Answer:
see explanation
Step-by-step explanation:
9W + 1 = 19 ( subtract 1 from both sides )
9W = 18 ( divide both sides by 9 )
W = 2
On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles, find the total distance traveled in the two days.
The Total mileage is "400" and the further solution can be defined as follows:
Let t become the time he spent commuting on the first day of his vacation.
It is then calculated as [tex]t + 2[/tex].
[tex]\to 40\times(t+2) = 60(t) + 20 \\\\\to 40t+80 = 60t + 20 \\\\\to 80-20 = 60t + 40t \\\\\to 60 = 20t \\\\\to t=\frac{60}{20} \\\\\to t=\frac{6}{2} \\\\\to t= 3\\\\[/tex]
It traveled [tex]40\times (3 + 2) + 20 = 40\times 5 + 20 = 200+20=220[/tex] miles on its first day of operation.
The car traveled [tex]180\ miles[/tex] on the second day, which was [tex]60 \ miles \times 3[/tex].
So,
Total mileage= first day traveled + second day traveled [tex]= 220+ 180= 400 \miles[/tex]
Learn more:
Total distance traveled: brainly.com/question/20670144
A computer system uses passwords that are exactly six characters and each character is one of the 26 letters (a–z) or 10 integers (0–9). Suppose that 10,000 users of the system have unique passwords. A hacker randomly selects (with replace- ment) one billion passwords from the potential set, and a match to a user’s password is called a hit. (a) What is the distribution of the number of hits? (b) What is the probability of no hits? (c) What are the mean and variance of the number of hits?
Answer:
The number of hits would follow a binomial distribution with [tex]n =10,\!000[/tex] and [tex]p \approx 4.59 \times 10^{-6}[/tex].
The probability of finding [tex]0[/tex] hits is approximately [tex]0.955[/tex] (or equivalently, approximately [tex]95.5\%[/tex].)
The mean of the number of hits is approximately [tex]0.0459[/tex]. The variance of the number of hits is approximately [tex]0.0459\![/tex] (not the same number as the mean.)
Step-by-step explanation:
There are [tex](26 + 10)^{6} \approx 2.18 \times 10^{9}[/tex] possible passwords in this set. (Approximately two billion possible passwords.)
Each one of the [tex]10^{9}[/tex] randomly-selected passwords would have an approximately [tex]\displaystyle \frac{10,\!000}{2.18 \times 10^{9}}[/tex] chance of matching one of the users' password.
Denote that probability as [tex]p[/tex]:
[tex]p := \displaystyle \frac{10,\!000}{2.18 \times 10^{9}} \approx 4.59 \times 10^{-6}[/tex].
For any one of the [tex]10^{9}[/tex] randomly-selected passwords, let [tex]1[/tex] denote a hit and [tex]0[/tex] denote no hits. Using that notation, whether a selected password hits would follow a bernoulli distribution with [tex]p \approx 4.59 \times 10^{-6}[/tex] as the likelihood of success.
Sum these [tex]0[/tex]'s and [tex]1[/tex]'s over the set of the [tex]10^{9}[/tex] randomly-selected passwords, and the result would represent the total number of hits.
Assume that these [tex]10^{9}[/tex] randomly-selected passwords are sampled independently with repetition. Whether each selected password hits would be independent from one another.
Hence, the total number of hits would follow a binomial distribution with [tex]n = 10^{9}[/tex] trials (a billion trials) and [tex]p \approx 4.59 \times 10^{-6}[/tex] as the chance of success on any given trial.
The probability of getting no hit would be:
[tex](1 - p)^{n} \approx 7 \times 10^{-1996} \approx 0[/tex].
(Since [tex](1 - p)[/tex] is between [tex]0[/tex] and [tex]1[/tex], the value of [tex](1 - p)^{n}[/tex] would approach [tex]0\![/tex] as the value of [tex]n[/tex] approaches infinity.)
The mean of this binomial distribution would be:[tex]n\cdot p \approx (10^{9}) \times (4.59 \times 10^{-6}) \approx 0.0459[/tex].
