Two con terminal angles 3pi/4 negative and positive answer in radians

Answer:

The amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.

Step-by-step explanation:

Let the random variable X represent the amount of money that the family has invested in different real estate properties.

The random variable X follows a Normal distribution with parameters μ = $225,000 and σ = $50,000.

It is provided that the family has invested in n = 10 different real estate properties.

Then the mean and standard deviation of amount of money that the family has invested in these 10 different real estate properties is:

[tex]\mu_{\bar x}=\mu=\$225,000\\\\\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{50000}{\sqrt{10}}=15811.39[/tex]

Now the lowest 80% of the amount invested can be represented as follows:

[tex]P(\bar X<\bar x)=0.80\\\\\Rightarrow P(Z<z)=0.80[/tex]

The value of z is 0.84.

*Use a z-table.

Compute the value of the mean amount invested as follows:

[tex]\bar x=\mu_{\bar x}+z\cdot \sigma_{\bar x}[/tex]

   [tex]=225000+(0.84\times 15811.39)\\\\=225000+13281.5676\\\\=238281.5676\\\\\approx 238281.57[/tex]

Thus, the amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.

Answer:

[tex]solution \\ 3(5x) + 8(2x) \\ = 3 \times 5x + 8 \times 2x \\ = 15x + 16x \\ = 31x[/tex]

hope this helps...

Good luck on your assignment...

The expression  [tex]3(5x) + 8(2x)[/tex] in simplest form is 31x.

To simplify the expression [tex]3(5x) + 8(2x)[/tex], we can apply the distributive property:

[tex]3(5x) + 8(2x)[/tex]

[tex]= 15x + 16x[/tex]

Combining like terms, we have:

[tex]15x + 16x = 31x[/tex]

Therefore, the expression [tex]3(5x) + 8(2x)[/tex] simplifies to [tex]31x.[/tex]

To learn more on Expressions click:

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Answer:

  5.03

Step-by-step explanation:

Answer:

5.03 = AC

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp/ adj

tan 40 = AC /6

6 tan 40 = AC

5.034597787 = AC

To the nearest hundredth

5.03 = AC

Answer:

[tex]7.98 \:m[/tex]

Step-by-step explanation:

Area of a triangle is base times height divided by 2.

[tex]A= \frac{bh}{2}[/tex]

[tex]69.6= \frac{b \times 17.45}{2}[/tex]

[tex]69.6 \times 2= b \times 17.45[/tex]

[tex]139.2=b \times 17.45[/tex]

[tex]\frac{17.45b}{17.45}=\frac{139.2}{17.45}[/tex]

[tex]b=\frac{2784}{349}[/tex]

[tex]b=7.97707[/tex]

The appropriate unit is meters.

Answer:

7.98 m

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Answer:

X is 3 units.

Step-by-step explanation:

Volume of prism is cross sectional area multiplied by length. So 1/2 ×2× x ×2 into 3x, which is equal to 6x^2. So, 6x^2=54. Therefore, x=3.

Answer:

x = 3.6

Step-by-step explanation:

By the Postulate of intersecting chords inside a circle.

[tex]x \times 5 = 3 \times 6 \\ 5x = 18 \\ x = \frac{18}{5} \\ x = 3.6 \\ [/tex]

Answer:

probability of the event E = 1/5

Step-by-step explanation:

We are given;

Sample space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},

Number of terms in sample S is;

n(S) = 10

We are given the event; E = {1, 2}

Thus, number of terms in event E is;

n(E) = 2

Now, Probability = favorable outcomes/total outcomes

Thus, the probability of the event E is;

P(E) = n(E)/n(S)

P(E) = 2/10

P(E) = 1/5

Answer:

(x+9)/(8x+4)

Step-by-step explanation:

Step-by-step explanation:

1. 35 + 65

2. y - 14

3. (6 x s) - 3

4. 10(x+8.5).. 6-p

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

An insurance company examines its pool of auto insurance customers and gathers the following information: (i) All customers insure at least one car. (ii) 70% of the customers insure more than one car. (iii) 20% of the customers insure a sports car. (iv) Of those customers who insure more than one car, 15% insure a sports car. Calculate the probability that a randomly selected customer insures exactly one car, and that car is not a sports car?

