[tex]f(s)\Longrightarrow L^{-1}=\{\frac{-4s-9}{s^2+25-8}\}[/tex]
First dismantle,
[tex]L^{-1}=\{-\frac{4s}{s^2+25-8}-\frac{9}{s^2+25-8}\}[/tex]
Now use the linearity property of Inverse Laplace Transform which states,
For functions [tex]f(s),g(s)[/tex] and constants [tex]a, b[/tex] rule applies,
[tex]L^{-1}=\{a\cdot f(s)+b\cdot g(s)\}=aL^{-1}\{f(s)\}+bL^{-1}\{f(s)\}[/tex]
Hence,
[tex]-4L^{-1}\{\frac{s}{s^2+25-8}\}-9L^{-1}\{\frac{1}{s^2+25-8}\}[/tex]
The first part simplifies to,
[tex]
L^{-1}\{\frac{s}{s^2+25-8}\} \\
\frac{d}{dt}(\frac{1}{\sqrt{17}}\sin(t\sqrt{17})) \\
\cos(t\sqrt{17})
[/tex]
The second part simplifies to,
[tex]
L^{-1}\{\frac{1}{s^2+25-8}\} \\
\frac{1}{\sqrt{17}}\sin(t\sqrt{17})
[/tex]
And we result with,
[tex]\boxed{-4\cos(t\sqrt{17})-\frac{9}{\sqrt{17}}\sin(t\sqrt{17})}[/tex]
Hope this helps.
If you have any additional questions please ask. I made process of solving as quick as possible therefore you might be left over with some uncertainty.
Hope this helps.
r3t40
Calculate the standard deviation for the following set of numbers: 73, 76, 79, 82, 84, 84, 97
Answer:
Standard deviation is 7.16
Step-by-step explanation:
We have given a set of numbers :
73, 76, 79, 82, 84, 84, 97
To calculate the standard deviation of the given data set, first we have to work out the mean.
Mean = [tex]\frac{(73+76+79+82+84+84+97)}{7}[/tex]
= [tex]\frac{575}{7}[/tex] = 82.14
Now for each number subtract the mean and square the result
(73 - 82.14)² = (-9.14)² = 83.54
(76 - 82.14)² = (-6.14)² = 37.70
(79 - 82.14)² = (-3.14)² = 9.86
(82 - 82.14)² = (0.14)² = 0.02
(84 - 82.14)² = (1.86)² = 3.46
(84 - 82.14)² = (1.86)² = 3.46
(97 - 82.14)² = (14.86)²= 220.82
Now we calculate the mean from of those squared differences :
Mean = [tex]\frac{83.54+37.70+9.86+0.02+3.46+3.46+220.82}{7}[/tex]
= [tex]\frac{358.86}{7}[/tex]
= 51.27
Now square root of this mean = standard deviation = √51.27 = 7.16
Therefore, Standard deviation is 7.16
Jayanta is raising money for the? homeless, and discovers each church group requires 2 hr of letter writing and 1 hr of? follow-up calls, while each labor union needs 2 hr of letter writing and 3 hr of ?follow-up. She can raise ?$150 from each church group and ?$175 from each union. She has a maximum of 20 hours of letter writing and 14 hours of ?follow-up available each month. Determine the most profitable mixture of groups she should contact and the most money she can raise in a month.
Answer:
Jayanta needs 2 labor union groups and 8 church groups.
Step-by-step explanation:
Let c denotes churches .
Let l denotes labor unions.
We know that Jayanta can only spend 20 hours letter writing and 14 hour of follow-up.
So, equations becomes:
[tex]2c+2l=20[/tex]
[tex]c+3l=14[/tex]
And total money raised can be shown by = [tex]150c+175l[/tex]
We have to maximize [tex]150c+175l[/tex] keeping in mind that [tex]2c+2l \leq 20[/tex] and [tex]c+3l \leq 14[/tex]
We will solve the two equations: [tex]2c+2l=20[/tex] and [tex]c+3l=14[/tex]
We get l = 2 and c = 8
And total money raised is [tex]150\times8 + 175\times2[/tex] = [tex]1200+350=1550[/tex] dollars.
Hence, Jayanta needs 2 labor union groups and 8 church groups.
Find the m∠p.
54
90°
27°
36°
please reply...wasn't taught this
Refer the attachment...
Hope it helps you...
Answer:
ABDE is a parallelogram.
CE=2.7 cm Reason: The diagonals bisect each other. CE=AC.
DC=3.08 cm Reason: The diagonals bisect each other. This is half of DB. Just like CB is half of DB.
mAngleABE=104 degrees Reason: Opposite angles are congruent in a parallelogram.
mAngleDEB=76 degrees Reason: Consecutive angles in a parallelogram are supplementary.
Step-by-step explanation:
You have a parallelogram because both of your pairs of opposite sides are parallel.
CE=2.7 cm because CE is congruent to AC. The diagonals of a parallelogram bisect each other. Bisect means to cut into equal halves.
DC=3.08 cm because DC is congruent to CB which means the measurement of DC is half the length of DB.
mAngleABE=mAngleADE because opposite angles of a parallelogram are congruent. So the mAgnleABE=104 degrees.
AngleDEB is supplementary to AngleADE because they are consecutive angles in a parallelogram. This means mAngleDEB=180-104=76 degrees.
Find all the solutions for the equation:
2y2 dx - (x+y)2 dy=0
(Introduction to Differential Equations)
[tex]2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0[/tex]
Divide both sides by [tex]x^2\,\mathrm dx[/tex] to get
[tex]2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}[/tex]
Substitute [tex]v(x)=\dfrac{y(x)}x[/tex], so that [tex]\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}[/tex]. Then
[tex]x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}[/tex]
[tex]x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}[/tex]
[tex]x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}[/tex]
The remaining ODE is separable. Separating the variables gives
[tex]\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x[/tex]
Integrate both sides. On the left, split up the integrand into partial fractions.