The variance of this binomial distribution would be:
[tex]\begin{aligned}& n \cdot p \cdot (1 - p)\\ & \approx(10^{9}) \times (4.59 \times 10^{-6}) \times (1- 4.59 \times 10^{-6})\\ &\approx 4.59 \times 10^{-6}\end{aligned}[/tex].
Rope pieces of lengths 45 cm, 75 cm and 81 cm have to be cut into same size pieces. What is the smallest piece length possible?
Answer:
2025 cm
Step-by-step explanation:
Given the length of pieces - 45 cm, 75 cm and 81 cm
To find the length of the rope we have to find the L.C.M. of 45, 75 and 81 :
3 | 45, 75, 81
| ________________
3 | 15, 25, 27
|________________
3 | 5, 25, 9
|________________
3 | 5, 25, 3
|________________
5 | 5, 25, 1
|________________
5 | 1, 5, 1
|________________
| 1, 1, 1
L.C.M. = 3 × 3 × 3 × 3 × 5 × 5
= 2025 cm
So, the least length of the rope should be 2025 cm which can be cut into a whole number of pieces of length 45 cm, 75 cm and 81 cm.
how long does it take for a deposit of $900 to double at 2% compounded continuously?
how many years does it take to double ? ___ years __ days
9514 1404 393
Answer:
34.6574 years34 years, 239.94 daysStep-by-step explanation:
For continuous compounding the "rule of 69" applies. That is the doubling time can be found from ...
t = 69.3147/r . . . . where r is the interest rate in percent.
Here, r=2, so ...
t = 69.3147/2 = 34.6574 . . . years
That's 34 years and 240 days.
I need help guys thanks so much
Answer: C
Step-by-step explanation:
Which number line represents the solutions to 1-2x = 4?
Answer:
The third choice down
Step-by-step explanation:
|-2x| = 4
There are two solutions, one positive and one negative
-2x = 4 and -2x = -4
Divide by -2
-2x/-2 = 4/-2 -2x/-2 = -4/-2
x = -2 and x = 2
The median for the given set of six ordered data values is 29.5
9 12 25_ 41 50
What is the missing value?
Answer:
34
Step-by-step explanation:
let the missing value is x
(25+x) /2 = 29.5
25+x = 29.5(2)
25+x = 59
x = 59-25
x = 34
Which expression is equivalent to
128xy
5 ? Assume x > 0 and y> 0.
2xy5
Moto
8
yax
8
BV
y
8.VY
X
Answer:
[tex]\sqrt{128x^8y^3} = 8 x^4 y \sqrt{2y}[/tex]
Step-by-step explanation:
Given
[tex]\sqrt{128x^8y^3}[/tex] --- the complete expression
Required
The equivalent expression
We have:
[tex]\sqrt{128x^8y^3}[/tex]
Expand
[tex]\sqrt{128x^8y^3} = \sqrt{128* x^8 * y^3}[/tex]
Further expand
[tex]\sqrt{128x^8y^3} = \sqrt{64 * 2* x^8 * y^2 * y}[/tex]
Rewrite as:
[tex]\sqrt{128x^8y^3} = \sqrt{64 * x^8 * y^2* 2 * y}[/tex]
Split
[tex]\sqrt{128x^8y^3} = \sqrt{64 * x^8 * y^2} * \sqrt{2 * y}[/tex]
Express as:
[tex]\sqrt{128x^8y^3} = (64 * x^8 * y^2)^\frac{1}{2} * \sqrt{2y}[/tex]
Remove bracket
[tex]\sqrt{128x^8y^3} = (64)^\frac{1}{2} * (x^8)^\frac{1}{2} * (y^2)^\frac{1}{2} * \sqrt{2y}[/tex]
[tex]\sqrt{128x^8y^3} = 8 * x^\frac{8}{2} * y^\frac{2}{2} * \sqrt{2y}[/tex]
[tex]\sqrt{128x^8y^3} = 8 * x^4 * y * \sqrt{2y}[/tex]
[tex]\sqrt{128x^8y^3} = 8 x^4 y \sqrt{2y}[/tex]
I need help
With these
Answer:
"A"
Step-by-step explanation:
a+b >c
a+c>b
b+c>a
~~~~~~~~~~~~
A. T,T,T
B. T,T,F
C. T,F,T
What is the equation of a line that passes through the point (1,8) and is perpendicular to the line whose equation is y=x/2+3?