Answer:

P( X' ∩ Y' ) = 0.205

Step-by-step explanation:

Let X is the event that the customer insures more than one car.

Let X' is the event that the customer insures exactly one car.

Let Y is the event that customer insures a sport car.

Let Y' is the event that customer insures not a sport car.

From the given information we have

70% of customers insure more than one car.

P(X) = 0.70

20% of customers insure a sports car.

P(Y) = 0.20

Of those customers who insure more than one car, 15% insure a sports car.

P(Y | X) = 0.15

We want to find out the probability that a randomly selected customer insures exactly one car, and that car is not a sports car.

P( X' ∩ Y' ) = ?

Which can be found by

P( X' ∩ Y' ) = 1 - P( X ∪ Y )

From the rules of probability we know that,

P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y )    (Additive Law)

First, we have to find out P( X ∩ Y )

From the rules of probability we know that,

P( X ∩ Y ) = P(Y | X) × P(X)       (Multiplicative law)

P( X ∩ Y ) = 0.15 × 0.70

P( X ∩ Y ) = 0.105

So,

P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y )

P( X ∪ Y ) = 0.70 + 0.20 - 0.105

P( X ∪ Y ) = 0.795

Finally,

P( X' ∩ Y' ) = 1 - P( X ∪ Y )

P( X' ∩ Y' ) = 1 - 0.795

P( X' ∩ Y' ) = 0.205

Therefore, there is 0.205 probability that a randomly selected customer insures exactly one car, and that car is not a sports car.

Answer:

(a) The value of E (X) is 4/7.

    The value of V (X) is 3/98.

(b) The value of P (X ≤ 0.5) is 0.3438.

Step-by-step explanation:

The random variable X is defined as the proportion of surface area in a randomly selected quadrant that is covered by a certain plant.

The random variable X follows a standard beta distribution with parameters α = 4 and β = 3.

The probability density function of X is as follows:

[tex]f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)} ; \hspace{.3in}0 \le x \le 1;\ \alpha, \beta > 0[/tex]

Here, B (α, β) is:

[tex]B(\alpha,\beta)=\frac{(\alpha-1)!\cdot\ (\beta-1)!}{((\alpha+\beta)-1)!}[/tex]

            [tex]=\frac{(4-1)!\cdot\ (3-1)!}{((4+3)-1)!}\\\\=\frac{6\times 2}{720}\\\\=\frac{1}{60}[/tex]

So, the pdf of X is:

[tex]f(x) = \frac{x^{4-1}(1-x)^{3-1}}{1/60}=60\cdot\ [x^{3}(1-x)^{2}];\ 0\leq x\leq 1[/tex]

(a)

Compute the value of E (X) as follows:

[tex]E (X)=\frac{\alpha }{\alpha +\beta }[/tex]

         [tex]=\frac{4}{4+3}\\\\=\frac{4}{7}[/tex]

The value of E (X) is 4/7.

Compute the value of V (X) as follows:

[tex]V (X)=\frac{\alpha\ \cdot\ \beta}{(\alpha+\beta)^{2}\ \cdot\ (\alpha+\beta+1)}[/tex]

         [tex]=\frac{4\cdot\ 3}{(4+3)^{2}\cdot\ (4+3+1)}\\\\=\frac{12}{49\times 8}\\\\=\frac{3}{98}[/tex]

The value of V (X) is 3/98.

(b)

Compute the value of P (X ≤ 0.5) as follows:

[tex]P(X\leq 0.50) = \int\limits^{0.50}_{0}{60\cdot\ [x^{3}(1-x)^{2}]} \, dx[/tex]

                    [tex]=60\int\limits^{0.50}_{0}{[x^{3}(1+x^{2}-2x)]} \, dx \\\\=60\int\limits^{0.50}_{0}{[x^{3}+x^{5}-2x^{4}]} \, dx \\\\=60\times [\dfrac{x^4}{4}+\dfrac{x^6}{6}-\dfrac{2x^5}{5}]\limits^{0.50}_{0}\\\\=60\times [\dfrac{x^4\left(10x^2-24x+15\right)}{60}]\limits^{0.50}_{0}\\\\=[x^4\left(10x^2-24x+15\right)]\limits^{0.50}_{0}\\\\=0.34375\\\\\approx 0.3438[/tex]

Thus, the value of P (X ≤ 0.5) is 0.3438.

