2. A multiple-choice test contains 10 questions. There are four possible answers for each question lo how many ways can a student answer the questions on the test if the student answers every question? the test if the studest

Answer: 0.5

Step-by-step explanation:

Binomial probability formula :-

[tex]P(x)=^nC_x\ p^x(q)^{n-x}[/tex], where P(x) is the probability of getting success in x trials , n is the total trials and p is the probability of getting success in each trial.

Given : The probability that the adults follow more than one game = 0.30

Then , q= 1-p = 1-0.30=0.70

The number of adults surveyed : n= 15

Let X be represents the adults who follow more than one sport.

Then , the probability that fewer than 4 of them will say that football is their favorite sport,

[tex]P(X\leq4)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)\\\\=^{15}C_{0}\ (0.30)^0(0.70)^{15}+^{15}C_{1}\ (0.30)^1(0.70)^{14}+^{15}C_{2}\ (0.30)^2(0.70)^{13}+^{15}C_{3}\ (0.30)^3(0.70)^{12}+^{15}C_{4}\ (0.30)^4(0.70)^{11}\\\\=(0.30)^0(0.70)^{15}+15(0.30)^1(0.70)^{14}+105(0.30)^2(0.70)^{13}+455(0.30)^3(0.70)^{12}+1365(0.30)^4(0.70)^{11}\\\\=0.515491059227\approx0.5[/tex]

Hence, the probability rounded to the nearest tenth that fewer than 4 of them will say that football is their favorite sport =0.5

Final answer:

To find the probability that the mean weight of a sample of 107 babies differs from the population mean by less than 40 grams, we can use the Central Limit Theorem. The probability is approximately 0.0441.

Explanation:

To find the probability that the mean weight of a sample of 107 babies differs from the population mean by less than 40 grams, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means will be approximately normal as long as the sample size is large enough. In this case, our sample size is 107, which is considered large enough.

The formula to calculate the standard deviation of the sampling distribution of the sample mean is sigma/sqrt(n), where sigma is the population standard deviation and n is the sample size. Plugging in the values, we have 446/sqrt(107) = 42.96 grams.

To find the probability, we need to calculate the z-score for a difference of 40 grams from the population mean. The z-score formula is (x - mu) / (sigma / sqrt(n)), where x is the desired difference, mu is the population mean, sigma is the population standard deviation, and n is the sample size.

Plugging in the values, we have (40 - 3242) / (446/sqrt(107)) = -1.7. We want to find the probability that the z-score is less than -1.7. Using a standard normal distribution table or calculator, we find that the probability is approximately 0.0441.

To solve this question, we can use the Central Limit Theorem (CLT), which implies that the sampling distribution of the sample means will be normally distributed if the sample size is large enough (usually n > 30 is considered sufficient). Since we are dealing with a sample of 107 babies, which is quite large, we can safely use the CLT.
According to the CLT, the mean of the sample means will be equal to the population mean (μ), and the standard deviation of the sample means (often called the standard error, SE) will be equal to the population standard deviation (σ) divided by the square root of the sample size (n).
Given:
- The population mean (μ) = 3242 grams
- The population standard deviation (σ) = 446 grams
- The sample size (n) = 107 babies
First, we must calculate the standard error (SE):
SE = σ / √n
SE = 446 grams / √107
SE ≈ 446 grams / 10.344
SE ≈ 43.13 grams
Now, we want to find the probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams. This means we are looking at weights from (μ - 40) grams to (μ + 40) grams.
So we will find the z-scores corresponding to (μ - 40) grams and (μ + 40) grams:
Z = (X - μ) / SE
For the lower limit (μ - 40 grams = 3202 grams):
Z Lower = (3202 - 3242) / 43.13
Z Lower ≈ -40 / 43.13
Z Lower ≈ -0.927
For the upper limit (μ + 40 grams = 3282 grams):
Z Upper = (3282 - 3242) / 43.13
Z Upper ≈ 40 / 43.13
Z Upper ≈ 0.927
Using the standard normal distribution (Z-distribution), we can find the probability that a Z-score falls between -0.927 and 0.927. These values correspond to the area under the standard normal curve between these two Z-scores.
If we look these values up in a Z-table, or use a calculator:
- The probability of Z being less than 0.927 is approximately 0.8238.
- The probability of Z being less than -0.927 is approximately 0.1762.
So, the probability of the mean weight being between μ - 40 grams and μ + 40 grams is the area between these two Z-scores, which can be found by subtracting the lower probability from the upper probability:
P(-0.927 < Z < 0.927) = P(Z < 0.927) - P(Z < -0.927)
P(-0.927 < Z < 0.927) = 0.8238 - 0.1762
P(-0.927 < Z < 0.927) = 0.6476
Rounding to four decimal places:
P(-0.927 < Z < 0.927) ≈ 0.6476
Therefore, the probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams is approximately 0.6476, or 64.76%.