Answer:
m=1/2
y-8=1/2(x-1)
y-8=1/2x-1/2
multiply through by 2
2y-16=x-1
2y-16+1-x=0
2y-15-x=0
2y-x-15=0
There is a sales tax of S6 on an item that costs 888 before tax. The sales tax on a second item is $21. How much does the second item cost before tax?
Step-by-step explanation:
before Tax
Coast = 888
in 2ND Item = $21
• 888/21
= $42.28
are ratios 2:3 and 8:12 equalvelent to eachother
Answer:
2:3 is equal to 8:12
Step-by-step explanation:
2:3
To get the first number to 8
8/2 = 4
Multiply by all terms 4
2*3 : 3*4
8:12
2:3 is equal to 8:12
8:12 = 8/12
= 2/3
= 2:3
Therefore 2:3 and 8:12 are equalent to each other.
Answered by Gauthmath must click thanks and mark brainliest
lim ₓ→∞ (x+4/x-1)∧x+4
It looks like the limit you want to find is
[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4}[/tex]
One way to compute this limit relies only on the definition of the constant e and some basic properties of limits. In particular,
[tex]e = \displaystyle\lim_{x\to\infty}\left(1+\frac1x\right)^x[/tex]
The idea is to recast the given limit to make it resemble this definition. The definition contains a fraction with x as its denominator. If we expand the fraction in the given limand, we have a denominator of x - 1. So we rewrite everything in terms of x - 1 :
[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\dfrac{x-1+5}{x-1}\right)^{x-1+5} \\\\ = \left(1+\dfrac5{x-1}\right)^{x-1+5} \\\\ =\left(1+\dfrac5{x-1}\right)^{x-1} \times \left(1+\dfrac5{x-1}\right)^5[/tex]
Now in the first term of this product, we substitute y = (x - 1)/5 :
[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(1+\dfrac1y\right)^{5y} \times \left(1+\dfrac5{x-1}\right)^5[/tex]
Then use a property of exponentiation to write this as
[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\left(1+\dfrac1y\right)^y\right)^5 \times \left(1+\dfrac5{x-1}\right)^5[/tex]
In terms of end behavior, (x - 1)/5 and x behave the same way because they both approach ∞ at a proportional rate, so we can essentially y with x. Then by applying some limit properties, we have
[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty} \left(\left(1+\dfrac1x\right)^x\right)^5 \times \left(1+\dfrac5{x-1}\right)^5 \\\\ = \lim_{x\to\infty}\left(\left(1+\dfrac1x\right)^x\right)^5 \times \lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)^5 \\\\ =\left(\lim_{x\to\infty}\left(1+\dfrac1x\right)^x\right)^5 \times \left(\lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)\right)^5[/tex]
By definition, the first limit is e and the second limit is 1, so that
[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = e^5\times1^5 = \boxed{e^5}[/tex]
You can also use L'Hopital's rule to compute it. Evaluating the limit "directly" at infinity results in the indeterminate form [tex]1^\infty[/tex].
Rewrite
[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \exp\left((x+4)\ln\dfrac{x+4}{x-1}\right)[/tex]
so that
[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty}\exp\left((x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ = \exp\left(\lim_{x\to\infty}(x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ =\exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right)[/tex]
and now evaluating "directly" at infinity gives the indeterminate form 0/0, making the limit ready for L'Hopital's rule.
We have
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\ln\dfrac{x+4}{x-1}\right] = -\dfrac5{(x-1)^2}\times\dfrac{1}{\frac{x+4}{x-1}} = -\dfrac5{(x-1)(x+4)}[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{x+4}\right]=-\dfrac1{(x+4)^2}[/tex]
and so
[tex]\displaystyle \exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right) = \exp\left(\lim_{x\to\infty}\frac{-\dfrac5{(x-1)(x+4)}}{-\dfrac1{(x+4)^2}}\right) \\\\ = \exp\left(5\lim_{x\to\infty}\frac{x+4}{x-1}\right) \\\\ = \exp(5) = \boxed{e^5}[/tex]
What is the equation of a circle with center (1, -4) and radius 2?