[tex]\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}[/tex]
[tex]\implies v^2+2v+1=a(v^2+1)+(bv+c)v[/tex]
[tex]\implies v^2+2v+1=(a+b)v^2+cv+a[/tex]
[tex]\implies a=1,b=0,c=2[/tex]
Then
[tex]\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v[/tex]
On the right, we have
[tex]\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C[/tex]
Solving for [tex]v(x)[/tex] explicitly is unlikely to succeed, so we leave the solution in implicit form,
[tex]\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C[/tex]
and finally solve in terms of [tex]y(x)[/tex] by replacing [tex]v(x)=\dfrac{y(x)}x[/tex]:
[tex]\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C[/tex]
[tex]\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C[/tex]
[tex]\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}[/tex]
A study was conducted to measure the effectiveness of hypnotism in reducing pain. The measurements are centimeters on a pain scale before and after hypnosis. Assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. Construct 95% confidence interval for the mean of the "before−after" differences. Does hypnotism appear to be effective in reducing pain?
Before 6.4 2.6 7.7 10.5 11.7 5.8 4.3 2.8
After 6.7 2.4 7.4 8.1 8.6 6.4 3.9 2.7
The confidence interval for mean of the "before-after" differences is [tex]\fbox{(-0.4037,1.8037)}[/tex]
Further explanation:
Find the difference between the before pain and the after pain.
Difference = before-after
Kindly refer to the Table for the difference of between the before and after pain.
Sum of difference = [tex]5.6[/tex]
Total number of observation = [tex]8[/tex]
Mean of difference = [tex]0.7[/tex]
Sample standard deviation [tex]s[/tex] = [tex]1.3201[/tex]
Level of significance = [tex]5\%[/tex]
Formula for confidence interval = [tex]\left( \bar{X} \pm t_{n-1, \frac{\alpha}{2}\%} \frac{s}{\sqrt{n}} \right)[/tex]
confidence interval = [tex]\left( 0.7 \pm t_{8-1, \frac{5}{2}\%} \frac{1.3201}{\sqrt{8}} \right)[/tex]
confidence interval = [tex]\left( 0.7 \pm t_{7, \frac{5}{2}\%} \frac{1.3201}{\sqrt{8}} \right)[/tex]
From the t-table.
The value of [tex]t_{7, \frac{5}{2}\%[/tex]=[tex]2.365[/tex]
Confidence interval = [tex]( 0.7 \pm 2.365}\times \frac{1.3201}{\sqrt{8}}) \right)[/tex]
Confidence interval = [tex]\left( 0.7 - 2.365}\times \0.4667,0.7 + 2.365}\times \0.4667) \right[/tex]
Confidence interval = [tex](0.7-1.1037,0.7+1.1037)[/tex]
Confidence interval = [tex]\fbox{(-0.4037,1.8037)}[/tex]
The [tex]95\%[/tex] confidence interval tells us about that [tex]95\%[/tex] chances of the true mean or population mean lies in the interval.
Yes, the hypnotism appear to be effective in reducing pain as confidence interval include includes the positive deviation from the mean.
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Answer Details:
Grade: College Statistics
Subject: Mathematics
Chapter: Confidence Interval
Keywords:
Probability, Statistics, Speed dating, Females rating, Confidence interval, t-test, Level of significance , Normal distribution, Central Limit Theorem, t-table, Population mean, Sample mean, Standard deviation, Symmetric, Variance.
Answer:
The 95% confidence interval for the mean of the “before-after” difference is (-0.4039,1.8039)
No, Hypnotism doesn’t appear to be effective in reducing pain.
Further explanation:
Given: The table of measure the effectiveness of hypnotism in reducing pain.
Before : 6.4 2.6 7.7 10.5 11.7 5.8 4.3 2.8
After : 6.7 2.4 7.4 8.1 8.6 6.4 3.9 2.7
we make the table of difference between “before-after”
(Before-After) :
6.4-6.7 2.6-2.4 7.7-7.4 10.5-8.1 11.7-8.6 5.8-6.4 4.3-3.9 2.8-2.7
-0.3 0.2 0.3 2.4 3.1 -0.6 0.4 0.1
Now, we find the sample mean and sample standard deviation of above table.
[tex]\text{Sample Mean, }\bar{x}=\dfrac{\text{Sum of number}}{\text{number of observation}}[/tex]
[tex]=\dfrac{-0.3+0.2+0.3+2.4+3.1-0.6+0.4+0.1}{8}[/tex]
[tex]=\dfrac{5.6}{8}=0.7[/tex]
[tex]\text{Sample Standard deviation, s} = \sqrt{\dfrac{(x-\bar{x})^2}{n-1}}[/tex]
[tex]=\sqrt{\dfrac{(-0.3-0.7)^2+(0.2-0.7)^2+(0.3-0.7)^2 ...+(0.1-0.7)^2}{8-1}}[/tex]
[tex]=\dfrac{12.2}{7}=1.3201[/tex]
For 95% confidence interval [tex]\mu_d[/tex] using t-distribution
[tex]\text{Marginal Error, E}=t_{\frac{\alpha}{2},df}\times \dfrac{s}{\sqrt{n}}[/tex]
Where,
[tex]t_{\frac{\alpha}{2},df}[/tex] is critical value.alpha is significance level, [tex]\alpha=1-0.95=0.05 [/tex]df is degree of freedom for t-distribution, df=n-1 =7s is sample standard deviation, s=1.3201n is sample size, n=8For critical value,
[tex]\Rightarrow t_{\frac{\alpha}{2},df}[/tex]
[tex]\Rightarrow t_{\frac{0.05}{2},7}[/tex]
[tex]\Rightarrow t_{0.025,7}[/tex]
using t-distribution two-tailed table,
[tex]t_{0.025,7}=2.365[/tex]
Substitute the values into formula and calculate E
[tex]E=2.65\times \dfrac{1.301}{\sqrt{8}}[/tex]
Therefore, Marginal error, E=1.1039
95% confidence interval given by:
[tex]=\bar{x}\pm E[/tex]
[tex]=0.7\pm 1.1039[/tex]
For lowest value of interval: 0.7-1.1039 = -0.4039For largest value of interval: 0.7+1.1039 = 1.8039Therefore, 95% confidence interval using t-distribution: (-0.4039,1.8039)
This interval contains 0
Therefore, Hypnotism doesn’t appear to be effective in reducing pain.
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Keywords:
T-distribution, Sample mean, sample standard deviation, Critical value of t, degree of freedom, t-test, confidence interval, significance level.
