Find the length of the second base of a trapezoid with one base measuring 12 feet,a height of 4,and an area of 58 square feet.

Answer:

B. 550

Step-by-step explanation:

550 is the smallest number that becomes 600 when rounded to the nearest hundred

the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.

The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.

In terms of hyperbola, F1F2=2c, c=20.

The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.

Use formula c^2=a^2+b^2c

2

=a

2

+b

2

to find b:

\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}

(20)

2

=(18)

2

+b

2

,

b

2

=400−324=76

.

The branches of hyperbola go in y-direction, so the equation of hyperbola is

\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1

b

2

y

2

−

a

2

x

2

=1 .

Substitute a and b:

\dfrac{y^2}{76}- \dfrac{x^2}{324}=1

76

y

2

−

324

x

2

=1 .

Answer:

[tex]\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}[/tex]

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:                                                                     [tex]\displaystyle \lim_{x \to c} x = c[/tex]

L'Hopital's Rule

Differentiation

Basic Power Rule:

Derivative Rule [Chain Rule]:                                                                                    [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Step-by-step explanation:

We are given the limit:

[tex]\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}[/tex]

When we directly plug in x = 0, we see that we would have an indeterminate form:

[tex]\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}[/tex]

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

[tex]\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}[/tex]

Plugging in x = 0 again, we would get:

[tex]\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}[/tex]

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

[tex]\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}[/tex]

Substitute in x = 0 once more:

[tex]\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}[/tex]

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

Answer:

29.1 Is the answer for 2009

Step-by-step explanation:

Answer:

The minimum number of drivers you would need to survey is 601.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].  

What is the minimum number of drivers you would need to survey to be 95% confident that the population proportion is estimated to within 0.04?

The number is n for which M = 0.04.

We don't have an estimate for the proportion, so we use [tex]\pi = 0.5[/tex]. Then

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]

[tex]0.04\sqrt{n} = 1.96*0.5[/tex]

[tex]\sqrt{n} = \frac{1.96*0.5}{0.04}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2[/tex]

[tex]n = 600.25[/tex]

Rounding up:

The minimum number of drivers you would need to survey is 601.

Answer:

B. Yes. By SSS~

Step-by-step explanation:

From the diagram given, we have the corresponding sides of both triangles as follows:

RQ/KL = 24/20 = 6/5

QP/LM = 18/15 = 6/5

RP/KM = 12/10 = 6/5

From the above, we can see that the ratio of the corresponding side lengths of both triangles are equal. This means that all three sides of one triangle are proportional to all corresponding sides of the other triangle.

The SSS similarity theorem states that if all sides of one triangle are proportional to all corresponding sides of another, then both triangles are similar to each other.

Therefore, ∆KLM ~ ∆RQP by SSS similarity.

Answer:

Explanation:

The first figure has 2 blocks.

The second figure has 4 blocks.

The third figure has 8 blocks

The pattern 2,4,8,... follows the rule "double the current value to get the next one". Because we have this going on, we have an exponential pattern here.

A linear pattern would be something like 2,4,6,8,10,... showing that we add on 2 each time, rather than multiply by 2 each time. So as you can guess, or already know, exponential patterns grow much quicker compared to linear ones.

36 different value meal packages are possible

Step-by-step explanation:

To answer this question, multiply all given numbers together.

4*3*3

12*3

36

Answer:

5π/4 radians or 225°

Step-by-step explanation:

Subtract the difference form the total:

717 - 67 = 650

Divide the remaining amount by 2:

650/2 = 375

The bench cost$375

Answer:

A, B, and E.

Step-by-step explanation:

A. 5^x * 5^x

= 5^x+x

=5^(2)(x)

=25^x

B. 5^2x

=5^(2)(x)

=25^x

C. 5*5^2x

=5^1+2x

D. 5*5^x

=5^1+x

E. (5*5)^x

=5^x*5^x

=5^(2)(x)

=25^x

F. 5^2*5^x

=5^2+x

Answer:

a) The sample mean is 1260 and the standard deviation is 48.

b) The 90% confidence interval for the mean of all tree-ring dates from this archaeological site is (1230, 1290).

Step-by-step explanation:

Question a:

Mean is the sum of all values divided by the number of values. So

[tex]\overline{x} = \frac{1285 + 1187 + 1222 + 1194 + 1268 + 1316 + 1275 + 1317 + 1275}{9} = 1260[/tex]

Standard deviation is the square root of the sum of the differences squared between each value and the mean, divided by the one less than the sample size. So

[tex]s = \sqrt{\frac{(1285-1260)^2 + (1187-1260)^2 + (1222-1260)^2 + (1194-1260)^2 + (1268-1260)^2 + ...}{8}} = 48[/tex]

The sample mean is 1260 and the standard deviation is 48.

Question b:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So

df = 9 - 1 = 8

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.8595

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 1.8595\frac{48}{\sqrt{9}} = 30[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1260 - 30 = 1230

The upper end of the interval is the sample mean added to M. So it is 1260 + 30 = 1290

The 90% confidence interval for the mean of all tree-ring dates from this archaeological site is (1230, 1290).

Answer:

A) Exponential growth

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

I explained why 5 minutes ago on a different question

Answer:

I can say for sure that the answer is not segment GH ≅ segment FH because the tangents that create the segment FG share a common endpoint.  I believe the answer is segment GH ≅ segment FH because arc EF ≅ arc GF.

Step-by-step explanation:

Again, I'm not sure about the correct answer but I know for sure it isn't segment GH ≅ segment FH because the tangents that create the segment FG share a common endpoint.  

The segment GH and the segment FH are equal to each other because the line AH is coming from the center of the circle and is bisecting the line GF.

The circumcenter of a triangle can be constructed by drawing any two of the three perpendicular bisectors. For three non-collinear points, these two lines cannot be parallel, and the circumcenter is the point where they cross.

Any point on the bisector is equidistant from the two points that it bisects, from which it follows that this point, on both bisectors, is equidistant from all three triangle vertices

Hence the segment GH and the segment FH are equal to each other because the line AH is coming from the center of the circle and is bisecting the line GF.

To know more about a Circumscibed circle follow

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Given:

d = 2

f = 4

To find:

Value of  [tex]\frac{14(7)-d}{2f}[/tex]

Steps:

we need to substitute and then find the value,

[tex]= \frac{14(7)-2}{2(4)}\\ \\=\frac{98-2}{8} \\\\=\frac{96}{8}\\\\=12[/tex]

Therefore, the answer is option C) 12

Happy to help :)

If you need help, feel free to ask

Note: you may need to delete the comma

============================================================

Explanation:

The info "5 months" is never used to compute the mean. We could easily replace it with "6 months" or "7 months" or any stretch of time, and still get the same answer. So we'll ignore this value.

What we'll do is add up the 21 items given to us, and then divide by 21.

Because there are so many values, and it's easy to get lost, I'm going to add up across the rows

Those subtotals then add up to this

1,499,116+1,586,126+1,591,076+842,147 = 5,518,465

This is the same as adding up all 21 values.

Finally, we divide that sum over 21 as there are 21 values in this list

(5,518,465)/21 = 262,784.047619048

That value then rounds up to 262,785

If your teacher just wanted things to the nearest whole number (without rounding up), then the answer would be 262,784

Side note: using a spreadsheet program would probably be the most efficient/fastest method for this type of problem.

Answers:

===========================================

Explanation:

Side MO is twice as long as the midsegment XY. Note how XY and MO are parallel.

This makes

MO = 2*XY = 2*10 = 20

Side MO breaks into two equal halves MZ and ZO

Each of MZ and ZO are 20/2 = 10 units long.

Put another way: XY, MZ and ZO are all the same length (all 10 units long).

---------------

The diagram shows that segment NO is 18 units long, which cuts in half to 18/2 = 9. This is the length of NY, YO and XZ

Also, MN = 14 which cuts in half to 7. This means MX, XN and YZ are all 7 units each.

Answer:

the answer is D

Step-by-step explanation:

v=πr²h

divide both side by πh

r²=v/πh

square both sides

r=√v/πh

The actual labor cost is $600 over the targeted labor cost.

Given that,

Work done by each person per week = 40 hours

Required labor hours per week = 600 hours

No. of workers in the team = 13

To find,

Actual weekly labor cost = ?

Procedure:

Actual weekly labor cost = No. of workers * no. of hours performed by them

[tex]= 13 * 40[/tex]

[tex]= 520 hours[/tex]

Given that,

[tex]Regular rate = $ 15.00 per hour[/tex]

[tex]Overtime rate = $ 22.50 per hour[/tex]

Thus,

Actual labor cost = (regular hours worked * regular rate) + (overtime * overtime rate)

[tex]= (520 * 15) + ([600 - 520] * 22.50)[/tex]

[tex]= $ 7,800 + $ 1,800[/tex]

[tex]= $ 9600[/tex]

Targeted Labor cost = $ 9,000 per week

Thus, option C i.e. the actual labor cost is $ 600 over the targeted labor cost.

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Answer:

a) 22.4 pounds.

b) 0.17 pounds.

c) 0.0127 = 1.27% approximate probability the sample mean weight will be less than 22.02.

d) c = 22.62

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

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.

Average 22.4 pounds with a standard deviation of 1.36 pounds.

This means that [tex]\mu = 22.4, \sigma = 1.36[/tex]

Consider the sample mean weight of 64 watermelons of this variety.

This means that [tex]n = 64, s = \frac{1.36}{\sqrt{64}} = 0.17[/tex]

a. What is the expected value of the sample mean weight?

By the Central Limit Theorem, 22.4 pounds.

b. What is the standard deviation of the sample mean weight?

By the Central Limit Theorem, 0.17 pounds.

c. What is the approximate probability the sample mean weight will be less than 22.02?

This is the p-value of Z when X = 22.02. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{22.02 - 22.4}{0.17}[/tex]

[tex]Z = -2.235[/tex]

[tex]Z = -2.235[/tex] has a p-value of 0.0127.

0.0127 = 1.27% approximate probability the sample mean weight will be less than 22.02.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.9?

This is the 90th percentile, that is, [tex]X = c[/tex] when z has a p-value of 0.9, so X when Z = 1.28.

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]1.28 = \frac{c - 22.4}{0.17}[/tex]

[tex]c - 22.4 = 1.28*0.17[/tex]

[tex]c = 22.62[/tex]

The domain of a set is the possible input values the set can take.

It is true that the domain of ∀x ∃y(x+y≥0) is the set of real numbers

Given that: ∀x ∃y(x+y≥0)

Considering x+y ≥ 0, it means that the values of x + y are at least 0.

Make y the subject in x+y ≥ 0

So, we have:

[tex]\mathbf{y \le -x}[/tex]

There is no restriction as to the possible values of x.

This means that x can take any real number.

Hence, it is true that the domain of ∀x ∃y(x+y≥0) is the set of real numbers.

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Answer:

Step-by-step explanation:

8/5 = 16/10 = 24/15

8:5 = 16:10 = 24:15

Answer:

x = 6

z = 60

Step-by-step explanation:

Solve for x

(6x + 84) = 120

     -  84     -84

6x = 36

6x/6 = 36/6

x = 6

Then solve for z

120 + z = 180

-120        -120

z = 60

Answer:

0

Step-by-step explanation:

Answer:

Following are the solution to the given points:

Step-by-step explanation:

As a result, Poisson for each driver seems to be the number of accidents.

Let X be the random vector indicating accident frequency.

Let, [tex]\lambda=[/tex]Expected accident frequency

[tex]P(X=0) = e^{-\lambda}[/tex]

For class 1:

[tex]P(X=0) = \frac{1}{(1-0.2)} \int_{0.2}^{1} e^{-\lambda} d\lambda \\\\P(X=0) = \frac{1}{0.8} \times [-e^{-1}-(-e^{-0.2})] = 0.56356[/tex]

For class 2:

[tex]P(X=0) = \frac{1}{(2-0.4)} \int_{0.4}^{2} e^{-\lambda} d\lambda\\\\P(X=0) = \frac{1}{1.6} \times [-e^{-2}-(-e^{-0.4})] = 0.33437[/tex]

For class 3:

[tex]P(X=0) = \frac{1}{(3-0.6)} \int_{0.6}^{3} e^{-\lambda} d\lambda\\\\P(X=0) = \frac{1}{2.4} \times [-e^{-3}-(-e^{-0.6})] = 0.20793[/tex]

The population is equally divided into three classes of drivers.

Hence, the Probability

[tex]\to P(X=0) = \frac{1}{3} \times 0.56356+\frac{1}{3} \times 0.33437+\frac{1}{3} \times 0.20793=0.36862[/tex]

RHS

[tex]\\ \sf\longmapsto \frac{2 \tan( \theta) }{ \sin(2 \theta) } \\ \\ \sf\longmapsto \frac{ \frac{2 \sin( \theta) }{2 \cos( \theta) } }{ \sin(2 \theta) } \\ \\ \sf\longmapsto \frac{1}{ \cos {}^{2} ( \theta) } \\ \\ \sf\longmapsto {sec}^{2} \theta[/tex]