Answer:
(x-1)^2 + (y+4)^2 = 4
Step-by-step explanation:
The equation for a circle is given by
(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius
(x-1)^2 + (y- -4)^2 = 2^2
(x-1)^2 + (y+4)^2 = 4
SOMEONE HELP ME PLEASE
Answer:
9/25
Step-by-step explanation:
3 novels , 1 bio , 1 poetry = 5 books
P( novel) = novels / books
= 3/5
Book is returned
3 novels , 1 bio , 1 poetry = 5 books
P( novel) = novels / books
= 3/5
P(novel, return, novel) = 3/5 * 3/5 = 9/25
A store is having a sale on chocolate chips and walnuts. For 8 pounds of chocolate chips and 3 pounds of walnuts, the total cost is $34. For 2 pounds of chocolate chips and 5 pounds of walnuts, the total cost is $17. Find the cost for each pound of chocolate chips and each pound of walnuts.
Answer:
chocolate chips are $2.00 per pound.
nd walnuts must be $3.50 per pound.
Step-by-step explanation:
Let x be the price of walnuts and y the price of chocolate chips.
2x + 5y = 17 (i)
8x + 3y = 34 (ii)
Multiply (i) by 4 to get
8x + 20y = 68
Subtract (ii) to get
17y = 34
Dividing by 17, we see that chocolate chips are $2.00 per pound.
Substituting y=2 in (i) or (ii), walnuts must be $3.50 per pound.
Jagroop is building a dock at his cottage. The length of the doc is 3 times the width, plus 2. Determine a simplified expression for the perimeter of the doc
Answer:
Step-by-step explanation:
Let length = y width = x
y = 3x + 2
Perimeter = Sum of all sides (or sum of both lengths and both widths)
2y + 2x
2(3x + 2) + 2x
6x + 4 + 2x
8x + 4
Dogsled drivers, known as mushers, use several different breeds of dogs to pull their sleds. One proponent of Siberian Huskies believes that sleds pulled by Siberian Huskies are faster than sleds pulled by other breeds. He times 47 teams of Siberian Huskies on a particular short course, and they have a mean time of 5.2 minutes. The mean time on the same course for 39 teams of other breeds of sled dogs is 5.5 minutes. Assume that the times on this course have a population standard deviation of 1.4 minutes for teams of Siberian Huskies and 1.1 minutes for teams of other breeds of sled dogs. Let Population 1 be sleds pulled by Siberian Huskies and let Population 2 be sleds pulled by other breeds. Step 1 of 2 : Construct a 95% confidence interval for the true difference between the mean times on this course for teams of Siberian Huskies and teams of other breeds of sled dogs
Answer:
The 95% confidence interval for the true difference between the mean times on this course for teams of Siberian Huskies and teams of other breeds of sled dogs is (-0.8276, 0.2276).
Step-by-step explanation:
Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
Siberian Huskies:
Sample of 47, mean of 5.2 minutes, standard deviation of 1.4. So
[tex]\mu_1 = 5.2[/tex]
[tex]s_1 = \frac{1.4}{\sqrt{47}} = 0.2042[/tex]
Others:
Sample of 39, mean of 5.5 minutes, standard deviation of 1.1. So
[tex]\mu_2 = 5.5[/tex]
[tex]s_2 = \frac{1.1}{\sqrt{39}} = 0.1761[/tex]
Distribution of the difference:
[tex]\mu = \mu_1 - \mu_2 = 5.2 - 5.5 = -0.3[/tex]
[tex]s = \sqrt{s_1^2+s_2^2} = \sqrt{0.2042^2+0.1761^2} = 0.2692[/tex]
Confidence interval:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]
Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
Now, find the margin of error M as such
[tex]M = zs[/tex]
In which s is the standard error. So
[tex]M = 1.96(0.2692) = 0.5276[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is -0.3 - 0.5276 = -0.8276.