The manager of a fashionable restaurant open Wednesday through Saturday says that the restaurant does about 29 percent of its business on Friday night, 31 percent on Saturday night, and 21 percent on Thursday night. What seasonal relatives would describe this situation?(Round your answers to 2 decimal places.)
Wednesday
Thursday
Friday
Saturday
Answer: The seasonal relatives is calculated are as follows:
Step-by-step explanation:
Given that,
restaurant only open from Wednesday to Saturday,
29 percent of its business on Friday31 percent on Saturday night21 percent on Thursday night.∴ The remaining 19% of its business he does on Wednesday
Now, suppose that total production of sales in a given week be 'y'
So, average sales in a week = [tex]\frac{y}{4}[/tex]
If we assume that y = 1
hence, average sales in a week = [tex]\frac{1}{4}[/tex]
= 0.25
Now, we have to calculate the seasonal relatives,
that is,
= [tex]\frac{Sales in a given day}{average sales in a week}[/tex]
Wednesday:
= [tex]\frac{0.19}{0.25}[/tex]
= 0.76
Thursday:
= [tex]\frac{0.21}{0.25}[/tex]
= 0.84
Friday:
= [tex]\frac{0.29}{0.25}[/tex]
= 1.16
Saturday:
= [tex]\frac{0.31}{0.25}[/tex]
= 1.24
- Wednesday:0.76
- Thursday:0.84
- Friday:1.16
- Saturday:1.24
To determine the seasonal relatives for each night, we need to express the business done each night as a percentage of the total business for the week. The given percentages are:
- Friday: 29%
- Saturday: 31%
- Thursday: 21%
First, let's find the total percentage accounted for by Wednesday, Thursday, Friday, and Saturday. Since we're missing Wednesday's percentage, we can sum the given percentages and subtract from 100%.
[tex]\[\text{Total percentage} = 29\% + 31\% + 21\% = 81\%\][/tex]
The remaining percentage for Wednesday is:
[tex]\[\text{Wednesday's percentage} = 100\% - 81\% = 19\%\][/tex]
Now, we'll convert these percentages into seasonal relatives. Seasonal relatives are the ratios of each night's business to the average nightly business across the four nights.
First, compute the average nightly business percentage:
[tex]\[\text{Average nightly business percentage} = \frac{100\%}{4} = 25\%\][/tex]
Next, calculate the seasonal relatives by dividing each night's percentage by the average nightly business percentage:
1. Wednesday:
[tex]\[ \text{Wednesday's seasonal relative} = \frac{19\%}{25\%} = 0.76 \][/tex]
2. Thursday:
[tex]\[ \text{Thursday's seasonal relative} = \frac{21\%}{25\%} = 0.84 \][/tex]
3. Friday:
[tex]\[ \text{Friday's seasonal relative} = \frac{29\%}{25\%} = 1.16 \][/tex]
4. Saturday:
[tex]\[ \text{Saturday's seasonal relative} = \frac{31\%}{25\%} = 1.24 \][/tex]
Rounded to two decimal places, the seasonal relatives are:
- Wednesday:0.76
- Thursday:0.84
- Friday:1.16
- Saturday:1.24
Find the p-value in a test of the claim that the mean College Algebra final exam score of engineering majors equal to 88, given that the test statistic is zequals1.50.
The p-value in a test of the claim that the mean College Algebra final exam score is 0.1336.
Given data:
To find the p-value in a test of the claim that the mean College Algebra final exam score of engineering majors is equal to 88, use the test statistic and the standard normal distribution.
The test statistic z = 1.50 represents how many standard deviations the sample mean is away from the hypothesized population mean.
The p-value is the probability of observing a test statistic as extreme or more extreme than the observed test statistic, assuming the null hypothesis is true.
Since the alternative hypothesis is not specified, assume a two-tailed hypothesis.
To find the p-value, we need to calculate the probability of observing a test statistic as extreme or more extreme than z = 1.50 in a standard normal distribution.
For a two-tailed test, we will find the probability in both tails.
The probability in the right tail is given by:
P(Z > 1.50) = 1 - P(Z < 1.50)
Using a standard normal distribution table, we find that P(Z < 1.50) is approximately 0.9332.
Therefore, P(Z > 1.50) = 1 - 0.9332 = 0.0668.
To find the p-value for the left tail, use the symmetry of the standard normal distribution.
P(Z < -1.50) = P(Z > 1.50) = 0.0668.
Since this is a two-tailed test, sum the probabilities of both tails to find the p-value:
p-value = 2 * 0.0668
p-value = 0.1336.
Hence, the p-value in this test is approximately 0.1336.
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In the given question, the p-value for a test statistic (z) of 1.50 can be computed as 0.0668. This means that if the null hypothesis is true (the mean final exam score being 88), there is a 6.68 percent chance we would observe a test statistic this extreme or more. This p-value is computed by subtracting the left tailed probability for z=1.50 from 1.
Explanation:The p-value (Probability Value) is a statistical measure used in hypothesis testing to determine the significance of the obtained results. It denotes the probability of obtaining the observed sample data given that the null hypothesis is true. In this case, we are interested in the p-value associated with a test statistic z=1.50 under the claim that the mean College Algebra final exam score for engineering majors is 88.
To find this p-value, we reference a standard normal (z) distribution table or use statistical software. Look up the value corresponding to z=1.50 which will give us the cumulative probability P(Z ≤ 1.50). However, since we want P(Z > 1.50), we subtract the obtained value from 1. This is due to the fact that the total probability under the curve of the standard normal distribution equals 1.
For a z score of 1.50 the standard z table gives a left tailed probability of approximately 0.9332. Therefore, P(Z > 1.50) = 1 - P(Z ≤ 1.50) = 1 - 0.9332 = 0.0668. So the p-value is approximately 0.0668.
The interpretation of this value would be: if the null hypothesis is true (the mean final exam score is 88), then there is a 0.0668 probability, or 6.68 percent, that we would observe a test statistic greater than or equal to 1.50.
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Conferences and conventions are resources that should be explored as part of a job search? True or False
Answer:
The given statement is true.
Step-by-step explanation:
Conferences and conventions are resources that should be explored as part of a job search.
Yes this statement is true.
One can choose the various conferences being held around them through sources like internet, social media, newspapers etc. These are very helpful for job seekers as one can get a lot of job related advice.