Answer:

Given

[tex]f(m+n)=f(m)f(n)[/tex]

If we assume

[tex]f(x)=ae^{x}\\\\f(m+n)=ae^{m+n}\\\\\therefore f(m+n)=ae^{m}\times ae^{n}(\because x^{a+b}=x^{a}\times x^{b})\\\\\Rightarrow f(m+n)=f(m)\times f(n)[/tex]

Similarly

We can generalise the result for

[tex]f(x)=am^{x }[/tex] where a,m are real numbers

2)

[tex]f(m\cdot n)=f(m)+f(n)\\\\let\\f(x)=log(x)\\\therefore f(mn)=log(mn)=log(m)+log(n)\\\\\therefore f(mn)=f(m)+f(n)[/tex]

[tex]x^2-6x + 8=x^2-6x+9-1=(x-3)^2-1[/tex]

Answer:

(x - 3)^2 - 1.

Step-by-step explanation:

x2 – 6x + 8.

We divide the coefficient of x by 2:

-6 / 2 = -3 so the contents of the parentheses is x - 3:

x^2 - 6x + 8

= (x - 3)^2 - (-3)^2 + 8

= (x - 3)^2 - 1.

Answer: True, hope this helps! :)

Answer: Probability that the proportion of students who graduated is greater than 0.743 is P = 0.4755

Step-by-step explanation:

Given that,

Probability of freshmen entering public high schools in 2006 graduated with their class in 2010, p = 0.74

Random sample of freshman, n = 81

Utilizing central limit theorem,

[tex]P(\hat{p}<p) = P(Z<\hat{p} - \frac{p}{\sqrt{\frac{p(1-p)}{n} }  } )[/tex]

So,

[tex](P(\hat{p}>0.743) = P(Z>0.743 - \frac{0.74}{\sqrt{\frac{0.74(1-0.74)}{81} }  } )[/tex]

= P( Z > 0.0616)

= 0.4755 ⇒ probability that the proportion of students who graduated is greater than 0.743.

Answer:

y > -7

Step-by-step explanation:

Isolate the variable, y. Treat the > as a equal sign, what you do to one side, you do to the other. Subtract 7 from both sides:

y + 7 (-7) > 0 (-7)

y > 0 - 7

y > -7

y > -7 is your answer.

~

Answer:

[tex]\huge \boxed{y>-7}[/tex]

Step-by-step explanation:

Switch sides.

[tex]\displaystyle y+7>0[/tex]

Subtract by 7 from both sides of equation.

[tex]\displaystyle y+7-7>0-7[/tex]

Simplify, to find the answer.

[tex]\displaystyle 0-7=-7[/tex]

[tex]\huge \boxed{y>-7}[/tex], which is our answer.

Answer:  0.9612

Step-by-step explanation:

The binomial distribution formula is given by :-

[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of x successes out of n trials, p is the probability of success on a particular trial.

Given : The probability of a baby being a boy= 0.522.

Then the probability of a girl : [tex]p=1-0.522=0.478[/tex]

Number of trials  : n= 5

Now, the required probability will be :

[tex]P(x \geq1)=1-P(0)\\\\=1-[^{5}C_0(0.478)^{0}(1-0.478)^{5-0}]\\\\=1-[(1)(0.522)^{5}]=0.961242789206\approx0.9612[/tex]

Thus,  the probability that at least one of them is a girl​ = 0.9612

The probability that at least one of them is a girl​ is approximately 0.9565.

Probability of At Least One Girl in Five Births

To find the probability that at least one of the next five randomly selected births in the country will be a girl, we can use the complement rule. The complement rule states that the probability of at least one girl is equal to 1 minus the probability of no girls (i.e., all boys).

Given:

→ Probability of a boy (P(B)) = 0.522

→ Probability of a girl (P(G)) = 1 - P(B)

                                            = 1 - 0.522

                                            = 0.478

Step-by-step calculation:

1. Calculate the probability of all five births being boys: (P(B))⁵

2. Substitute the values: (0.522)⁵

3. Calculate: (0.522)⁵ ≈ 0.0435

Therefore, the probability of all five births being boys is 0.0435.

Using the complement rule:

→ Probability of at least one girl = 1 - Probability of all boys

                                                    = 1 - 0.0435

                                                    ≈ 0.9565

Thus, the probability that at least one of the next five births is a girl is approximately 0.9565.

The values of 'a' and 'b' for which the line 4x + y = b is tangent to the parabola[tex]y = ax^2[/tex] at x = 4 are a = -1/2 and b = 8.

To find the values of a and b for which the line 4x + y = b is tangent to the parabola [tex]y = ax^2[/tex] when x = 4, we need to ensure that the line and the parabola have the same slope at the point of tangency.

Convert the equation to the slope-intercept form (y = mx + c), where m is the slope and c is the y-intercept.

4x + y = b

y = -4x + b

The slope (m) of the line is -4.

Now find the slope of the parabola y = ax^2 when x = 4:

To do this, find the derivative of the parabola with respect to x and then evaluate it at x = 4.

[tex]y = ax^2[/tex]

dy/dx = 2ax (derivative of ax^2 with respect to x)

Now, evaluate the derivative at x = 4:

dy/dx = 2a(4)

= 8a

The slope (m) of the parabola [tex]y = ax^2[/tex] when x = 4 is 8a.