The upper end of the interval is the sample mean added to M. So it is -0.3 + 0.5276 = 0.2276
The 95% confidence interval for the true difference between the mean times on this course for teams of Siberian Huskies and teams of other breeds of sled dogs is (-0.8276, 0.2276).
use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x
First check the characteristic solution: the characteristic equation for this DE is
r ² - 3r + 2 = (r - 2) (r - 1) = 0
with roots r = 2 and r = 1, so the characteristic solution is
y (char.) = C₁ exp(2x) + C₂ exp(x)
For the ansatz particular solution, we might first try
y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)
where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).
However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :
y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)
Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.
y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)
… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)
y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)
… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)
Substituting every relevant expression and simplifying reduces the equation to
(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)
… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]
… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]
= 2x + (1 + 2x) exp(x) + 4 exp(3x)
… … …
2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)
= 2x + (1 + 2x) exp(x) + 4 exp(3x)
Then, equating coefficients of corresponding terms on both sides, we have the system of equations,
x : 2a = 2
1 : -3a + 2b = 0
exp(x) : 2c - d = 1
x exp(x) : -2c = 2
exp(3x) : 2e = 4
Solving the system gives
a = 1, b = 3/2, c = -1, d = -3, e = 2
Then the general solution to the DE is
y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)
Which ratio is equal to 27 : 81?
Answer:
1:3
Step-by-step explanation:
27 : 81
Divide each side by 27
27/27 : 81/27
1:3
What is the area of this triangle?
Enter your answer in the box.
units2
Answer:
8 units^2
Step-by-step explanation:
The area of a tringle is 1/2 bh. The base, LK, measures 4 while the height is also 4(you can get these values by counting the squares). This means the area is:
1/2 * (4)(4) = 1/2 * 16 = 8 units^2
Which of the following behaviors would best describe someone who is listening and paying attention? a) Leaning toward the speaker O b) Interrupting the speaker to share their opinion c) Avoiding eye contact d) Asking questions to make sure they understand what's being said
The answer is A and D
good luck
Most of the heat loss for outdoor swimming pools is due to surface
evaporation. So, the greater the area of the surface of the pool, the greater
the heat loss. For a given perimeter, which surface shape would be more
efficient at retaining heat: a circle or a rectangle? Justify your answer.
Answer:
rectangle
Step-by-step explanation:
Perimeter of 20 feet
rectangle (square is technically a rectangle):
sides 5 and 5
5*5 = 25ft²
Circle:
20/(2π) = 3.18309...
3.1809...²π = 31.831ft²
Max area of rectangle (i.e. square) has a smaller area than a circle.
I NEED HELP THANK YOU!!
Answer:
rt3/2
Step-by-step explanation:
first off cosine is the x coordinate
now if you do't want to use a calculator, you can use use the unit circle.
360 - 330 = 30 (360 degrees is a whole circle)
a 30 60 90 triangle is made, then use the law for 30 60 90 triangles:
if the shortest leg is x, the other leg is x*rt3 and the hypotenuse is 2x.
Answer:
D
Step-by-step explanation:
cos 330 = cos (360-330)
= cos 30
= √3 /2
Using a profit P1 of $5,000, a profit P2 of $4,500, and a profit P3 of $4,000, calculate a 95% confidence interval for the mean profit per customer that SoftBus can expect to obtain. (Round your answers to one decimal place.) Lower Limit Upper Limit
Answer:
Confidence Interval
Lower Limit = $4,233.3
Upper Limit = $4,766.7
With 95% confidence, the mean profit per customer that SoftBus can expect to obtain is between $4,233.30 and $4,766.7 based on the given sample data.
Step-by-step explanation:
The z-score of 95% = 1.96
Profit Mean Square Root
Difference of MD
P1 $5,000 $500 $250,000
P2 4,500 0 0
P3 4,000 -500 $250,000
Total $13,500 $500,000
Mean $4,500 ($13,500/3) $166,667 ($500,000/3)
Standard Deviation = Square root of $166,667 = 408.2
Margin of error = (z-score * standard deviation)/n
= (1.96 * 408.2)/3
= 266.7
= $266.7
Confidence Interval = Sample mean +/- Margin of error
= $4,500 +/- 266.7
Lower Limit = $4,233.3 ($4,500 - $266.7)
Upper Limit = $4,766.7 ($4,500 + $266.7)
A survey of 30-year-old males provided data on the number of auto accidents in the previous 5 years. The sample mean is 1.3 accidents per male. Test the hypothesis that the number of accidents follows a Poisson distribution at the 5% level of significance.