Find the GCF and LCM of 217x328x11 and 213x345x74and explain your reasoning.
Answer:
Given expressions are,
217 x 328 x 11
213 x 345 x 74,
Since, 217 = 7 × 31
328 = 2 × 2 × 2 × 41,
11 = 1 × 11,
So, we can write, 217 x 328 x 11 = 7 × 31 × 2 × 2 × 2 × 41 × 1 × 11
Now, 213 = 3 × 71
345 = 3 × 5 × 23,
74 = 2 × 37,
So, 213 x 345 x 74 = 3 × 71 × 3 × 5 × 23 × 3 × 5 × 23
Thus, GCF ( greatest common factor ) of the given expressions = 1 ( because there are no common factors )
We know that if two numbers have GCF 1 then their LCM is obtained by multiplying them,
Hence, LCM ( least common multiple ) of the given expressions = 217 x 328 x 11 x 213 x 345 x 74
A tank contains 300 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.
Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
___
The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
The number of grams of salt in the tank at any time t is 40 grams. The inflow and outflow of brine do not affect the amount of salt in the tank because the solution is well-mixed, and the salt concentration remains constant.
To solve this problem, we need to set up a differential equation that describes the rate of change of salt in the tank over time. Let A(t) represent the number of grams of salt in the tank at time t.
Let's break down the components affecting the rate of change of salt in the tank:
Salt inflow rate: The brine is being pumped into the tank at a constant rate of 6 liters per minute, and it contains 1 gram of salt per liter. So, the rate of salt inflow is 6 grams per minute.
Salt outflow rate: The solution in the tank is being pumped out at the same rate of 6 liters per minute, which means the rate of salt outflow is also 6 grams per minute.
Mixing of the solution: Since the tank is well-mixed, the concentration of salt remains uniform throughout the tank.
Now, let's set up the differential equation for A(t):
dA/dt = Rate of salt inflow - Rate of salt outflow
dA/dt = 6 grams/min - 6 grams/min
dA/dt = 0
The above equation shows that the rate of change of salt in the tank is constant and equal to zero. This means the number of grams of salt in the tank remains constant over time.
Now, let's find the constant value of A(t) using the initial condition where the tank initially contains 40 grams of salt.
When t = 0, A(0) = 40 grams
Since the rate of change is zero, A(t) will be the same as the initial amount of salt in the tank at any time t:
A(t) = 40 grams
So, the number of grams of salt in the tank at any time t is 40 grams. The inflow and outflow of brine do not affect the amount of salt in the tank because the solution is well-mixed, and the salt concentration remains constant.
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If x, y, a and b are greater than zero and x/y lessthanorequalto a/b, prove that x+a/y+b lessthanorequalto a/b
Answer:
Step-by-step explanation:
Given x,y,a&b are greater than zero
also [tex]\frac{x}{y}[/tex][tex]\leq [/tex][tex]\frac{a}{b}[/tex]
since x,y,a&b are greater than zero therefore we can cross multiply them without changing the inequality
therefore
[tex]\frac{x}{a}[/tex][tex]\leq [/tex][tex]\frac{y}{b}[/tex]
adding 1 on both sides we get
[tex]\frac{x}{a}[/tex]+1[tex]\leq [/tex][tex]\frac{y}{b}[/tex]+1
[tex]\frac{x+a}{a}[/tex][tex]\leq [/tex][tex]\frac{y+b}{b}[/tex]
rearranging
[tex]\frac{x+a}{y+b}[/tex][tex]\leq [/tex][tex]\frac{a}{b}[/tex]
The number of diners at a restaurant each day is recorded and a daily average is calculated every month (assume 30 days in a month). The number of diners each day has a mean of 109 and a standard deviation of 53, but does not necessarily follow a normal distribution. The probability that a daily average over a given month is greater than x is 2.5%. Calculate x. Give your answer to 3 decimal places.
Answer: 212.88
Step-by-step explanation:
Given : The probability that a daily average over a given month is greater than x = [tex]2.5\%=0.025[/tex]
The probability that corresponds to 0.025 from a Normal distribution table is 1.96.
Mean : [tex]\mu = 109[/tex]
Standard deviation : [tex]\sigma = 53[/tex]
The formula for z-score : -[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
[tex]\Rightarrow\ 1.96=\dfrac{x-109}{53}\\\\\Rightarrow\ x=53\times1.96+109\\\\\Rightarrow\ x=212.88[/tex]
Z scores (converted value in standard normal distribution) can be mapped to probabilities by z tables. The value of x is 212.88 approx.
How to get the z scores?If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
For the given case, let the random variable X tracks the number of dinners at given restaurant. Assuming normal distribution being pertained by X, we get:
[tex]X \sim N(109, 53)[/tex]
The given data shows that:
2.5% of all daily averages records lie bigger than value X = x
or
P(X > x) = 2.5% 0.025
Converting it to standard normal distribution(so that we can use z tables and p values to get the unknown x), we get:
[tex]z = \dfrac{x-\mu}{\sigma} = \dfrac{x - 109}{53}[/tex]
The given probability statement is expressed as:
[tex]P(Z > z) = 2.5\% = 0.025\\P(Z \leq z) = 1 - 0.025 = 0.975[/tex]
Seeing the z tables, we will try to find at what value of z, the p value is obtained near to 0.975
We get z = 1.96.
Thus,
[tex]z = 1.96 = \dfrac{x - 109}{53}\\\\x = 1.96 \times 53} + 109 = 212.88[/tex]
Thus,
The value of x is 212.88 approx.
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4. Fraction: Explain what 5/6 means. Write an explanation of the term fraction that should work with 5/6 and %.
Answer:
See below.
Step-by-step explanation:
5/6 is a fraction. The 5 is in the numerator, and the 6 is in the denominator.
The denominator is the number of parts the unit was divided into. In this case, the denominator is 6. That means one unit, 1, was divided into 6 equal parts. Each part is one-sixth.
The numerator is the number of those parts that you use. 5 in the numerator means to use 5 of those parts, each of which is 1/6 of 1.
In other words, 5/6 means divide 1 into 6 equal parts, and take 5 of those parts.
5/6 is the same as 5 divided by 6, so as a decimal it is 0.8333...
