Answer:
down b3low
Step-by-step explanation:
The point where the two lines intersect is the only solution. An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect. A dependent system has infinitely many solutions.
Which of the following situations may be modeled by the equation y = 2x +20
A. Carlos has written 18 pages of his article. He plans to write an
additional 2 pages per day.
B. Don has already sold 22 vehicles. He plans to sell 2 vehicles per
week.
C. Martin has saved $2. He plans to save $20 per month.
D. Eleanor has collected 20 action figures. She plans to collect 2
additional figures per month
Answer:
D.
m = 2 = figures/month
b = 20 = # of action figures
Crime and Punishment: In a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison. Among those sent to prison, 40% chose to plead guilty. Among those not sent to prison, 55% chose to plead guilty.
(A) If one of the study subjects is randomly selected, find the probability of getting someone who was not sent to prison.
(B) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, find the probability that this person was not sent to prison.
Answer:
(a) The probability of getting someone who was not sent to prison is 0.55.
(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is 0.63.
Step-by-step explanation:
We are given that in a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison. Among those sent to prison, 40% chose to plead guilty. Among those not sent to prison, 55% chose to plead guilty.
Let the probability that subjects studied were sent to prison = P(A) = 0.45
Let G = event that subject chose to plead guilty
So, the probability that the subjects chose to plead guilty given that they were sent to prison = P(G/A) = 0.40
and the probability that the subjects chose to plead guilty given that they were not sent to prison = P(G/A') = 0.55
(a) The probability of getting someone who was not sent to prison = 1 - Probability of getting someone who was sent to prison
P(A') = 1 - P(A)
= 1 - 0.45 = 0.55
(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is given by = P(A'/G)
We will use Bayes' Theorem here to calculate the above probability;
P(A'/G) = [tex]\frac{P(A') \times P(G/A')}{P(A') \times P(G/A') +P(A) \times P(G/A)}[/tex]
= [tex]\frac{0.55 \times 0.55}{0.55\times 0.55 +0.45 \times 0.40}[/tex]
= [tex]\frac{0.3025}{0.4825}[/tex]
= 0.63
PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS
You want to end up with [tex]A\sin(\omega t+\phi)[/tex]. Expand this using the angle sum identity for sine:
[tex]A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi[/tex]
We want this to line up with [tex]2\sin(4\pi t)+5\cos(4\pi t)[/tex]. Right away, we know [tex]\omega=4\pi[/tex].
We also need to have
[tex]\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}[/tex]
Recall that [tex]\sin^2x+\cos^2x=1[/tex] for all [tex]x[/tex]; this means
[tex](A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}[/tex]
Then
[tex]\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)[/tex]
So we end up with
[tex]2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)[/tex]
Answer:
y(t) = √29·sin(4πt +1.1903)amplitude: √29angular frequency: 4πphase shift: 1.1903 radiansStep-by-step explanation:
In the form ...
y(t) = Asin(ωt +φ)
you have ...
Amplitude = Aangular frequency = ωphase shift = φThe translation from ...
y(t) = 2sin(4πt) +5cos(4πt)
is ...
A = √(2² +5²) = √29 . . . . the amplitude
ω = 4π . . . . the angular frequency in radians per second
φ = arctan(5/2) ≈ 1.1903 . . . . radians phase shift
Then, ...
y(t) = √29·sin(4πt +1.1903)
_____
Comment on the conversion
You will notice we used "2" and "5" to find the amplitude and phase shift. In the generic case, these are "coefficient of sin( )" and "coefficient of cos( )". When determining phase shift, pay attention to whether your calculator is giving you degrees or radians. (Set the mode to what you want.)
If you have a negative coefficient for sin( ), you will need to add 180° (π radians) to the phase shift value given by the arctan( ) function.
a 12- inch ruler is duvided into 3 parts. the large part is 3 times longer than the small. the meddium part is times longer than then small, the medium part is 2 times long as the smallest .how long is the smallest part?
Answer:
2 inches
Step-by-step explanation:
x= smallest
3x=largest
2x=medium
x+3x+2x=12
6x=12
x=2
so smallest is 2
largest is 6 (3x)
medium is 4 (2x)
2+6+4=12
Assume that there is a 6% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive? b. If copies of all your computer data are stored on independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive? four a. With two hard disk drives, the probability that catastrophe can be avoided is . (Round to four decimal places as needed.) b. With four hard disk drives, the probability that catastrophe can be avoided is . (Round to six decimal places as needed.)