Answer:

Step-by-step explanation:

Given that;

the following procedure for accomplishing our task are:

1. Flip the coin.

2. Flip the coin again.

From here will know that the coin is first flipped twice

3. If both flips land on heads or both land on tails, it implies that we return to step 1 to start again. this makes the flip to be insignificant since both flips land on heads or both land on tails

But if the outcomes of the two flip are different i.e they did not land on both heads or both did not land on tails , then we will consider such an outcome.

Let the probability of head = p

so P(head) = p

the probability of tail be = (1 - p)

This kind of probability follows a conditional distribution and the probability  of getting heads is :

[tex]P( \{Tails, Heads\})|\{Tails, Heads,( Heads ,Tails)\})[/tex]

[tex]= \dfrac{P( \{Tails, Heads\}) \cap \{Tails, Heads,( Heads ,Tails)\})}{ {P( \{Tails, Heads,( Heads ,Tails)\}}}[/tex]

[tex]= \dfrac{P( \{Tails, Heads\}) }{ {P( \{Tails, Heads,( Heads ,Tails)\}}}[/tex]

[tex]= \dfrac{P( \{Tails, Heads\}) } { {P( Tails, Heads) +P( Heads ,Tails)}}[/tex]

[tex]=\dfrac{(1-p)*p}{(1-p)*p+p*(1-p)}[/tex]

[tex]=\dfrac{(1-p)*p}{2(1-p)*p}[/tex]

[tex]=\dfrac{1}{2}[/tex]

Thus; the probability of getting heads is [tex]\dfrac{1}{2}[/tex] which typically implies that the coin is fair

(b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

For a fair coin (0<p<1) , it's certain that both heads and tails at the end of the flip.

The procedure that is talked about in (b) illustrates that the procedure gives head if and only if the first flip comes out tail with probability 1 - p.

Likewise , the procedure gives tail if and and only if the first flip comes out head with probability of  p.

In essence, NO, procedure (b) does not give a fair coin flip outcome.

Answer:

38 units

Step-by-step explanation:

We can find the perimeter of the shaded figure be finding out the number of unit lengths we have along the boundary of the given figure.

Thus, see attachment below for the number of units of each length of the figure that we have counted.

The perimeter of the figure = sum of all the lengths = 7 + 7 + 10 + 2 + 2 + 6 + 2 + 2 = 38

Perimeter of the shaded figure = 38 units

Answer:

y = 2x+4

Step-by-step explanation:

First we need to find the slope using two points

(-2,0) and (0,4)

m = (y2-y1)/(x2-x1)

m = (4-0)/(0--2)

   = 4/+2

   = 2

we have the y intercept  which is 4

Using the slope intercept form of the line

y = mx+b where m is the slope and b is the y intercept

y = 2x+4

Answer:

1937.98

Step-by-step explanation:

In the given question, to find the value to be added per year we will use the formula

P= A. r/n/ (1 +r/n)ⁿ - 1

Here A = 50,000

r (rate of interest) = 5 % or 0.05.

n = 1

t = 17

P = value deposit per year

therefore, P = (50,000 X 0.05)/ (1 +0.05)¹⁷ - 1

P =   2500 / 2.29- 1

= 1937.98 $.

therefore, person has to deposit 1937.98 $ per month.

Answer:

The 88% confidence interval for the proportion of defectives today is (0.053, 0.123)

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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 160, \pi = \frac{14}{160} = 0.088[/tex]

88% confidence level

So [tex]\alpha = 0.12[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.12}{2} = 0.94[/tex], so [tex]Z = 1.555[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.088 - 1.555\sqrt{\frac{0.088*0.912}{160}} = 0.053[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.088 + 1.555\sqrt{\frac{0.088*0.912}{160}} = 0.123[/tex]

The 88% confidence interval for the proportion of defectives today is (0.053, 0.123)

Answer:

see below

Step-by-step explanation:

y = -sqrt(x) +1

We know that the domain is from 0 to infinity

The range is from 1 to negative infinity

Answer:

b

Step-by-step explanation:

e2020

Answer:

The full name of the place is the "Danat Jebel Dhanna".  The Jebel Dhanna is currently located in the Abu Dhabi.  It is said that it is one of the most best beach in the UAE, they also say that it is the biggest resort, of course, with a bunch of hotels.

hope this helps ;)

best regards,

`FL°°F~` (floof)

Answer:

  7

Step-by-step explanation:

The number of cells in a tile is 4. If colored alternately, there are 3 of one color and 1 of the alternate color. To balance the coloring, an even number of tiles is needed. Hence the board dimensions must be multiples of 4.

In the given range, there are 7 such boards:

  4×4, 8×8, 12×12, 16×16, 20×20, 24×24, and 28×28

Answer:

x^2 - 10x

Step-by-step explanation:

2x^2 - 4x - x^2 +6x

You subtract x^2 from 2x^2 and you get x^2

Then you add 6x and 4x together and get 10x

So then you have x^2 - 10x

(plus I took the test and this was the correct answer.)

Answer:145

Step-by-step explanation: $19=20 76=80 53=50 23=20 12=10 total = 180 325-180 =145

Answer:

The probability that the sample proportion will differ from the population proportion by greater than 0.03 is 0.009.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 [tex]\mu_{\hat p}=p[/tex]

The standard deviation of this sampling distribution of sample proportion is:

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

As the sample size is large, i.e. n = 492 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by the normal distribution.

The mean and standard deviation of the sampling distribution of sample proportion are:

[tex]\mu_{\hat p}=p=0.07\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.07(1-0.07)}{492}}=0.012[/tex]

Compute the probability that the sample proportion will differ from the population proportion by greater than 0.03 as follows:

[tex]P(|\hat p-p|>0.03)=P(|\frac{\hat p-p}{\sigma_{\hat p}}|>\frac{0.03}{0.012})[/tex]

                           [tex]=P(|Z|>2.61)\\\\=1-P(|Z|\leq 2.61)\\\\=1-P(-2.61\leq Z\leq 2.61)\\\\=1-[P(Z\leq 2.61)-P(Z\leq -2.61)]\\\\=1-0.9955+0.0045\\\\=0.0090[/tex]

Thus, the probability that the sample proportion will differ from the population proportion by greater than 0.03 is 0.009.

Answer:

Domain: All real numbers or (negative infinity, positive infinity)

Range: [0, positive infinity)

Step-by-step explanation:

Domain; Since all values of x would work for this equation, simply any number could be plugged in. That means the domain would stretch to infinity because  there are an infinite amount of inputs and outputs

Range; Even though we have an infinite amount of domain, when we plug in a negative x, anything inside the absolute value will turn positive. Therefore, no output (y) value will ever go below zero, and we have [0, positive infinity).

Answer:

x ≈ 6.7 ft

Step-by-step explanation:

We are going to use tan∅ to find our answer:

tan66° = 15/x

xtan66° = 15

x = 15/tan66°

x = 6.67843 ft

Answer:

0.989

Step-by-step explanation:

For each graduate, there are only two possible outcomes. Either they find a job in their chosen field within a year after graduation, or they do not. The probability of a graduate finding a job is independent of other graduates. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

A study conducted at a certain college shows that "53%" of the school's graduates find a job in their chosen field within a year after graduation.