As a percent it is 83.333...%
A large fishing farm with thousands of fish has been treating its fish to stop a spreading fungal infection. The owner of the fishing farm claims that fewer than 10% of the fish are infected. A random sample of 50 fish is taken to determine the proportion p that is infected in this population. A careful examination determines that 6 of the fish sampled are infected. The test statistic for the above hypothesis test of proportion of fish that are infected is... (Round your answer to two decimal places.)
Answer:
The test statistic for the above hypothesis test of proportion of fish that are infected is 0.472.
Step-by-step explanation:
[tex]\text{consider the provided information}[/tex]
It is given that the total sample space is 50 fish. Out of which we have found that 6 of the fish sampled are infected. Therefore,
n is 50 and x = 6
[tex]\text{The hypotheses is}[/tex]
[tex]H_0: P=0.10, H_a: P<0.10[/tex]
Now, calculate the sample proportion by dividing the infected sample by sample space as shown:
[tex]\hat{p}=\frac{6}{50}=0.12[/tex]
The standard deviation of proportion can be calculated by using the formula:
[tex]\sigma=\sqrt{\frac{p(1-p)}{n}}[/tex]
[tex]\text{Now substitute the respective values in the above formula}[/tex]
[tex]\sigma=\sqrt{\frac{0.10(1-0.10)}{50}}[/tex]
[tex]\sigma=\sqrt{\frac{0.10(0.9)}{50}}[/tex]
[tex]\sigma=\sqrt{0.0018}[/tex]
[tex]\sigma=0.0424[/tex]
[tex]\text{The test statistic is:}[/tex]
[tex]z=\frac{\hat{p}-p}{\sigma}[/tex]
[tex]\text{Now substitute the respective values in the above formula}[/tex]
[tex]z=\frac{0.12-0.10}{0.0424}[/tex]
[tex]z=\frac{0.02}{0.0424}[/tex]
[tex]z=0.472\ approximately [/tex]
Hence, the test statistic for the above hypothesis test of proportion of fish that are infected is 0.472.
In a hypothesis test for a fishing farm regarding the proportion of fish that are infected, using the given sample size of 50 fish and the number of successes (infected fish) as 6, we calculate a sample proportion of 0.12. In using a normal approximation given the conditions are met, the test statistic (Z-score) is calculated as 0.67 when rounded to two decimal places.
Explanation:From your question, it seems we are testing a claim about a population proportion in a fishing farm. Here the null hypothesis (H0) would be that p ≥ 0.10 (i.e., 10% or more of the fish are infected) and the alternative hypothesis (H1) would be that p < 0.10 (i.e., less than 10% of the fish are infected).
We're given a sample (n) of 50 fish, of which 6 are infected (successes, x). Since both np and nq are larger than 5 (6 > 5 and 44 > 5), we can use the normal approximation to the binomial distribution. Thus, we calculate the sample proportion (p') as x/n = 6/50 = 0.12.
The test statistic is calculated using the formula Z = (p' - p) / sqrt [ p(1-p) / n ].
Substitute: Z = (0.12 - 0.10) / sqrt [ 0.10 * (1 - 0.10) / 50 ]Simplify: Z = 0.02 / sqrt(0.009)Calculate: Z = 0.67 (rounded to two decimal places)
Therefore, the test statistic for the above hypothesis test of the proportion of fish that are infected is 0.67.
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Solve one of the following non-homogeneous Cauchy-Euler equations using whatever technique you prefer. Put an "X" through whichever equations you would not put an "X" through either equation, I will grade whichever one I prefer. a) x^2 y" + 10xy' + 8y = x^2 b) x^2 y" - 3xy' + 13y = 4 + 3x
Answer:
a.[tex]y(x)=C_1x^{-1}+C_2x^{-8}+\frac{1}{30}x^2[/tex]
b.[tex]y(x)=x^2(C_1cos (3lnx)+C_2sin(3lnx))+\frac{4}{13}+\frac{3}{10}x[/tex]
Step-by-step explanation:
1.[tex]x^2y''+10xy'+8y =x^2[/tex]
It is Cauchy-Euler equation where [tex]x=e^t[/tex]
Then auxillary equation
[tex]D'(D'-1)+10D'+8=0[/tex]
[tex]D'^2+9D'+8=0[/tex]
[tex](D'+1)(D'+8)=0[/tex]
D'=-1 and D'=-8
Hence, C.F=[tex]C_1e^{-t}+C_2e^{-8t}[/tex]
C.F=[tex]C_1\frac{1}{x}+C_2\frac{1}{x^8}[/tex]
P.I=[tex]\frac{e^{2t}}{D'^2+9D'+8}=\frac{e^{2t}}{4+18+8}[/tex]
Where D'=2
[tex]P.I=\frac{1}{30}e^{2t}=\frac{1}{30}x^2[/tex]
[tex]y(x)=C_1x^{-1}+C_2x^{-8}+\frac{1}{30}x^2[/tex]
b.[tex]x^2y''-3xy'+13y=4+3x[/tex]
Same method apply
Auxillary equation
[tex] D'^2-D'-3D'+13=0[/tex]
[tex]D'^2-4D'+13=0[/tex]
[tex]D'=2\pm3i[/tex]
C.F=[tex]e^{2t}(C_1cos 3t+C_2sin 3t)[/tex]
C.F=[tex]x^2(C_1cos (3lnx)+C_2sin(3lnx))[/tex]
[tex]e^t=x[/tex]
P.I=[tex]\frac{4e^{0t}}{D'^2-4D'+13}+3\frac{e^t}{D'^2-4D'+13}[/tex]
Substitute D'=0 where[tex] e^{0t}[/tex] and D'=1 where [tex]e^t[/tex]
P.I=[tex]\frac{4}{13}+\frac{3}{10}e^t[/tex]
P.I=[tex]\frac{4}{13}+\frac{3}{10}x[/tex]
[tex]y(x)=C.F+P.I=x^2(C_1cos (3lnx)+C_2sin(3lnx))+\frac{4}{13}+\frac{3}{10}x[/tex]
Verify that y = c_1 + c_2 e^2x is a solution of the ODE y" - 2y' = 0 for all values of c_1 and c_2.
Answer:
For any value of C1 and C2, [tex]y = C1 + C2*e^{2x}[/tex] is a solution.
Step-by-step explanation:
Let's verify the solution, but first, let's find the first and second derivatives of the given solution:
[tex]y = C1 + C2*e^{2x}[/tex]
For the first derivative we have:
[tex]y' = 0 + C2*(2x)'*e^{2x}[/tex]
[tex]y' = C2*(2)*e^{2x}[/tex]
For the second derivative we have:
[tex]y'' = C2*(2)*(2x)'*e^{2x}[/tex]
[tex]y'' = C2*(2)*(2)*e^{2x}[/tex]
[tex]y'' = C2*(4)*e^{2x}[/tex]
Let's solve the ODE by the above equations:
[tex]y'' - 2y' = 0[/tex]
[tex]C2*(4)*e^{2x} - 2*C2*(2)*e^{2x} = 0[/tex]
[tex]C2*(4)*e^{2x} - C2*(4)*e^{2x} = 0[/tex]
From the above equation we can observe that for any value of C2 the equation is solved, and because the ODE only involves first (y') and second (y'') derivatives, C1 can be any value as well, because it does not change the final result.
What is the Z-score for the value that is two standard deviations away from the mean?
Answer:
2
Step-by-step explanation:
The z-score is the number of standard deviations the value is above the mean. If that number is 2, then Z=2. If that number is -2, then Z=-2.
___
Both Z=2 and Z=-2 are values that are two standard deviations from the mean.
Use Newton's method with initial approximation x1 = −2 to find x2, the second approximation to the root of the equation x3 + x + 6 = 0. (Round your answer to four decimal places.)
Answer:
The value of [tex]x_2=-1.6923[/tex].
Step-by-step explanation:
Consider the provided information.
The provided formula is [tex]f(x)=x^3+x+6[/tex]
Substitute [tex]x_1=-2[/tex] in above equation.
[tex]f(x_1)=(-2)^3+(-2)+6[/tex]
[tex]f(x_1)=-8-2+6[/tex]
[tex]f(x_1)=-4[/tex]
Differentiate the provided function and calculate the value of [tex]f'(x_1)[/tex]
[tex]f'(x)=3x^2+1[/tex]
[tex]f'(x)=3(-2)^2+1[/tex]
[tex]f'(x)=13[/tex]
The Newton iteration formula: [tex]x_2=x_1-\frac{f(x_1)}{f'(x_1)}[/tex]
Substitute the respective values in the above formula.
[tex]x_2=-2-\frac{(-4)}{13}[/tex]
[tex]x_2=-2+0.3077[/tex]
[tex]x_2=-1.6923[/tex]
Hence, the value of [tex]x_2=-1.6923[/tex].
To use Newton's method with an initial approximation of -2 on the equation x3 + x + 6 = 0, we identify the function and its derivative. We then substitute into Newton's method's formula, x_2 = x_1 - f(x_1) / f'(x_1). This will provide us with an approximate second root, which can then be refined further.
Explanation:The subject of the question pertains to Newton's method for root finding, which is a method for finding successively better approximations to the roots (or zeroes) of a real-valued function. The function in question is x3 + x + 6 = 0 and the initial approximation provided is -2.
First, in order to use Newton's method, we need to identify the function and its derivative. The function (f(x)) is x3 + x + 6. The derivative (f'(x)) would be 3x2 + 1.
Newton's method follows the formula: x_(n+1) = x_n - f(x_n) / f'(x_n).
With x_1 as -2, we substitute into the Newton method's formula to find x_2. Hence, x_2 = -2 - f(-2) / f'(-2), which results in an approximation of the root that can be further refined. Remember to round your answer to four decimal places after calculations.
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An oil refinery is located 1 km north of the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 9 km east of the refinery. The cost of laying pipe is $500,000/km over land to a point P on the north bank and $1,000,000/km under the river to the tanks. To minimize the cost of the pipeline, how far downriver from the refinery should the point P be located? (Round your answer to two decimal places.)
km
Answer:
7.86 km
Step-by-step explanation:
Let x represent the distance point P lies east of the refinery. (We assume this direction is downriver from the refinery.)
The cost of laying pipe to P from the refinery (in millions of $) will be ...
0.5√(1² +x²)
The cost of laying pipe under the river from P to the storage facility will be ...
1.0√(2² +(9-x)²) = √(85 -18x +x²)
We want to minimize the total cost c. That total cost is ...
c = 0.5√(x² +1) +√(x² -18x +85)
The minimum value is best found using technology. (Differentiating c with respect to x results in a messy radical equation that has no algebraic solution.) A graphing calculator shows it to be at about x ≈ 7.86 km.
Point P should be located about 7.86 km downriver from the refinery.
The cost minimization problem of constructing a pipeline from a refinery to the storage tanks on the other side of the river can be solved by differential calculus. The distance of point P down the river from the refinery can be determined by differentiating the total cost function and equating it to zero to minimize the cost.
Explanation:This is a math problem related to cost minimization and deals with the principles of trigonometry. Let's denote the distance downriver from the refinery to point P as x. Since the refinery, point P, and the oil tank form a right triangle, we can apply the Pythagorean theorem. The length of the pipeline from point P to the tanks is the hypotenuse of the triangle, which is sqr((9-x)^2+2^2). Hence, the total cost of the pipeline is C = $500,000*x + $1,000,000*sqr((9-x)^2+2^2).
To find the minimum cost, we need to differentiate the total cost function C with respect to x and equate it to zero. By solving the derivative of C equals to zero, we can find the optimal distance x value. Therefore, solving this derivative can give us the solution in terms of miles.
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Find the slope and the y -intercept of the line.
Write your answers in simplest form.
9x - 3y = -2
Answer:
The slope is: 3
The y-intercept is: [tex]\frac{2}{3}[/tex] or [tex]0.66[/tex]
Step-by-step explanation:
The equation of the line in Slope-Intercept form is:
[tex]y=mx+b[/tex]
Where "m" is the slope of the line and "b" is the y-intercept.
To write the given equation in this form, we need to solve for "y":
[tex]9x - 3y = -2\\\\- 3y = -9x-2\\\\y=3x+\frac{2}{3}[/tex]
Therefore, you can identify that the slope of this line is:
[tex]m=3[/tex]
And the y-intercept is:
[tex]b=\frac{2}{3}=0.66[/tex]
Solve for the indicated variable.