Answer: 0.9964
Step-by-step explanation:
Consider,
P (disk failure) = 0.06
q = 0.06
p = 1- q
p = 1- 0.06,
p = 0.94
Step 2
Whereas p represents the probability that a disk does not fail. (i.e. working entire year).
a)
Step 3
a)
n = 2,
let x be a random variable for number...
Continuation in the attached document
Please help. I’ll mark you as brainliest if correct!
Answer:
8lb of the cheaper Candy
17.5lb of the expensive candy
Step-by-step explanation:
Let the cheaper candy be x
let the costly candy be y
X+y = 25.5....equation one
2.2x +7.3y = 25.5(5.7)
2.2x +7.3y = 145.35.....equation two
X+y = 25.5
2.2x +7.3y = 145.35
Solving simultaneously
X= 25.5-y
Substituting value of X into equation two
2.2(25.5-y) + 7.3y = 145.35
56.1 -2.2y +7.3y = 145.35
5.1y = 145.35-56.1
5.1y = 89.25
Y= 89.25/5.1
Y= 17.5
X= 25.5-y
X= 25.5-17.5
X= 8
1000 randomly selected Americans were asked if they believed the minimum wage should be raised. 600 said yes. Construct a 95% confidence interval for the proportion of Americans who believe that the minimum wage should be raised.
a. Write down the formula you intend to use with variable notation).
b. Write down the above formula with numeric values replacing the symbols.
c. Write down the confidence interval in interval notation.
Answer:
a. p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]
b.0.6 ± 1.96 [tex]\sqrt \frac{0.6* 0.4}{1000}[/tex]
c. { -1.96 ≤ p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex] ≥ 1.96} = 0.95
Step-by-step explanation:
Here the total number of trials is n= 1000
The number of successes is p` = 600/1000 = 0.6. The q` is 1 - p`= 1- 0.6 = 0.4
The degree of confidence is 95 % therefore z₀.₀₂₅ = 1.96 ( α/2 = 0.025)
a. The formula used will be
p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex] ( z with the base alpha by 2 (α/2 = 0.025))
b. Putting the values
0.6 ± 1.96 [tex]\sqrt \frac{0.6* 0.4}{1000}[/tex]
c. Confidence Interval in Interval Notation.
{ -1.96 ≤ p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex] ≥ 1.96} = 0.95
{ -z( base alpha by 2) ≤ p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex] ≥ z( base alpha by 2) } = 1- α
An experiment involves 17 participants. From these, a group of 3 participants is to be tested under a special condition. How many groups of 3 participants can
be chosen, assuming that the order in which the participants are chosen is irrelevant?
Answer: 680
Step-by-step explanation:
When order doesn't matter,then the number of combinations of choosing r things out of n = [tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]
Given: Total participants = 17
From these, a group of 3 participants is to be tested under a special condition.
Number of groups of 3 participants chosen = [tex]^{17}C_3=\dfrac{17!}{3!(17-3)!}\[/tex]
[tex]^{17}C_3=\dfrac{17!}{3!(17-3)!}\\\\=\dfrac{17\times16\times15\times14!}{3\times2\times14!}\\\\=680[/tex]
Hence, there are 680 groups of 3 participants can be chosen,.
A company had a market price of $38.50 per share, earnings per share of $1.75, and dividends per share of $0.90. its price-earnings ratio equals:
Answer: Price-earnings ratio= 22.0
Step-by-step explanation:
Given: A company had a market price of $38.50 per share, earnings per share of $1.75, and dividends per share of $0.90
To find: price-earnings ratio
Required formula: [tex]\text{price-earnings ratio }=\dfrac{\text{ Market Price per Share}}{\text{Earnings Per Share}}[/tex]
Then, Price-earnings ratio = [tex]\dfrac{\$38.50}{\$1.75}[/tex]
⇒Price-earnings ratio = [tex]\dfrac{22}{1}[/tex]
Hence, the price-earnings ratio= 22.0
A gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than work for someone else. A random sample of 600 18-29 year-olds is obtained today. What is the probability that no more than 70% would prefer to start their own business?
Answer:
The probability that no more than 70% would prefer to start their own business is 0.1423.
Step-by-step explanation:
We are given that a Gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than work for someone else.
Let [tex]\hat p[/tex] = sample proportion of people who prefer to start their own business
The z-score probability distribution for the sample proportion is given by;
Z = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, p = population proportion who would prefer to start their own business = 72%
n = sample of 18-29 year-olds = 600
Now, the probability that no more than 70% would prefer to start their own business is given by = P( [tex]\hat p[/tex] [tex]\leq[/tex] 70%)
P( [tex]\hat p[/tex] [tex]\leq[/tex] 70%) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.70-0.72}{\sqrt{\frac{0.70(1-0.70)}{600} } }[/tex] ) = P(Z [tex]\leq[/tex] -1.07) = 1 - P(Z < 1.07)
= 1 - 0.8577 = 0.1423
The above probability is calculated by looking at the value of x = 1.07 in the z table which has an area of 0.8577.
WILL GIVE BRAINLIEST IF CORRECT!! Please help ! -50 POINTS -
Answer:
i think (d) one i think it will help you
HELPNEEDED.Two boys and three girls are auditioning to play the piano for a school production. Two students will be chosen, one as the pianist, the other as the alternate.
What is the probability that the pianist will be a boy and the alternate will be a girl?
30%
40%
50%
60%
TRIANGLE ABC IS DILATED BY A SCALE FACTOR OF 0.5 WITH THE ORIGIN AS THE CENTER OF DILATION, RESULTING IN THE IMAGE TRIANGLE A'B'C. IF A=(2,2). IF A (2,2), B= (4,3) AND C=(6,3), WHAT IS THE LENGTH OF LINE B'C'?
Answer: The length of the line B'C" is 1 unit.
Step-by-step explanation:
Given: Triangle ABC is dilated by a scale factor of 0.5 with the origin as the center of dilation , resulting in the image Triangle A'B'C'.
If A (2,2), B= (4,3) and C=(6,3).
Distance between (a,b) and (c,d): [tex]D=\sqrt{(d-b)^2+(c-b)^2}[/tex]
Then, BC [tex]=\sqrt{(3-3)^2+(6-4)^2}[/tex]
[tex]\\\\=\sqrt{0+2^2}\\\\=\sqrt{4}\\\\=2\text{ units}[/tex]
Length of image = scale factor x length in original figure
B'C' = 0.5 × BC
= 0.5 × 2
= 1 unit
Hence, the length of the line B'C" is 1 unit.
A coin is thrown at random into the rectangle below.
A rectangle is about 90 percent white and 10 percent green.
What is the likelihood that the coin will land in the green region?
It is certain.
It is impossible.
It is likely.
It is unlikely.
Answer:
It is unlikely.
Step-by-step explanation:
Certain = 100%
Impossible = 0%
Likely = more than 50%
Unlikely = less than 50%
It is less than 50%, so it is unlikely.
Answer:
(A) it is likely
Step-by-step explanation:
i took the test on edge
ASAP PLEASE HELP!!!!!! Find the y-intercept of the rational function. A rational function is graphed in the first quadrant, and in the second, third and fourth quadrants are other pieces of the graph. The graph crosses the x axis at negative 10 and crosses the y axis at negative 2.
Answer:
(0,-2)
Step-by-step explanation:
The y-intercept is simply when the function touches or crosses the y-axis.
We're told that the graph crosses the y-axis at -2. In other words, the y-intercept is at -2.
The ordered pair would be (0,-2)
A regression model between sales (y in $1000), unit price (x1 in dollars), and television advertisement (x2 in dollars) resulted in the following function: Ŷ = 7 - 3x1 + 5x2 For this model, SSR = 3500, SSE = 1500, and the sample size is 18. If we want to test for the significance of the regression model, the critical value of F at the 5% level of significance is a. 3.29. b. 3.24. c. 3.68. d. 4.54.
Answer: C. 3.68
Step-by-step explanation:
Given that;
Sample size n = 18
degree of freedom for numerator k = 2
degree of freedom for denominator = n - k - 1 = (18-2-1) = 15
level of significance = 5% = 5/100 = 0.05
From the table values,
the critical value of F at 0.05 significance level with (2, 18) degrees of freedom is 3.68
Therefore option C. 3.68 is the correct answer
Find the common ratio of the following geometric sequence:
11,55, 275, 1375, ....
Answer:
Hey there!
The common ratio is 5, because you multiply by 5 to get from one term to the next.
Hope this helps :)
Answer:
5
Step-by-step explanation:
To find the common ratio take the second term and divide by the first term
55/11 = 5
The common ratio would be 5
g The average salary in this city is $45,600. Is the average different for single people? 53 randomly selected single people who were surveyed had an average salary of $46,356 and a standard deviation of $15,930. What can be concluded at the α α = 0.05 level of significance?
Answer:
Step-by-step explanation:
The average salary in this city is $45,600.
Using the formula
z score = x - u /(sd/√n)
Where x is 46,356, u is 45,600 sd is 15,930 and n is 53.
z = 46,356 - 45600 / (15930/√53)
z = 756/(15930/7.2801)
z = 756/(2188.1568)
z = 0.3455
To draw a conclusion, we have to determine the p value, at 0.05 level of significance for a two tailed test, the p value is 0.7297. The p value is higher than the significance level, thus we will fail to reject the null and can conclude that there is not enough statistical evidence to prove that the average is any different for single people.
Historically, the proportion of students entering a university who finished in 4 years or less was 63%. To test whether this proportion has decreased, 114 students were examined and 51% had finished in 4 years or less. To determine whether the proportion of students who finish in 4 year or less has statistically significantly decreased (at the 5% level of signficance), what is the critical value
Answer:
z(c) = - 1,64
We reject the null hypothesis
Step-by-step explanation:
We need to solve a proportion test ( one tail-test ) left test
Normal distribution
p₀ = 63 %
proportion size p = 51 %
sample size n = 114
At 5% level of significance α = 0,05, and with this value we find in z- table z score of z(c) = 1,64 ( critical value )
Test of proportion:
H₀ Null Hypothesis p = p₀
Hₐ Alternate Hypothesis p < p₀
We now compute z(s) as:
z(s) = ( p - p₀ ) / √ p₀q₀/n
z(s) =( 0,51 - 0,63) / √0,63*0,37/114
z(s) = - 0,12 / 0,045
z(s) = - 2,66
We compare z(s) and z(c)
z(s) < z(c) - 2,66 < -1,64
Therefore as z(s) < z(c) z(s) is in the rejection zone we reject the null hypothesis
let x = the amoun of raw sugar in tons a procesing plant is a sugar refinery process in one day . suppose x can be model as exponetial distribution with mean of 4 ton per day . The amount of raw sugar (x) has
Answer:
The answer is below
Step-by-step explanation:
A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?
Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:
[tex]f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\[/tex]
a) P(x > 5) = [tex]\int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865[/tex]
b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.
That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757
c) Let b be the amount of raw sugar should be stocked for the plant each day.
P(x > a) = [tex]\int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}[/tex]
But P(x > a) = 0.05
Therefore:
[tex]e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98[/tex]
a ≅ 12
Refer to the following wage breakdown for a garment factory:
Hourly Wages Number of employees
$4 up to $7 18
7 up to 10 36
10 up to 13 20
13 up to 16 6
What is the class interval for the preceding table of wages?
A. $4
B. $2
C. $5
D. $3
Answer:
The class interval is $3Step-by-step explanation:
The class interval is simply the difference between the lower or upper class boundary or limit of a class and the lower or upper class boundary or limit of the next class.
In this case for the class
$4 up to $7 18 and
$7 up to $10 36
The lower class boundary of the first class is $4 and the lower class boundary of the second class is $7
Hence the class interval = $7-$4= $3If w'(t) is the rate of growth of a child in pounds per year, what does 7 w'(t)dt 4 represent? The change in the child's weight (in pounds) between the ages of 4 and 7. The change in the child's age (in years) between the ages of 4 and 7. The child's weight at age 7. The child's weight at age 4. The child's initial weight at birth.
Complete Question
If w'(t) is the rate of growth of a child in pounds per year, what does
[tex]\int\limits^{7}_{4} {w'(t)} \, dt[/tex] represent?
a) The change in the child's weight (in pounds) between the ages of 4 and 7.
b) The change in the child's age (in years) between the ages of 4 and 7.
c) The child's weight at age 7.
d) The child's weight at age 4. The child's initial weight at birth.
Answer:
The correct option is option a
Step-by-step explanation:
From the question we are told that
[tex]w'(t)[/tex] represents the rate of growth of a child in [tex]\frac{pounds}{year}[/tex]
So [tex]{w'(t)} \, dt[/tex] will be in [tex]pounds[/tex]
Which then mean that this [tex]\int\limits^{7}_{4} {w'(t)} \, dt[/tex] the change in the weight of the child between the ages of [tex]4 \to 7[/tex] years
what is the answer to 100×338
Answer:
33800
Step-by-step explanation:
100 x 338 = 33800
Answer:
33800
Step-by-step explanation:
338x10=3380 then 3380x10=33800
-------------------------------------------------------
Good luck with your assignment...
Which correlation coefficient could represent the relationship in the scatterpot. Beach visitors
Answer:
A. 0.89.
Step-by-step explanation:
The value of correlation coefficient ranges from -1 to 1. Any value outside this range cannot possibly be correlation coefficient of a scatter plot representing relationship between two variables.
The scatter plot given shows a positive correlation between average daily temperatures and number of visitors, as the trend shows the two variables are moving in the same direction. As daily temperature increases, visitors also increases.
From the options given, the only plausible correlation that can represent this positive relationship is A. 0.89.
A company has five employees on its health insurance plan. Each year, each employee independently has an 80% probability of no hospital admissions. If an employee requires one or more hospital admissions, the number of admissions is modeled by a geometric distribution with a mean of 1.50. The numbers of hospital admissions of different employees are mutually independent. Each hospital admission costs 20,000.
Calculate the probability that the company's total hospital costs in a year are less than 50,000.
Answer:
the probability that the company's total hospital costs in a year are less than 50,000 = 0.7828
Step-by-step explanation:
From the given information:
the probability that the company's total hospital costs in a year are less than 50,000 will be the sum of the probability of the employees admitted.
If anyone is admitted to the hospital, they have [tex]\dfrac{1}{3}[/tex] probability of making at least one more visit, and a [tex]\dfrac{2}{3}[/tex] probability that this is their last visit.
If zero employee was admitted ;
Then:
Probability = (0.80)⁵
Probability = 0.3277
If one employee is admitted once;
Probability = [tex](0.80)^4 \times (0.20)^1 \times (^5_1) \times (\dfrac{2}{3})[/tex]
Probability = [tex](0.80)^4 \times (0.20)^1 \times (\dfrac{5!}{(5-1)!}) \times (\dfrac{2}{3})[/tex]
Probability = 0.2731
If one employee is admitted twice
Probability = [tex](0.80)^3 \times (0.20)^2 \times (^5_2) \times (\dfrac{2}{3})^2[/tex]
Probability = [tex](0.80)^3 \times (0.20)^2 \times (\dfrac{5!}{(5-2)!}) \times (\dfrac{2}{3})^2[/tex]
Probability = 0.1820
If two employees are admitted once
Probability = [tex](0.80)^4\times (0.20)^1 \times (^5_1) \times (\dfrac{1}{3}) \times (\dfrac{2}{3})[/tex]
Probability = [tex](0.80)^4 \times (0.20)^1 \times (\dfrac{5!}{(5-1)!}) \times (\dfrac{1}{3}) \times (\dfrac{2}{3})[/tex]
Probability = 0.0910
∴
the probability that the company's total hospital costs in a year are less than 50,000 = 0.3277 + 0.2731 + 0.1820
the probability that the company's total hospital costs in a year are less than 50,000 = 0.7828
Please help asap.
A pizza is cut into six unequal slices (each cut starts at the center). The largest slice measures $90$ degrees If Larry eats the slices in order from the largest to the smallest, then the number of degrees spanned by a slice decreases at a constant rate. (So the second slice is smaller than the first by a certain number of degrees, then the third slice is smaller than the second slice by that same number of degrees, and so on.) What is the degree measure of the fifth slice Larry eats?
Answer:
The answer is 5th angle = [tex]\bold{42^\circ}[/tex]
Step-by-step explanation:
Given that pizza is divided into six unequal slices.
Largest slice has an angle of [tex]90^\circ[/tex].
He eats the pizza from largest to smallest.
Let the difference in angles in each slice = [tex]d^\circ[/tex]
1st angle = [tex]90^\circ[/tex]
2nd angle = 90-d
3rd angle = 90-d-d = 90 - 2d
4th angle = 90-2d-d = 90 - 3d
5th angle = 90-3d-d = 90 - 4d
6th angle = 90-4d -d = 90 - 5d
We know that the sum of all the angles will be equal to [tex]360^\circ[/tex] (The sum of all the angles subtended at the center).
i.e.
[tex]90+90-d+90-2d+90-3d+90-4d+90-5d=360\\\Rightarrow 540 - 15d = 360\\\Rightarrow 15d = 540 -360\\\Rightarrow 15d = 180\\\Rightarrow d = 12^\circ[/tex]
So, the angles will be:
1st angle = [tex]90^\circ[/tex]
2nd angle = 90- 12 = 78
3rd angle = 78-12 = 66
4th angle = 66-12 = 54
5th angle = 54-12 = 42
6th angle = 42 -12 = 30
So, the answer is 5th angle = [tex]\bold{42^\circ}[/tex]
A car travels 133 mi averaging a certain speed. If the car had gone 30 mph faster, the trip would have taken 1 hr less. Find the car's average speed.
Answer:
49.923 mph
Step-by-step explanation:
we know that the car traveled 133 miles in h hours at an average speed of x mph.
That is, xh = 133.
We can also write this in terms of hours driven: h = 133/x.
If x was 30 mph faster, then h would be one hour less.
That is, (x + 30)(h - 1) = 133, or h - 1 = 133/(x + 30).
We can rewrite the latter equation as h = 133/(x + 30) + 1
We can then make a system of equations using the formulas in terms of h to find x:
h = 133/x = 133/(x + 30) + 1
133/x = 133/(x + 30) + (x + 30)/(x + 30)
133/x = (133 + x + 30)/(x + 30)
133 = x*(133 + x + 30)/(x + 30)
133*(x + 30) = x*(133 + x + 30)
133x + 3990 = 133x + x^2 + 30x
3990 = x^2 + 30x
x^2 + 30x - 3990 = 0
Using the quadratic formula:
x = [-b ± √(b^2 - 4ac)]/2a
= [-30 ± √(30^2 - 4*1*(-3990))]/2(1)
= [-30 ± √(900 + 15,960)]/2
= [-30 ± √(16,860)]/2
= [-30 ± 129.846]/2
= 99.846/2 ----------- x is miles per hour, and a negative value of x is neglected, so we'll use the positive value only)
= 49.923
Check if the answer is correct:
h = 133/49.923 = 2.664, so the car took 2.664 hours to drive 133 miles at an average speed of 49.923 mph.
If the car went 30 mph faster on average, then h = 133/(49.923 + 30) = 133/79.923 = 1.664, and 2.664 - 1 = 1.664.
Thus, we have confirmed that a car driving 133 miles at about 49.923 mph would have arrive precisely one hour earlier by going 30 mph faster
In which table does y vary inversely with x? A. x y 1 3 2 9 3 27 B. x y 1 -5 2 5 3 15 C. x y 1 18 2 9 3 6 D. x y 1 4 2 8 3 12
Answer:
In Table C, y vary inversely with x.
1×18 = 18
2×9 = 18
3×6 = 18
18 = 18 = 18
Step-by-step explanation:
We are given four tables and asked to find out in which table y vary inversely with x.
We know that an inverse relation has a form given by
y = k/x
xy = k
where k must be a constant
Table A:
x | y
1 | 3
2 | 9
3 | 27
1×3 = 3
2×9 = 18
3×27 = 81
3 ≠ 18 ≠ 81
Hence y does not vary inversely with x.
Table B:
x | y
1 | -5
2 | 5
3 | 15
1×-5 = -5
2×5 = 10
3×15 = 45
-5 ≠ 10 ≠ 45
Hence y does not vary inversely with x.
