
Find x and round to the nearest tenth.
Answer:
83.0°
Step-by-step explanation:
Given ∆XYZ, with 3 known sides, to find angle X, apply the Law of Cosines, c² = a² + b² - 2ab*cos(C).
For convenience sake, this formula can be rewritten to make the angle we are looking for the subject of the formula.
Thus, we would have this following:
[tex] cos(C) = \frac{a^2 + b^2 - c^2}{2ab} [/tex]
Where,
C = X = ?
a = 8 ft
b = 16 ft
c = 17 ft
Plug in the stated values into the formula and solve for X
[tex] cos(X) = \frac{8^2 + 16^2 - 17^2}{2*8*16} [/tex]
[tex] cos(X) = \frac{320 - 289}{256} [/tex]
[tex] cos(X) = \frac{31}{256} [/tex]
[tex] cos(X) = 0.1211 [/tex]
[tex] X = cos^{-1}(0.1211) [/tex]
[tex] X = 83.0 [/tex] (to nearest tenth)
Answer:
its actually 83 not 83.0
Step-by-step explanation:
im only saying this bc i know people with type 83.0 in the box
Help with 5 questions frequency table
Answer:
The given data is:
30, 32, 11, 14, 40, 37, 16, 26, 12, 33, 13, 19, 38, 12, 25, 15, 39, 11, 37, 17, 27, 14, 36
We will fill the table with the relevant information:
Question 1: 21 - 25 (because the previous range stops at 20 and the following range starts at 26)
Question 2: III (write 3 as a tally)
Question 3: II (write 2 as a tally)
Question 4: 8 (write the tally as a number)
Question 5: 4 (write IIII as a number)
Please answer this correctly without making mistakes
Answer:
3/11
Step-by-step explanation:
There are eleven equal parts.
So the denominator is 11.
He copies 8 parts on Sunday.
11-8=3.
He copied 3 parts on Saturday.
Hope this helps ;) ❤❤❤
Which of the following can be calculated using the formula ?
A.
Area of a circle
B.
Circumference of a circle
C.
Arc length of a circle
D.
Diameter of a circle
The formula C = π2r
Is used for the circumference.
Which of the following can be calculated using the formula?We know that the number π is defined as the quotient between the circumference of a circle and its diameter, so we can write:
C/d = π
And remember that the diameter is twice the radius, so we can write:
d = 2r
Then we can rewrite the equation for the circumference as:
C = π2r
Then we conclude that the correct option is B.
If you want to learn more about circles:
https://brainly.com/question/1559324
#SPJ1
Which, if any, of the following proofs are correct demonstrations of the validity of this argument? A ⊃ (B ⊃ C) B ⊃ (~C ⊃ ~A) Proof 1 (1) A ⊃ (B ⊃ C) /B ⊃ (~C ⊃ ~A) Premise/Conclusion (2) (A • B) ⊃ C 1 Exp (3) (B • A) ⊃ C 2 Com (4) B ⊃ (A ⊃ C) 3 Exp (5) B ⊃ (~C ⊃ ~A) 4 Contra Proof 2 (1) A ⊃ (B ⊃ C) /B ⊃ (~C ⊃ ~A) Premise/Conclusion (2) B Assumption (3) A Assumption (4) B ⊃ C 1, 3 MP (5) C 2, 4 MP (6) A ⊃ C 3–5 CP (7) B ⊃ (A ⊃ C) 2–6 CP (8) B ⊃ (~C ⊃ ~A) 7 Contra
Answer
Step-by-step explanation:
Answer:
See the argument below
Step-by-step explanation:
I will give the argument in symbolic form, using rules of inference.
First, let's conclude c.
(1)⇒a by simplification of conjunction
a⇒¬(¬a) by double negation
¬(¬a)∧(2)⇒¬(¬c) by Modus tollens
¬(¬c)⇒c by double negation
Now, the premise (5) is equivalent to ¬d∧¬h which is one of De Morgan's laws. From simplification, we conclude ¬h. We also concluded c before, then by adjunction, we conclude c∧¬h.
An alternative approach to De Morgan's law is the following:
By contradiction proof, assume h is true.
h⇒d∨h by addition
(5)∧(d∨h)⇒¬(d∨h)∧(d∨h), a contradiction. Hence we conclude ¬h.
3. Write an equation of a line that is perpendicular to the line x – 2y = 8.
Answer:
y=0.5x+40
Step-by-step explanation:
Copy the equation.
x-2y=8
Subtract x from both sides.
-2y=-x-8
Divide both sides by -2.
y=0.5x+4
Now we know the slope is 0.5.
Any line with a slope of 0.5 will be perpendiculr to the original line.
One that you can use is y=0.5x+40.
A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 4 m from the dock
Answer:
-1.031 m/s or [tex]\frac{-\sqrt{17} }{4}[/tex]
Step-by-step explanation:
We take the length of the rope from the dock to the bow of the boat as y.
We take x be the horizontal distance from the dock to the boat.
We know that the rate of change of the rope length is [tex]\frac{dy}{dt}[/tex] = -1 m/s
We need to find the rate of change of the horizontal distance from the dock to the boat = [tex]\frac{dx}{dt}[/tex] = ?
for x = 4
Applying Pythagorean Theorem we have
[tex]1^{2} +x^{2} =y^{2}[/tex] .... equ 1
solving, where x = 4, we have
[tex]1^{2} +4^{2} =y^{2}[/tex]
[tex]y^{2} = 17[/tex]
[tex]y = \sqrt{17}[/tex]
Differentiating equ 1 implicitly with respect to t, we have
[tex]2x\frac{dx}{dt} = 2y\frac{dy}{dt}[/tex]
substituting values of
x = 4
y = [tex]\sqrt{17}[/tex]
[tex]\frac{dy}{dt}[/tex] = -1
into the equation, we get
[tex]2(4)\frac{dx}{dt} = 2(\sqrt{17} )(-1)[/tex]
[tex]\frac{dx}{dt} = \frac{-\sqrt{17} }{4}[/tex] = -1.031 m/s
A catering service offers 11 appetizers, 12 main courses, and 8 desserts. A customer is to select 9 appetizers, 2 main courses, and 3 desserts for a banquet. In how many ways can this be done?
Answer: 203,280
Step-by-step explanation:
Given: A catering service offers 11 appetizers, 12 main courses, and 8 desserts.
Number of combinations of choosing r things out of n = [tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]
A customer is to select 9 appetizers, 2 main courses, and 3 desserts for a banquet.
Total number of ways to do this: [tex]^{11}C_9\times ^{12}C_2\times^{8}C_3[/tex]
[tex]=\dfrac{11!}{9!2!}\times\dfrac{12!}{2!10!}\times\dfrac{8!}{3!5!}\\\\=\dfrac{11\times10}{2}\times\dfrac{12\times11}{2}\times\dfrac{8\times7\times6}{3\times2}\\\\= 203280[/tex]
hence , this can be done in 203,280 ways.
Is this equation linear or nonlinear?
y =x/2
Answer:
linear
Step-by-step explanation: