Using the diagram, which of the following choices represent alternate exterior angles

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.

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Answer:

Two tailed test

Test statistic = 3.20

Pvalue = 0.001

H1 : p ≠ 0.91

Step-by-step explanation:

Given :

Test of p=0.91 vs p≠0.91

The use if not equal to ≠ sign in the null means we have a tow tailed test, which means a difference exists in the proportion which could be lesser or greater than the stated population proportion.

The test statistic :

This is the Z value from the table given = 3.20

The Pvalue = 0.001

Since Pvalue < α ;Reject H0

Answer:

-14 is the answer for the second term (?)

Answer:

B. 550

Step-by-step explanation:

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

Answer:

A) Exponential growth

Step-by-step explanation:

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:

the answer is 89

Step-by-step explanation:

this is a hard one to salve but basically if you know and lern hout to do it it is not that hard

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]

Answer:

0.9378 = 93.78% probability that a randomly selected person has diabetes, given that his test is positive.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive test

Event B: Person has diabetes.

Probability of a positive test:

0.95 out of 0.1(person has diabetes).

0.007 out of 1 - 0.1 = 0.9(person does not has diabetes). So

[tex]P(A) = 0.95*0.1 + 0.007*0.9 = 0.1013[/tex]

Probability of a positive test and having diabetes:

0.95 out of 0.1. So

[tex]P(A \cap B) = 0.95*0.1 = 0.095[/tex]

What is the probability that a randomly selected person has diabetes, given that his test is positive?

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.095}{0.1013} = 0.9378[/tex]

0.9378 = 93.78% probability that a randomly selected person has diabetes, given that his test is positive.

Answer:

5π/4 radians or 225°

Step-by-step explanation:

Answer:

−4304

Step-by-step explanation:

1. The given determinant is :

[tex]\begin{vmatrix}7 &31 \\ 142& 14\end{vmatrix}[/tex]

We need to find its determinant . It can be solved as follows :

[tex]\begin{vmatrix}7 &31 \\ 142& 14\end{vmatrix}=7(14)-142(31)\\\\=-4304[/tex]

So, the value of determinant is equal to −4304.

Answer:

A= -4269

B= 1768

C= 647.36

Step-by-step explanation:

Answer:

The acceleration is 1.8 m/s^2.

Step-by-step explanation:

initial velocity, u = 30 mph = 13.4 m/s

Final velocity, v = 50 mph = 22.4 m/s

time, t = 5 s

Let the acceleration is a.

Use first equation of motion

v = u + at

22.4 = 13.4 + 5 a

a = 1.8 m/s^2

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]

Answer:

X-intercept = -3 and y-intercept = 6

Step-by-step explanation:

We can start off by isolating the y term. To do that, we must add 2y to both sides to get

[tex]4x=2y-12[/tex]

Now, we must add 12 to both sides and the y term will be all alone on the right side:

[tex]4x+12=2y[/tex]

Now, to have only y on the right side, we must divide by 2 to get:

[tex]y=2x+6[/tex]

In slope-intercept form, b is the y-intercept, and 'b' in this equation is 6. We have our y-intercept.

To find our x-intercept, y must be equal to zero. We can plug in that value for y and solve for x:

[tex]0=2x+6[/tex]

We can start off by subtracting 6 from both sides to get:

[tex]2x=-6[/tex]

We can then divide both sides to get [tex]x=-3[/tex] when y is equal to 0. Thus, we have our x-intercept.

Answer:

y-intercept= -6

x-intercept= 3

Step-by-step explanation:

First, rearrange the equation to be in y=mx+b.

4x-2y=12

4x-12=2y

(1/2)(4x-12)=y

y=2x-6

From here, we know that the 'b' in an equation in form y=mx+b is the y-intercept, which is -6.

To find the x intercept make y=0 and solve.

You can also solve without rearranging the equation and simply making x=0 and solving to find the y-intercept. and making y=0 and solving to find the x-intercept.

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:

46%

Step-by-step explanation:

Divde the smaller # by the bigger # to get the precentage

An average San Francisco customer uses what percent of electricity used by an average Houston customer?

In other words, San Francisco is what part of Houston?

---Just like, 7 is what part of 49? These are the same questions and would be solved in the same way

San Francisco / Houston

6753 / 14542

0.4644 = 46.44%

ANSWER: 46%

Hope this helps!

Answer:

D

Step-by-step explanation:

I explained why 5 minutes ago on a different question

Answer:

z(max) = 16

z(min) = -24

Step-by-step explanation:

2x - y = 12  multiply by 2

4x - 2y = 24  (1)

4x + 2y = 0  add equations

8x = 24

x = 3

4(3) + 2y = 0

y = -6

so (3, -6) is a common point on these two lines

z = 2(3) + 5(-6) = -24

4x - 2y = 24   (1)

x + 2y = 6   add equations

      5x = 30

        x = 6

6 + 2y = 6

         y = 0

so (6, 0) is a common point on these two lines

z = 2(6) + 5(0) = 12

4x + 2y = 0    multiply by -1

-4x - 2y = 0

  x + 2y = 6  add equations

       -3x = 6

         x = -2

-2 + 2y = 6

         y = 4

so (-2, 4) is a common point on these two lines

z = 2(-2) + 5(4) = 16

Answer:

Step-by-step explanation:

[tex]\begin{bmatrix}3 & 6 & 1\\ 2 & 4& 0\\ 0 & 6 & 2\end{bmatrix}\times\begin{bmatrix}2\\ 0\\ 1\end{bmatrix}[/tex]

Multiply the terms of the rows of the first matrix with the terms given in the column of the second matrix.

[tex]=\begin{bmatrix}(3\times 2+6\times 0+1\times 1)\\ (2\times 2+4\times 0+0\times1)\\ (0\times 2+6\times 0+2\times 1)\end{bmatrix}[/tex]

[tex]=\begin{bmatrix}7\\ 4\\ 2\end{bmatrix}[/tex]

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.

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

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]

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:

Step-by-step explanation:

8/5 = 16/10 = 24/15

8:5 = 16:10 = 24:15

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

Step-by-step explanation:

         1                                          2                                       3  

days hours minutes       days   hours   minutes     days    hours  minutes  

  7        8         20             6       24+8       20              6        31        60+20

  4       10         30

 -

_______________

         4

days     hours    minutes

  6           31           80

  4            10          30

-

____________________

  2             21          50

___________________

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

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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