Answer:
[tex]y=\frac{2x}{9}-2[/tex]
Step-by-step explanation:
The given equation is:
2x-9y=18
To solve for y means we need to isolate y on one side of the equation, carrying all the other variables and terms to the other side so that we get a formula for y. This can be done as shown below:
2x - 9y = 18
Subtracting 2x from both sides, we get:
-9y = 18 - 2x
Dividing both sides by -9, we get:
[tex]\frac{-9y}{-9}=\frac{18}{-9}-\frac{2x}{-9}\\\\ y=-2+\frac{2x}{9}\\\\ y=\frac{2x}{9}-2[/tex]
Answer:
[tex] y = \frac { 2 ( x - 9 ) } { 9 } [/tex]
Step-by-step explanation:
We are given the following expression and we are to solve it for the indicated variable (y):
[tex] 2 x - 9 y = 1 8 [/tex]
Making [tex] y [/tex] the subject of the equation and simplifying it to get:
[tex] 2 x - 1 8 = 9 y \\\\ 9 y = 2 x - 1 8 \\\\ y = \frac { 2 x - 1 8 } { 9 } \\\\ y = \frac { 2 ( x - 9 ) } { 9 } [/tex]
For the following velocity field: V = (u,v)=(1 -0.5x)i+ (-1.5 - 2y), find out if there astagnation point. If so, calculate its location in coordinates.r and y (20 pts)
Answer:
[tex]\overrightarrow{r}=2\widehat{i}-0.75\widehat{j}[/tex]
Step-by-step explanation:
Stagnation point is point of zero velocity thus each component of the velocity must also be zero
[tex]Given\\u=(1-0.5x)[/tex]
[tex]\therefore[/tex] u=0 at x=2m
Similarly
[tex]\\v=(-1.5-2y)\\\\\therefore v=0 \\\\y=\frac{1.5}{-2}\\\\y=-0.75[/tex]
Thus point of stagnation is (2,-0.75)
Thus it's position vector is given by
[tex]\overrightarrow{r}=2\widehat{i}-0.75\widehat{j}[/tex]
Which of the following is an example of qualitative data? a. Average rainfall on 25 days of the month b. Number of accidents occurred in the month of September c. Weight of 35 students in a class d. Height of 40 students in a class e. Political affiliation of 2,250 randomly selected voters
Answer: Option 'e' is correct.
Step-by-step explanation:
Qualitative data refers to those data which does not include any numerical value.
Average rainfall on 25 days of the month is a quantitative data as it would represented as numerical value.
Number of accidents occurred in the month of September is quantitative data.
Height of 40 students in a class is quantitative data too.
Weight of 35 students in a class is quantitative data too.
But political affiliation of 2250 randomly selected voters is qualitative data as it has not included any numerical value.
Hence, option 'e' is correct.
Option e, 'Political affiliation of 2,250 randomly selected voters,' represents qualitative data because it categorizes voters based on a non-numeric characteristic, their political affiliation.
The question 'Which of the following is an example of qualitative data? e. Political affiliation of 2,250 randomly selected voters' is asking us to identify a type of data among the given options. Qualitative data refers to information that is categorized based on attributes or qualities rather than numerical values. In contrast, quantitative data involve numbers and can either be discrete, which are countable, or continuous, which can take on any value within a range.
To answer the question, option e, 'Political affiliation of 2,250 randomly selected voters,' represents qualitative data because it describes a characteristic (political affiliation) that is non-numeric and is usually expressed in words or categories. Other options such as average rainfall, number of accidents, weight, and height involve numerical measurements and are therefore examples of quantitative data, which can either be discrete or continuous depending on the nature of the measurement.
Option e is unique as the only example of qualitative data in the options provided, because political affiliation does not result from a measurement or count, but rather, it is a category used to describe a voter's preferred political party or stance.
Farmer Jones, and his wife, Dr. Jones, decide to build a fence in their field, to keep the sheep safe. Since Dr. Jones is a mathematician, she suggests building fences described by y=6x2 and y=x2+2. Farmer Jones thinks this would be much harder than just building an enclosure with straight sides, but he wants to please his wife. What is the area of the enclosed region?
Answer:[tex]\frac{8}{3}\times \sqrt{\frac{2}{5}}[/tex]
Step-by-step explanation:
Given two upward facing parabolas with equations
[tex]y=6x^2 & y=x^2+2[/tex]
The two intersect at
[tex]6x^2=x^2+2[/tex]
[tex]5x^2=2[/tex]
[tex]x^2[/tex]=[tex]\frac{2}{5}[/tex]
x=[tex]\pm \sqrt{\frac{2}{5}}[/tex]
area enclosed by them is given by
A=[tex]\int_{-\sqrt{\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left [ \left ( x^2+2\right )-\left ( 6x^2\right ) \right ]dx[/tex]
A=[tex]\int_{\sqrt{-\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left ( 2-5x^2\right )dx[/tex]
A=[tex]4\left [ \sqrt{\frac{2}{5}} \right ]-\frac{5}{3}\left [ \left ( \frac{2}{5}\right )^\frac{3}{2}-\left ( -\frac{2}{5}\right )^\frac{3}{2} \right ][/tex]
A=[tex]\frac{8}{3}\times \sqrt{\frac{2}{5}}[/tex]
The area of the enclosed region is -√2/15 square units.
Explanation:To find the area of the enclosed region, we need to find the points of intersection between the two equations y=6x^2 and y=x^2+2. Setting them equal to each other:
6x^2 = x^2 + 2
5x^2 = 2
x^2 = 2/5
x = ±√(2/5)
Substituting these values of x back into one of the equations, we can find the corresponding y values:
For x = √(2/5), y = 6(√(2/5))^2 = 6(2/5) = 12/5
For x = -√(2/5), y = 6(-√(2/5))^2 = 6(2/5) = 12/5
Now we can find the area of the enclosed region by calculating the definite integral of y=6x^2 - (x^2+2) from x = -√(2/5) to x = √(2/5). This can be done using the fundamental theorem of calculus:
∫[√(2/5), -√(2/5)] [6x^2 - (x^2+2)] dx = ∫[√(2/5), -√(2/5)] (5x^2 - 2) dx = [5/3x^3 - 2x] [√(2/5), -√(2/5)] = 2(√(2/5))^3 - 5/3(√(2/5))^3 = 4√2/15 - 5√2/15 = -√2/15
Therefore, the area of the enclosed region is -√2/15 square units.
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Convert the measurement as indicated. 52 ft. = _____ yd. _____ ft.
options:
17, 1
16, 1
17, 2
18, 2
Answer:
17, 1
Step-by-step explanation:
You can find the yards by multiplying by the conversion factor, then determining what the fraction or remainder means. Since there are 3 ft in 1 yd, 1/3 yd is 1 ft.
52 ft = (52 ft) × (1 yd)/(3 ft) = 52/3 yd = 17 1/3 yd = 17 yd 1 ft
To convert feet to yards, you divide by 3. Therefore, 52 feet is equivalent to 17 yards and 1 foot.
Explanation:In mathematics, particularly in the subject of measurement conversions, it is useful to know that 1 yard (yd) is equal to 3 feet (ft). So, to convert from feet to yards, you divide the number of feet by 3. Applying this principle to the question, you will take the 52 feet and divide it by 3.
52 feet ÷ 3 = 17.3 repeating
However, since the options provided do not include a decimal, we take the whole number part of the answer, which is 17 yards. There is a remainder when you divide 52 by 3, which indicates there are additional feet not making up a full yard. In this case, it is 1 foot (the decimal part times 3), making the final conversion 17 yards 1 foot.
So, 52 ft equals 17 yards and 1 foot.
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The sales decay for a product is given by S = 70000e^ -0.8x where S is the monthly sales and x is the number of months that have passed since the end of a promotional campaign. How many months after the end of the campaign will sales drop below 1000, if no new campaign is initiated? (Round your answers to two decimal places.) What will be the sales 5 months after the end of the campaign?
Answer:
After 5.31 months sales will drop below 1000 and 5 months after the end of the campaign sales will be 1282.09
Step-by-step explanation:
Let's find the solutions for the two questions.
First question: How many months after the end of the campaign will sales drop below 1000.
Because the problem asks for how many months, and since 'x' represents month variable, then the problem is asking for 'x'.
Using the same equation for sales we can observe the following:
[tex]S=70000*e^{-0.8X}[/tex], but we have S which is 1000, so:
[tex]1000=70000*e^{-0.8X}[/tex] which is equal to:
[tex]1000/70000=e^{-0.8X}[/tex] which is equal to:
[tex]1/70=e^{-0.8X}[/tex] by applying ln(x) properties:
[tex]ln(1/70)=ln(e^{-0.8X})[/tex] which is equal to:
[tex]ln(1/70)=-0.8X[/tex] which is equal to:
[tex]ln(1/70)/(-0.8)=X[/tex] so:
[tex]X=5.31 months[/tex]
Second question: what will be the sales 5 months after the end of the campaign.
Because the problem asks for what will be the sales, and since 'S' represents the sales, then the problem is asking for 'S'.
Using the same equation for sales we can observe the following:
[tex]S=70000*e^{-0.8X}[/tex], but we have x which is 5 months, so:
[tex]S=70000*e^{-0.8*5}[/tex] which is equal to
[tex]S=1282.09[/tex]
In conclusion, after 5.31 months sales will drop below 1000 and 5 months after the end of the campaign sales will be 1282.09.
h6) Introducing different sized atoms into the crystal or grain can help prevent: Slip Corrosion Hardness Conductivity
5. Dislocation densities for ceramics are very low and usually as low as 10Pmm A. True B. False
Answer:
Step-by-step explanation:94
You purchase boxes of cereal until you obtain one with the collector's toy you want. If, on average, you get the toy you want in every 49th cereal box, what is the probability of getting the toy you want in any given cereal box?
Answer:
The probability of getting the toy in any given cereal box is [tex]\frac{1}{49}[/tex].
Step-by-step explanation:
Given,
On average, we get a toy in every 49th cereal box,
That is, in every 49 boxes there is a toy,
So, the total outcomes = 49,
Favourable outcomes ( getting a toy ) = 1
Since, we know that,
[tex]\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex]
Hence, the probability of getting the toy in any given cereal box = [tex]\frac{1}{49}[/tex]
Answer:
The probability of getting the toy you want in any given cereal box is of 0.0204 = 2.04%.
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected number of trials for r sucesses is:
[tex]E = \frac{r}{p}[/tex]
If, on average, you get the toy you want in every 49th cereal box, what is the probability of getting the toy you want in any given cereal box?
This means that [tex]E = 49, r = 1[/tex]
So
[tex]49 = \frac{1}{p}[/tex]
[tex]49p = 1[/tex]
[tex]p = \frac{1}{49}[/tex]
[tex]p = 0.0204[/tex]
The probability of getting the toy you want in any given cereal box is of 0.0204 = 2.04%.
A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t) = 60t - 16t^2 . What is the maximum height that the ball will reach?
Do not round your answer.
Answer:
56.25 feet.
Step-by-step explanation:
h(t) = 60t - 16t^2
Differentiating to find the velocity:
v(t) = 60 -32t
This equals zero when the ball reaches its maximum height, so
60-32t = 0
t = 60/32 = 1.875 seconds
So the maximum height is h(1.875)
= 60* 1.875 - 16(1.875)^2
= 56.25 feet.
Answer: 56.25 feet.
Step-by-step explanation:
For a Quadratic function in the form [tex]f(x)=ax^2+bx+c[/tex], if [tex]a<0[/tex] then the parabola opens downward.
Rewriting the given function as:
[tex]h(t) = - 16t^2+60t[/tex]
You can identify that [tex]a=-16[/tex]
Since [tex]a<0[/tex] then the parabola opens downward.
Therefore, we need to find the vertex.
Find the x-coordinate of the vertex with this formula:
[tex]x=\frac{-b}{2a}[/tex]
Substitute values:
[tex]x=\frac{-60}{2(-16)}=1.875[/tex]
Substitute the value of "t" into the function to find the height in feet that the ball will reach. Then:
[tex]h(1.875)=- 16(1.875)^2+60(1.875)=56.25ft[/tex]