No. of accident No. of males
0 39
1 22
2 14
3 11
>=4 4
Required:
a. What's the Expected probability of finding males with 0 accidents?
b. What's the Expected probability of finding males with 4 or more accidents?
Answer:
0.2725
0.0431
Step-by-step explanation:
The distribution here is a poisson distribution :
λ = 1.3
The poisson distribution :
p(x) = [(e^-λ * λ^x)] ÷ x!
Expected probability of finding male with 0 accident ; x = 0
p(0) = [(e^-1.3 * 1.3^0)] ÷ 0!
p(0) = [0.2725317 * 1] ÷ 1
p(0) = 0.2725317
= 0.2725
2.)
P(x ≥ 4) = 1 - P(x < 4)
P(x < 4) = p(x = 0) + p(x. = 1) + p(x = 2) + p(x = 3)
p(x = 0) = p(0) = [(e^-1.3 * 1.3^0)] ÷ 0! = 0.2725
p(x = 1) = p(1) = [(e^-1.3 * 1.3^1)] ÷ 1! = 0.35429
p(x = 2) = p(2) = [(e^-1.3 * 1.3^2)] ÷ 2! = 0.23029 p(x = 3) = p(3) = [(e^-1.3 * 1.3^3)] ÷ 0! = 0.09979
P(x < 4) = 0.2725 + 0.35429 + 0.23029 + 0.09979 = 0.95687
P(x ≥ 4) = 1 - 0.95687 = 0.0431
Find the missing side lengths leave your answer as a racials simplest form
Answer:
m=[tex]7\sqrt3[/tex]
n=7
Step-by-step explanation:
Hi there!
We are given a right triangle (notice the 90°) angle, the measure of one of the acute angles as 60°, and the measure of the hypotenuse (the side OPPOSITE from the 90 degree angle) as 14
We need to find the lengths of m and n
Firstly, let's find the measure of the other acute angle
The acute angles in a right triangle are complementary, meaning they add up to 90 degrees
Let's make the measure of the unknown acute angle x
So x+60°=90°
Subtract 60 from both sides
x=30°
So the measure of the other acute angle is 30 degrees
This makes the right triangle a special kind of right triangle, a 30°-60°-90° triangle
In a 30°-60°-90° triangle, if the length of the hypotenuse is a, then the length of the leg (the side that makes up the right angle) opposite from the 30 degree angle is [tex]\frac{a}{2}[/tex], and the leg opposite from the 60 degree angle is [tex]\frac{a\sqrt3}{2}[/tex]
In this case, a=14, n=[tex]\frac{a}{2}[/tex], and m=[tex]\frac{a\sqrt3}{2}[/tex]
Now substitute the value of a into the formulas to find n and m to find the lengths of those sides
So that means that n=[tex]\frac{14}{2}[/tex], which is equal to 7
And m=[tex]\frac{14\sqrt3}{2}[/tex], which simplified, is equal to [tex]7\sqrt3[/tex]
Hope this helps!
A rocket is launched at t = 0 seconds. Its height, in meters above sea-level, is given by the equation
h = -4.9t2 + 112t + 395.
At what time does the rocket hit the ground? The rocket hits the ground after how many seconds
Answer:
Step-by-step explanation:
In order to find out how long it takes for the rocket to hit the ground, we only need set that position equation equal to 0 (that's how high something is off the ground when it is sitting ON the ground) and factor to solve for t:
[tex]0=-4.9t^2+112t+395[/tex]
Factor that however you are factoring in class to get
t = -3.1 seconds and t = 25.9 seconds.
Since time can NEVER be negative, it takes the rocket approximately 26 seconds to hit the ground.
If sin x = –0.1 and 270° < x < 360°, what is the value of x to the nearest degree?
Answer:
354°15'38.99''
Step-by-step explanation:
Which points lie on the graph of f(x) = loggx?
Check all that apply.
Step-by-step explanation:
f(x)=log(x)
=d(log(x)/dx)
=>y=1/x