Equate the slopes of the line and the parabola at x = 4:

We want the slopes of both the line and the parabola to be equal at x = 4:

-4 = 8a

Now, solve for 'a':

a = -4/8

a = -1/2

To find the value of 'b' by substituting 'a' and the given point (x = 4, [tex]y = ax^2[/tex]) into the equation of the line:

y = -4x + b

y = -4(4) + b

y = -16 + b

Now, set [tex]ax^2 = -16 + b[/tex], and substitute the value of 'a' we found earlier:

[tex]\frac{-1}{2}(4)^2 = -16 + b[/tex]

[tex]\frac{-1}{2}(16) = -16 + b[/tex]

[tex]-8 = -16 + b[/tex]

[tex]b=8[/tex]

Hence, the values of 'a' and 'b' for which the line 4x + y = b is tangent to the parabola[tex]y = ax^2[/tex] at x = 4 are a = -1/2 and b = 8.

To learn more on Parabola click here:

https://brainly.com/question/11911877

#SPJ4

Final answer:

The values of a and b are calculated to be 1/2 and 24, respectively, by solving simultaneous equations based on these conditions.

Explanation:

The question asks for the values of a and b where the line 4x + y = b is tangent to the parabola y = ax2 at x = 4.

To find these values, we need to satisfy two conditions: the line and the parabola must intersect at x = 4, and their slopes at this point must be equal, as this characterizes tangency.

First, solve the equation of the parabola for x = 4: y = a(4)2 = 16a.

Substituting x = 4 into the equation of the line gives 4(4) + y = b, or 16 + y = b.

Since the line and the parabola intersect at this point, their y-values must be equal, hence 16 + 16a = b.

To find the values of a, we must equate the slopes of the tangent line and the parabola at x = 4.

The slope of the line is the coefficient of x, which is 4. The slope of the parabola at any point x is derived by differentiating y = ax2, giving dy/dx = 2ax.

At x = 4, the slope is 2a(4) = 8a, which must equal the slope of the line, 4. Therefore, 8a = 4, and a = 1/2.

Substituting a = 1/2 into 16 + 16a = b gives b = 16 + 8 = 24.

Therefore, the values of a and b for which the line is tangent to the parabola at x = 4 are a = 1/2 and b = 24.

Parameterize [tex]S[/tex] by

[tex]\vec r(u,v)=\dfrac23u\cos v\,\vec\imath+\dfrac23u\sin v\,\vec\jmath+u\,\vec k[/tex]

with [tex]4\le u\le5[/tex] and [tex]0\le v\le2\pi[/tex]. Take the normal vector to [tex]S[/tex] to be

[tex]\vec r_u\times\vec r_v=-\dfrac23u\cos v\,\vec\imath-\dfrac23u\sin v\,\vec\jmath+\dfrac49u\,\vec k[/tex]

(orientation doesn't matter here)

Then the area of [tex]S[/tex] is

[tex]\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle\frac{2\sqrt{13}}9\int_0^{2\pi}\int_4^5u\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle\frac{4\sqrt{13}\,\pi}9\int_4^5u\,\mathrm du=\boxed{2\sqrt{13}\,\pi}[/tex]

Answer: 77.45 %

Step-by-step explanation:

We assume that the measurements are normally distributed.

Given : Mean : [tex]\mu=39[/tex]

Standard deviation : [tex]\sigma=4.0[/tex]

Let x be the randomly selected measurement.

Now we calculate z score for the normal distribution as :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x = 35, we have

[tex]z=\dfrac{35-39}{4}=-1[/tex]

For x = 45, we have

[tex]z=\dfrac{45-39}{4}=1.5[/tex]

Now, the p-value = [tex]P(35<x<45)=P(-1<z<1.5)[/tex]

[tex]=P(z<1.5)-P(z<-1)\\\\=0.9331927-0.1586553=0.7745374\approx0.7745[/tex]

In percent , [tex]0.7745\times100=77.45\%[/tex]

Hence, the percent of measurements should fall between 35 and 45 = 77.45 %

Final answer:

Approximately 68% of the hardness measurements on the Rockwell C scale should fall between 35 and 45, according to the empirical rule and the provided mean and standard deviation.

Explanation:

To determine the percentage of hardness measurements that fall between 35 and 45 on the Rockwell C scale, we utilize the concept of standard deviation and the empirical rule.

Given the average (or mean) Rockwell C value is 39 and the standard deviation is 4.0, the range from 35 to 45 encompasses from one standard deviation below to one standard deviation above the mean.

According to the empirical (68-95-99.7) rule, about 68% of data within a normal distribution falls within one standard deviation of the mean.

Since the range from 35 to 45 represents this bracket around the mean, we can conclude that approximately 68% of the hardness measurements should fall within this range, assuming a normal distribution of the measurements.

Answer:  

Sample Response: Determine what should go on each axis. Grades will go on the x-axis, and the number of students will be plotted on the y-axis. Find the number of different scores and range. Choose an interval – say, 10 percent. Find the frequency within each interval. Choose a scale – say, 0–3. Draw bars the height of the frequency for each interval.

Step-by-step explanation:

(Sample Response)

Creating a histogram is a way to visually represent the distribution of a dataset. Here's how you can create a histogram for today’s math quiz scores step by step:
### Step 1: Collect the Data
First, gather all the scores you want to include in the histogram. In this case, the scores are 75, 95, 60, 75, 95, and 80.
### Step 2: Determine the Number of Bins
Choose an appropriate number of bins (groups or class intervals) for your histogram. The number of bins can significantly affect the appearance of the histogram. There are different rules of thumb for this, such as using the square root of the number of data points. However, you often have to decide based on the range of your data and the level of granularity you want.
### Step 3: Determine the Bin Intervals
The bins are ranges of scores that your data will be split into. You need to ensure the bins collectively cover the entire range of your data. To do this, find the minimum and maximum scores. Then, decide how wide each bin should be. This can be done either by dividing the entire range by the number of bins or by choosing a bin width that makes sense practically.
### Step 4: Tally Scores in Each Bin
Go through your data and count how many scores fall into each bin range. For instance, if your first bin includes scores from 60 to 70, count all the scores that are within this range.
### Step 5: Draw the Histogram
On graph paper or using graphing software:
- Draw a horizontal axis (the x-axis) and a vertical axis (the y-axis).
- Label the horizontal axis with the score ranges for each bin.
- Label the vertical axis with the frequency (the number of scores in each bin).
- For each bin, draw a bar that reaches up to the frequency of scores in that bin. The width of each bar should correspond to the bin width, and there should be no space between bars if the data is continuous.
### Step 6: Label Your Histogram
Finally, provide a title for your histogram and label the axes to make it clear what is being represented.
Keep in mind that histograms should give a clear picture of how the data is distributed. If you have bins that are too large, you'll lose detail. If they're too small, the histogram may be too choppy to identify any trends. Adjust the number of bins and bin width accordingly if your first histogram doesn’t seem to represent your data clearly.

Answer:

a) Velocity of the galaxy=573.756 km/s

b) Distance to the galaxy=716 Mpc

Step-by-step explanation:

a) [tex]v=H_0 D\\Where\ v=Velocity\ of\ the\ galaxy\\H_0= Hubble\ constant=69.8\pm 1.9\ as\ of\ 16\ July\ 2019\\D= Proper\ distance=8.22\ Mpc\\\Rightarrow v=69.8\times 8.22\\\therefore v=573.756\ km/s\\[/tex]

b)[tex]v=H_0 D\\v=50000\ km/s\\H_0= Hubble\ constant=69.8\pm 1.9\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{50000}{69.8}\\\therefore D=716\ Mpc[/tex]

Answer: 1.8

Step-by-step explanation:

a. Calculate the probability that at most two accidents occur in any given week.

Probability of 0 accidents + Probability of 1 accident + Probability of 2 accidents = 0.20 + 0.30 + 0.20 = 0.70.

b. What is the probability that there are at least two weeks between any two accidents?

Probability of no accidents + Probability of 1 accident = 0.20 + 0.30 = 0.50.

Answer:

1.99 pounds per gallon of salt in t=2 in the tank.

Step-by-step explanation:

First we consider the matter balance equation that contemplates the input and output; the generation and consumption equal to the acomulation.  

[tex]  Acomulation = Input - Output + Generation - Consumption [/tex]

In this case we have no Generation neither do Consumption so, if we consider Acomulation = A(t), the rate of change of A(t) in time is given by:

[tex]\frac{dA}{dt}+R_{out}A(t)= CR_{s}[/tex] ---(1)

where C is th concentration, with the initial value statement that A(t=0) = 0 because there is no salt in the time cero in the tank, only water.

Given the integral factor -> [tex]u(t)= exp[R_{out}] [/tex] and multipying the entire (1) by it, we have:

[tex]\int \frac{d}{dt}[A(t) \exp[R_{out}t]] \, dt = CR_{in} \int \exp[R_{out}t] \, dt[/tex]

Solving this integrals we obtain:

[tex]A(t)=\frac{CR_{in}}{R_{out}}+Cte*\exp[-R_{out}t][/tex]

So given the initial value condition A(t=0)=0 we have:

[tex]Cte=- \frac{CR_{in}}{R_{out}}[/tex],

and the solution is,

[tex]A(t)=\frac{CR_{in}}{R_{out}}-\frac{CR_{in}}{R_{out}}\exp[-R_{out}t][/tex].

If we give the actual values we obtain then,

[tex]A(t)=2\frac{pounds}{gallon}-2\frac{pounds}{gallon} \exp[-3t][/tex].

So in t= 2 we have [tex]A(t)=2\frac{pounds}{gallon}[/tex].

Answer:

Yes! There are 3 different explanations I have. Pick your favorite.

Step-by-step explanation:

So we are asked to see if the following equation holds:

[tex]\frac{4 \text{ ft }}{6 \text{ ft}}=\frac{12 \text{ sec }}{18 \text{sec}}[/tex]

The units cancel out ft/ft=1 and sec/sec=1.

So we are really just trying to see if 4/6 is equal to 12/18

Or you could cross multiply and see if the products are the same on both sides:

[tex]\frac{4}{6}=\frac{12}{18}[/tex]

Cross multiply:

[tex]4(18)=6(12)[/tex]

[tex]72=72[/tex]

Since you have the same thing on both sides then the ratios given were proportional.

OR!

Put 4/6 and 12/18 in your calculator.  They both come out to have the same decimal expansion of .66666666666666666(repeating) so the ratios gives are proportional.

OR!

Reduce 12/18 and reduce 4/6 and see if the reduced fractions are same.

12/18=2/3  (I divided top and bottom by 6)

4/6=2/3  (I divided top and bottom by 2)

They are equal to the same reduced fraction so they are proportional.

Answer: The mean number of checks written per​ day = 0.3260

The standard deviation = 0.5710

The variance = 0.3260

Step-by-step explanation:

Given : The number of checks written by the person = 119

We assume that the year is not a leap year.

Thus, the number of days in the year must be 365.

Now, the mean number of checks written per​ day is given by :-

[tex]\lambda=\dfrac{119}{365}=0.3260273972\approx0.3260[/tex]

Also, in Poisson distribution , the variance is also equals to the mean value .

[tex]\text{Thus , Variance }=\sigma^2= 0.3260[/tex]

Then , [tex]\sigma= \sqrt{0.3260}=0.570964096945\approx0.5710[/tex]

Thus,  Standard deviation = 0.5710

Answer:

discount= $94.5

Proceeds=$2605.5

Step-by-step explanation:

amount Dawin has to pay= $2700

in 6 months time

at a discount rate of 7%

here future value = $2700

D= discount before m months is given by

[tex]D= \frac{FV\times r\times m}{1200}[/tex]

m= 6 months, r= 7% and FV= 2700

putting values

[tex]D= \frac{2700\times 7\times 6}{1200}[/tex]

solving we get

D= $94.5

now present value = future value - discount

=2700-94.5= $ 2605.5

here PV(present valve )= Proceeds= $2605.5

Answer: 91

Step-by-step explanation:

Given : The number of documentaries = 5

The number of comedies = 7

The number of mysteries = 4

The number of horror films =5

The total number of movies other than comedy = 14

Now, the number of possible combinations of 9 movies can he rent if he wants all 7 comedies is given by :-

[tex]^7C_7\times^{14}C_2\\\\\dfrac{7!}{7!(7-7)!}\times\dfrac{14!}{2!(14-2)!}\\\\=(1)\times\dfrac{14\times13}{2}\\\\=91[/tex]

Therefore, the number of possible combinations of 9 movies can he rent if he wants all 7 comedies is 91 .

Answer:

Step-by-step explanation:

We need to graph the functions

p(x) = x^2

q(x) = x^2 - 4

r(x) = x^2 +1

the table used to draw the graph is:

x    p(x)     q(x)      r(x)

-2    4         0         5

-1     1          -3        2      

0     0         -4        1  

1       1          -3       2

2      4          0        5

The graph is attached below.

Answer:

Option C (f(x) = [tex]-x^2 + 2x + 4[/tex])

Step-by-step explanation:

In this question, the first step is to write the general form of the quadratic equation, which is f(x) = [tex]ax^2 + bx + c[/tex], where a, b, and c are the arbitrary constants. There are certain characteristics of the values of a, b, and c which determine the nature of the function. If a is a positive coefficient (i.e. if a>0), then the quadratic function is a minimizing function. On the other hand, a is negative (i.e. if a<0), then the quadratic function is a maximizing function. Since the latter condition is required, therefore, the first option and the last option are incorrect. The features of the values of b are irrelevant in this question, so that will not be discussed here. The value of c is actually the y-intercept of the quadratic equation. Since the y-intercept is 4, the correct choice for this question will be Option C. In short, Option C fulfills both the criteria of the function which has a maximum and a y-intercept of 4!!!

Answer:

has only one solution: [tex]x=8, y=3[/tex]

Step-by-step explanation:

[tex]\left \{ {{\frac{5x}{4} -\frac{2y}{3}=8(equation 1)} \atop {\frac{x}{4} +\frac{5y}{3} =7(equation 2)}} \right\\[/tex]

substract  [tex]\frac{1}{5}[/tex]×(equation 1) from equation2

[tex]\left \{ {{\frac{5x}{4} -\frac{2y}{3}=8(equation 1)} \atop {0x +\frac{9y}{5} =\frac{27}{5}(equation 2)}} \right\\[/tex]

from equation 2 we obtain

[tex]y=\frac{27}{9} =3[/tex]

and we replace 3 in equation 1

[tex]\frac{5x}{4} -\frac{2(3)}{3} =8\\\\\frac{5x}{4} =10\\\\x=\frac{40}{5} =8[/tex]

Answer:

Lenny's tax payment for the year will be $7,752

(The third quarter payment would be $1,938)

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Problems.

Taxes are calculated once a year based on your yearly income. That being said we can look at Lenny's income for the year ($40,800) and multiply that by 0.19 (19%) to calculate his tax payment  for the year.

[tex]$40,800 * 0.19 = $7,752[/tex]

So Lenny's tax payment for the year will be $7,752 . The question asks for payment for the third quarter but as mentioned above tax is calculated yearly.

That being said if you would like to know how much it would be quarterly we can simply divide the tax payment by 4.

[tex]7,752 / 4 = 1,938[/tex]

So the third quarter payment would be $1,938

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Answer:  0.317

Step-by-step explanation:

Let A be the event that the selected household is prosperous and B the event that it is educated.

Given : P(A)=0.138,   P(B)=0.261

P(A and B)=0.082

We know that for any events M and N ,

[tex]\text{P(M or N)=P(M)+P(N)-P(M or N)}[/tex]

Thus , [tex]\text{P(A or B)=P(A)+P(B)-P(A or B)}[/tex]

[tex]\text{P(A or B)}=0.138+0.261-0.082\\\\\Rightarrow\text{ P(A or B)}=0.317[/tex]

Hence, the probability P(A or B) that the household selected is either prosperous or educated = 0.317

Final answer:

The probability that a randomly selected American household is either prosperous or educated is 0.317.

Explanation:

We are asked to find the probability P(A or B) that a randomly selected American household is either prosperous (income over $100,000) or educated (householder completed college). To calculate the probability of the union of two events A and B, we use the formula:

P(A or B) = P(A) + P(B) - P(A and B).

According to the given data:

P(A) is the probability that a household is prosperous, which is 0.138.

P(B) is the probability that a household is educated, which is 0.261.

P(A and B) is the probability that a household is both prosperous and educated, which is 0.082.

Now we apply the given probabilities to the formula:

P(A or B) = 0.138 + 0.261 - 0.082 = 0.317.

Therefore, the probability that a selected household is either prosperous or educated is 0.317.

Answer:

The list of bijections from A into B are shown below.

Step-by-step explanation:

A function f is called one-to-one or injective, if and only if

[tex]f(x)=f(y)\Rightarrow x = y[/tex]

for all x and y in the domain of f.

A function f from X to Y is called onto or surjective, if and only if

for every element y∈Y there is an element x∈X with f(x)=y.

If a function is one-one and onto, then it is called bijective.

Part (a):

A={q,r,s} and B={2,3,4}

We need to find all the bijections from A into B.

(1) [tex]A\rightarrow B=\{(q,2),(r,3),(s,4)\}[/tex]

(2) [tex]A\rightarrow B=\{(q,2),(r,4),(s,3)\}[/tex]

(3) [tex]A\rightarrow B=\{(q,3),(r,2),(s,4)\}[/tex]

(4) [tex]A\rightarrow B=\{(q,3),(r,4),(s,2)\}[/tex]

(5) [tex]A\rightarrow B=\{(q,4),(r,2),(s,3)\}[/tex]

(6) [tex]A\rightarrow B=\{(q,4),(r,3),(s,2)\}[/tex]

Part (b):

A={1,2,3,4} and B={5,6,7,8}

We need to find all the bijections from A into B.

(1) [tex]A\rightarrow B=\{(1,5),(2,6),(3,7),(4,8)\}[/tex]

(2) [tex]A\rightarrow B=\{(1,5),(2,6),(3,8),(4,7)\}[/tex]

(3) [tex]A\rightarrow B=\{(1,5),(2,7),(3,6),(4,8)\}[/tex]

(4) [tex]A\rightarrow B=\{(1,5),(2,7),(3,8),(4,6)\}[/tex]

(5) [tex]A\rightarrow B=\{(1,5),(2,8),(3,6),(4,7)\}[/tex]

(6) [tex]A\rightarrow B=\{(1,5),(2,8),(3,7),(4,6)\}[/tex]

(7) [tex]A\rightarrow B=\{(1,6),(2,5),(3,7),(4,8)\}[/tex]

(8) [tex]A\rightarrow B=\{(1,6),(2,5),(3,8),(4,7)\}[/tex]

(9) [tex]A\rightarrow B=\{(1,6),(2,7),(3,5),(4,8)\}[/tex]

(10) [tex]A\rightarrow B=\{(1,6),(2,7),(3,8),(4,5)\}[/tex]

(11) [tex]A\rightarrow B=\{(1,6),(2,8),(3,5),(4,7)\}[/tex]

(12) [tex]A\rightarrow B=\{(1,6),(2,8),(3,7),(4,5)\}[/tex]

(13) [tex]A\rightarrow B=\{(1,7),(2,6),(3,5),(4,8)\}[/tex]

(14) [tex]A\rightarrow B=\{(1,7),(2,6),(3,8),(4,5)\}[/tex]

(15) [tex]A\rightarrow B=\{(1,7),(2,5),(3,6),(4,8)\}[/tex]

(16) [tex]A\rightarrow B=\{(1,7),(2,5),(3,8),(4,6)\}[/tex]

(17) [tex]A\rightarrow B=\{(1,7),(2,8),(3,6),(4,5)\}[/tex]

(18) [tex]A\rightarrow B=\{(1,7),(2,8),(3,5),(4,6)\}[/tex]

(19) [tex]A\rightarrow B=\{(1,8),(2,6),(3,7),(4,5)\}[/tex]

(20) [tex]A\rightarrow B=\{(1,8),(2,6),(3,5),(4,7)\}[/tex]

(21) [tex]A\rightarrow B=\{(1,8),(2,7),(3,6),(4,5)\}[/tex]

(22) [tex]A\rightarrow B=\{(1,8),(2,7),(3,5),(4,6)\}[/tex]

(23) [tex]A\rightarrow B=\{(1,8),(2,5),(3,6),(4,7)\}[/tex]

(24) [tex]A\rightarrow B=\{(1,8),(2,5),(3,7),(4,6)\}[/tex]

Answer: (0.1828,0.4572)

Step-by-step explanation:

Given : A medical researcher wishes to estimate what proportion of babies born at a particular hospital are born by Caesarean section.

Sample size : n= 49

Proportion of babies were Caesarean sections : [tex]\hat{p}=0.32[/tex]

Significance level : [tex]\alpha =1-0.95=0.05[/tex]

Standard error : [tex]S.E.=\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

[tex]=\sqrt{\dfrac{0.32\times0.68}{49}}=0.06663945022\approx0.07[/tex]

Margin of error : [tex]E=z_{\alpha/2}\times S.E.[/tex]

[tex]=z_{0.025}\times0.07=1.96\times0.07=0.1372[/tex]

The confidence interval for the population proportion is given by :-

[tex]\hat{p}\pm E[/tex]

[tex]=0.32\pm0.1372=(0.1828,0.4572)[/tex]

The question pertains to calculating the 95% confidence interval for the proportion of babies born by Caesarean section at a specific hospital, given a sample size of 49 births and a sample proportion of 32%. The answer shows how to use the confidence interval formula for a proportion (p ± Z * √((p*(1-p))/n)) and provides a calculation.

The subject of this question involves using statistics to calculate a confidence interval for a population proportion. In this context, the population proportion is the proportion of babies born by Caesarean section at a particular hospital.

We are provided with a random sample of 49 births, and within this sample, 32% were Caesarean sections. To answer this question about confidence intervals, we need to use a formula for the confidence interval of a proportion. The formula is p ± Z * √((p*(1-p))/n), where p is the proportion in the sample (0.32 in this case), Z is the Z-score from the Z-table associated with the desired confidence level (Z=1.96 for 95% confidence), and n is the number of observations (49).

Plugging the values in the formula, we get 0.32 ± 1.96 * √((0.32*(1-0.32))/49). When you calculate that, you will get a 95% confidence interval for the population proportion of Caesarean sections at this hospital.

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Answer:

[tex]\cos(arcsin(\frac{1}{4}))=\frac{\sqrt{15}}{4}[/tex].

Step-by-step explanation:

We want to evaluate cos(arcsin(1/4)) probably without a calculator.

If you did want a calculator answer, that would be 0.968245837.

Alright so anyways, this is the way I begin these trig(arctrig( )) types of problems when the trig parts are different.

Let u=arcsin(1/4).

If u=arcsin(1/4) then sin(u)=1/4.

So we want to find cos(u) given sin(u)=1/4.  (I just replace arcsin(1/4) in cos(arcsin(1/4)) with u.)

Let's use a Pythagorean Identity:

[tex]\cos^2(u)+\sin^2(u)=1[/tex].

Let's plug in 1/4 for sin(u):

[tex]\cos^2(u)+(\frac{1}{4})^2=1[/tex]

Simplify a bit:

[tex]\cos^2(u)+\frac{1}{16}=1[/tex]

Subtract 1/16 on both sides:

[tex]\cos^2(u)=1-\frac{1}{16}[/tex]

Simplify the right hand side:

[tex]\cos^2(u)=\frac{15}{16}[/tex]

Take the square root of both sides:

[tex]\cos(u)=\pm \sqrt{\frac{15}{16}}[/tex]

Separate the square thing to the numerator and denominator:

[tex]\cos(u)=\pm \frac{\sqrt{15}}{\sqrt{16}}[/tex]

Replace [tex]\sqrt{16}[/tex] with 4 since [tex]4^2=16[/tex]:

[tex]\cos(u)=\pm \frac{\sqrt{15}}{4}[/tex]

Now how do we determine if the cosine should be positive or negative.

arcsin(1/4) is an angle that is going to be between -pi/2 and pi/2 due to restrictions upon the sine curve to be one to one.

cosine of an angle between -pi/2 and pi/2 is going to be positive because these are the 1st and 4th quadrant where the x-coordinate is positive (the cosine value is positive)

[tex]\cos(u)=\frac{\sqrt{15}}{4}[/tex]

So recall u=arcsin(1/4):

[tex]\cos(arcsin(\frac{1}{4}))=\frac{\sqrt{15}}{4}[/tex].

For fun, put [tex]\frac{\sqrt{15}}{4}[/tex].  If you don't get  0.968245837 then you made a mistake in the above reasoning. We do get that so the results of the calculator and our trigonometry/algebra confirm each other.

To find maximum and minimum values using Lagrange multipliers, set up a Lagrangian function, apply the technique of setting partial derivatives to zero, and solve the resulting system of equations. This technique is applied to both given functions to solve for the variables and the multiplier.

The subject of this question relates to the application of Lagrange multipliers in mathematics, specifically to find the maximum and minimum values of functions with constraints.

Starting with the first function, we set up the Lagrangian function L = f(x, y) - λ(g(x, y) - c) where λ is the Lagrange multiplier, and g(x, y) is the constraint. So, we get L = 81x^2 + y^2 - λ(4x^2 + y^2 -9). By setting the partial derivatives of L to zero and solving that system of equations, we can find x, y and λ.

For the second function, we apply the same process, the Lagrangian function will be L = y^2 - 10z - λ(x^2 + y^2 + z^2 - 36). Setting partial derivatives equal to zero and solving the resulting system will provide x, y, z, and λ.

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The maximum for the first problem is 182.25 and the minimum is 9, while the maximum for the second problem is 61.

To solve these problems using Lagrange multipliers, we need to find the maximum and minimum values of the given functions subject to the provided constraints.

(i) Function f(x,y) = 81x^2 + y² with Constraint 4x² + y² = 9

→ Define the Lagrangian function: L(x, y, λ) = 81x² + y² + λ(9 - 4x² - y²).

Compute the partial derivatives and set them to zero:

→ [tex]L_x = 162x - 8x\lambda = 0[/tex]

→ [tex]L_y = 2y - 2y\lambda = 0[/tex]

→ [tex]L\lambda = 9 - 4x^2 - y^2 = 0[/tex]

Solve the system of equations:

→ 162x = 8xλ

⇒ λ = 20.25

→ 2y = 2yλ

⇒ λ = 1 (y ≠ 0)

But we need a consistent value of λ; thus, solve considering different cases (y = 0):

→ If y = 0, constraint becomes 4x² = 9

⇒ x = ±√(9/4)

      = ±3/2

→ If x = 0, constraint becomes y² = 9

⇒ y = ±3

Evaluate f(x, y) at these points:

→ f(3/2, 0) = 81 (3/2)²

                = 81 * 2.25

                = 182.25 (Maximum)

→ f(0, 3) = 3²

            = 9

(ii) Function f(x, y, z) = y² - 10z with Constraint x² + y² + z² = 36

Define the Lagrangian function: L(x, y, z, λ) = y² - 10z + λ(36 - x² - y² - z²).

Compute the partial derivatives and set them to zero:

→ [tex]L_x = -2x\lambda = 0[/tex]

⇒ x = 0

→ [tex]L_y = 2y - 2y\lambda = 0[/tex]

⇒ y = 0 or λ = 1

→ [tex]L_z = -10 - 2z\lambda = 0[/tex]

⇒ λ = -5/z

→ [tex]L\lambda = 36 - x^2 - y^2 - z^2 = 0[/tex]

Solve the system of equations considering constraints:

Substituting values, x = 0 and λ = 1, we need to solve for y and z:

→ 36 = y² + z²

→ λ = -5/z;

Thus z = -5

Solving y² + 25 = 36, y² = 11

⇒ y = ±√11

Evaluate f(x, y, z) at these points:

→ f(0, √11, -5) = (√11)² - 10(-5)

                     = 11 + 50              

                     = 61 (Maximum)

→ f(0, -√11, -5) = 61

Answer:

Find the slope of the line that passes through the points shown in the table.

The slope of the line that passes through the points in the table is

.

Step-by-step explanation:

Answer:

The required cost function is [tex]C(x)=1036x+410[/tex].

Step-by-step explanation:

It is given that the cost function represents a linear relationship.

The fixed cost is $410 and the cost of 5 items is $5,590. It means the linear function passes through the points (0,410) and (5,5590).

If a line passes through two points then the equation of line is

[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

The equation of cost function is

[tex]y-410=\frac{5590-410}{5-0}(x-0)[/tex]

[tex]y-410=\frac{5180}{5}(x)[/tex]

[tex]y-410=36x[/tex]

[tex]y-410=1036x[/tex]

Add 410 on both the sides.

[tex]y=1036x+410[/tex]

The required cost function is

[tex]C(x)=1036x+410[/tex]

Therefore the required cost function is [tex]C(x)=1036x+410[/tex].

The required cost function is [tex]\rm C(x)= 1,036x +410[/tex].

Given

The relationship is linear.

Fixed cost, $410; 5 items cost $5,590 to produce.

An equation between two variables that gives a straight line when plotted on a graph.

The standard form represents the linear equation;

[tex]\rm y=mx+c[/tex]

The fixed cost is $410 and the cost of the 5 items is $5,590. It means the linear function passes through the points (0,410) and (5,5590).

If a line passes through two points then the equation of a line is;

[tex]\rm y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\y-410=\dfrac{5590-410}5-0}(x-0)\\\\y-410=\dfrac{5180}{5}x\\\\y-410=1036x\\\\y=1036x+410\\\\C(x)=1036x+410[/tex]

Hence, the required cost function is [tex]\rm C(x)= 1,036x +410[/tex].

To know more about the Linear equations click the link given below.

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