This means that [tex]p = 0.53[/tex]

6 randomly selected graduates

This means that [tex]n = 6[/tex]

Probability that at least one finds a job in his or her chosen field within a year of graduating:

Either none find a job, or at least one does. The sum of the probabilities of these outcomes is 1. So

[tex]P(X = 0) + P(X \geq 1) = 1[/tex]

We want [tex]P(X \geq 1)[/tex]

So

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{6,0}.(0.53)^{0}.(0.47)^{6} = 0.011[/tex]

So

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.011 = 0.989[/tex]

Answer:

a) P(x > 43) = 0.9599

b) P(x < 42) = 0.0228

c) P(x > 57.5) = 0.03

d) P(x = 42) = 0.

e) P(x<40 or x>55) = 0.1118

f) 43.42

g) Between 46.64 and 53.36.

h) Above 45.852.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

[tex]\mu = 50, \sigma = 4[/tex]

a) x>43

This is 1 subtracted by the pvalue of Z when X = 43. So

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

[tex]Z = \frac{43 - 50}{4}[/tex]

[tex]Z = -1.75[/tex]

[tex]Z = -1.75[/tex] has a pvalue of 0.0401

1 - 0.0401 = 0.9599

P(x > 43) = 0.9599

b) x<42

This is the pvalue of Z when X = 42.

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

[tex]Z = \frac{42 - 50}{4}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228

P(x < 42) = 0.0228

c) x>57.5

This is 1 subtracted by the pvalue of Z when X = 57.5. So

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

[tex]Z = \frac{57.5 - 50}{4}[/tex]

[tex]Z = 1.88[/tex]

[tex]Z = 1.88[/tex] has a pvalue of 0.97

1 - 0.97 = 0.03

P(x > 57.5) = 0.03

d) P(x = 42)

In the normal distribution, the probability of an exact value is 0. So

P(x = 42) = 0.

e) x<40 or x>55

x < 40 is the pvalue of Z when X = 40. So

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

[tex]Z = \frac{40 - 50}{4}[/tex]

[tex]Z = -2.5[/tex]

[tex]Z = -2.5[/tex] has a pvalue of 0.0062

x > 55 is 1 subtracted by the pvalue of Z when X = 55. So

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

[tex]Z = \frac{55 - 50}{4}[/tex]

[tex]Z = 1.25[/tex]

[tex]Z = 1.25[/tex] has a pvalue of 0.8944

1 - 0.8944 = 0.1056

0.0062 + 0.1056 = 0.1118

P(x<40 or x>55) = 0.1118

f) 5% of the values are less than what X value?

X is the 5th percentile, which is X when Z has a pvalue of 0.05, so X when Z = -1.645.

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

[tex]-1.645 = \frac{X - 50}{4}[/tex]

[tex]X - 50 = -1.645*4[/tex]

[tex]X = 43.42[/tex]

43.42 is the answer.

g) 60% of the values are between what two X values (symmetrically distributed around the mean)?

Between the 50 - (60/2) = 20th percentile and the 50 + (60/2) = 80th percentile.

20th percentile:

X when Z has a pvalue of 0.2. So X when Z = -0.84.

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

[tex]-0.84 = \frac{X - 50}{4}[/tex]

[tex]X - 50 = -0.84*4[/tex]

[tex]X = 46.64[/tex]

80th percentile:

X when Z has a pvalue of 0.8. So X when Z = 0.84.

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

[tex]0.84 = \frac{X - 50}{4}[/tex]

[tex]X - 50 = 0.84*4[/tex]

[tex]X = 53.36[/tex]

Between 46.64 and 53.36.

h) 85% of the values will be above what X value?

Above the 100 - 85 = 15th percentile, which is X when Z has a pvalue of 0.15. So X when Z = -1.037.

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

[tex]-1.037 = \frac{X - 50}{4}[/tex]

[tex]X - 50 = -1.037*4[/tex]

[tex]X = 45.852[/tex]

Above 45.852.

Answer:

Step-by-step explanation:

Answer:

there will be 9 id no. which it contains

Answer:

C =81.64 cm

Step-by-step explanation:

The circumference is given by

C = 2* pi *r

The radius is 13

C = 2 * 3.14 * 13

C =81.64 cm

Answer:

[tex]= 81.64cm \\ [/tex]

Step-by-step explanation:

[tex]c = 2\pi \: r \\ = 2 \times 3.14 \times 13 \\ = 81.64cm[/tex]

hope this helps

brainliest appreciated

good luck! have a nice day!