Table C:
x | y
1 | 18
2 | 9
3 | 6
1×18 = 18
2×9 = 18
3×6 = 18
18 = 18 = 18
Hence y vary inversely with x.
Table D:
x | y
1 | 4
2 | 8
3 | 12
1×4 = 4
2×8 = 16
3×12 = 36
4 ≠ 16 ≠ 36
Hence y does not vary inversely with x.
The automatic opening device of a military cargo parachute has been designed to open when the parachute is 155 m above the ground. Suppose opening altitude actually has a normal distribution with mean value 155 and standard deviation 30 m. Equipment damage will occur if the parachute opens at an altitude of less than 100 m. What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes
Answer:
the probability that one parachute of the five parachute is damaged is 0.156
Step-by-step explanation:
From the given information;
Let consider X to be the altitude above the ground that a parachute opens
Then; we can posit that the probability that the parachute is damaged is:
P(X ≤ 100 )
Given that the population mean μ = 155
the standard deviation σ = 30
Then;
[tex]P(X \leq 100 ) = ( \dfrac{X- \mu}{\sigma} \leq \dfrac{100- \mu}{\sigma})[/tex]
[tex]P(X \leq 100 ) = ( \dfrac{X- 155}{30} \leq \dfrac{100- 155}{30})[/tex]
[tex]P(X \leq 100 ) = (Z \leq \dfrac{- 55}{30})[/tex]
[tex]P(X \leq 100 ) = (Z \leq -1.8333)[/tex]
[tex]P(X \leq 100 ) = \Phi( -1.8333)[/tex]
From standard normal tables
[tex]P(X \leq 100 ) = 0.0334[/tex]
Hence; the probability of the given parachute damaged is 0.0334
Let consider Q to be the dropped parachute
Given that the number of parachute be n= 5
The probability that the parachute opens in each trail be p = 0.0334
Now; the random variable Q follows the binomial distribution with parameters n= 5 and p = 0.0334
The probability mass function is:
Q [tex]\sim[/tex] B(5, 0.0334)
Similarly; the event that one parachute is damaged is :
Q ≥ 1
P( Q ≥ 1 ) = 1 - P( Q < 1 )
P( Q ≥ 1 ) = 1 - P( Y = 0 )
P( Q ≥ 1 ) = 1 - b(0;5; 0.0334 )
P( Q ≥ 1 ) = [tex]1 -(^5_0)* (0.0334)^0*(1-0.0334)^5[/tex]
P( Q ≥ 1 ) = [tex]1 -( \dfrac{5!}{(5-0)!}) * (0.0334)^0*(1-0.0334)^5[/tex]
P( Q ≥ 1 ) = 1 - 0.8437891838
P( Q ≥ 1 ) = 0.1562108162
P( Q ≥ 1 ) [tex]\approx[/tex] 0.156
Therefore; the probability that one parachute of the five parachute is damaged is 0.156
Scores made on a certain aptitude test by nursing students are approximately normally distributed with a mean of 500 and a variance of 10,000. If a person is about to take the test what is the probability that he or she will make a score of 650 or more?
Answer:
0.0668 or 6.68%
Step-by-step explanation:
Variance (V) = 10,000
Standard deviation (σ) = √V= 100
Mean score (μ) = 500
The z-score for any test score X is:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
For X = 650:
[tex]z=\frac{650-500}{100}\\z=1.5[/tex]
A z-score of 1.5 is equivalent to the 93.32nd percentile of a normal distribution. Therefore, the probability that he or she will make a score of 650 or more is:
[tex]P(X\geq 650)=1-P(X\leq 650)\\P(X\geq 650)=1-0.9332\\P(X\geq 650)=0.0668=6.68\%[/tex]
The probability is 0.0668 or 6.68%
The probability that he or she will make a score of 650 or more is 0.0668.
Let X = Scores made on a certain aptitude test by nursing students
X follows normal distribution with mean = 500 and variance of 10,000.
So, standard deviation = [tex]\sqrt{10000}=100[/tex].
z score of 650 is = [tex]\frac{\left(650-500\right)}{100}=1.5[/tex].
The probability that he or she will make a score of 650 or more is:
[tex]P(X\geq 650)\\=P(z\geq 1.5)\\=1-P(z<1.5)\\=1-0.9332\\=0.0668[/tex]
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