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blagie [28]
1 year ago
5

ACD is a triangle and B is a point on AC. AB = 8cm and BC is 6cm. Angle BCD = 48° and angle BDC = 50°. (a) Find the length of BD

. (b) Find the length of AD. (c) Find the area of triangle ABD. (This is all one question)​

Mathematics
1 answer:
FromTheMoon [43]1 year ago
5 0

Answer:

  • 5.8206 cm
  • 10.528 cm
  • 23.056 cm^2

Step-by-step explanation:

(a) The Law of Sines can be used to find BD.

  BD/sin(48°) = BD/sin(50°)

  BD = (6 cm)(sin(48°)/sin(60°)) ≈ 5.82064 cm

__

(b) We can use the Law of Cosines to find AD.

  AD^2 = AB^2 +BD^2 -2·AB·BD·cos(98°) . . . . . angle ABD = 48°+50°

  AD^2 ≈ 110.841

  AD ≈ √110.841 ≈ 10.5281 . . . cm

__

(c) The area of ∆ABD can be found using the formula ...

  A = ab·sin(θ)/2 . . . . . where a=AB, b=BD, θ = 98°

  A = (8 cm)(5.82064 cm)sin(98°)/2 ≈ 23.0560 cm^2

_____

Angle ABD is the external angle of ∆BCD that is the sum of the remote interior angles BCD and BDC. Hence ∠ABD = 48° +50° = 98°.

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The time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. Round the
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Answer:

a) 0.019

b) 0.563

c) x = 1.966 hours

Step-by-step explanation:

E(X) = 1

Exponential random variable's probability function is given as

P(X=x) = λ e^(-λ.x)

The cumulative distribution function is given as

P(X ≤ x) = 1 - e^(-λ.x)

a) The time between the arrivals of small aircraft at a county airport that is exponentially distributed.

But the number of planes that land every hour will be obtained using the Poisson distribution formula.

It is the best for discrete systems.

Poisson distribution formula is given as

P(X = x) = (e^-λ)(λˣ)/x!

λ = 1 aircraft per hour.

The probability that more than three aircraft arrive within an hour = P(X > 3)

P(X > 3) = 1 - P(X ≤ 3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]

P(X > 3) = 1 - 0.98101 = 0.01899 = 0.019 to 3 d.p

b) If 30 separate one-hour intervals are chosen, what Is the probability that no interval contains more than three arrivals

Probability of one 1-hour interval not containing more than 3 arrivals = 1 - P(X > 3)

= 1 - 0.01899 = 0.98101

Probability that thirty 1-hour intervals will not contain more than 3 arrivals = (0.98101)³⁰ = 0.5626 = 0.563 to 3 d.p

c) Determine the length of an interval of time (In hours) such that the probability that no arrivals occur during the interval is 0.14

We can now use the cumulative distribution function for exponential random variable for this

P(X ≤ x) = 1 - e^(-λ.x)

P(X > x) = 1 - P(X ≤ x)

P(X > x) = e^(-λ.x)

λ = 1, x = ?,

0.14 = e⁻ˣ

e⁻ˣ = 0.14

In e⁻ˣ = In 0.14 = -1.966

-x = -1.966

x = 1.966 hours

Hope this Helps!!!

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1 year ago
Jane is one of 50 students to take a standardized math test that includes 100 multiple choice questions. If she has the highest
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Solution:
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Olaf needs a total of 3 cups of sugar to make 4 cakes. Write and solve an equation to find the number of cups of sugar he needs
irinina [24]

Answer:

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Step-by-step explanation:

7 0
2 years ago
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A random sample of 20 individuals who graduated from college five years ago were asked to report the total amount of debt (in $)
Gekata [30.6K]

Answer:

a. As college debt increases current investment decreases.

b. Y= 68778.2406 - 1.9112X

Every time the college debt increases one dollar, the estimated mean of the current investments decreases 1.9112 dollars.

c. There is a significant linear relationship between college debt and current investment because the P-value is less than 0.1.

d. Y= $59222.2406

e. R²= 0.9818

Step-by-step explanation:

Hello!

You have the information on a random sample of 20 individuals who graduated from college five years ago. The variables of interest are:

Y: Current investment of an individual that graduated from college 5 years ago.

X: Total debt of an individual when he graduated from college 5 years ago.

a)

To see the relationship between the information about the debt and the investment is it best to make a scatterplot with the sample information.

As you can see in the scatterplot (attachment) there is a negative relationship between the current investment and the debt after college, this means that the greater the debt these individuals had, the less they are currently investing.

The statement that best describes it is: As college debt increases current investment decreases.

b)

The population regression equation is Y= α + βX +Ei

To develope the regression equation you have to estimate alpha and beta:

a= Y[bar] -bX[bar]

a= 44248.55 - (-1.91)*12829.70

a= 68778.2406

b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }

b=\frac{9014653088-\frac{(256594)(884971)}{20} }{4515520748-\frac{(256594)^2}{20} }

b= -1.9112

∑X= 256594

∑X²= 4515520748

∑Y= 884971

∑Y²= 43710429303

∑XY= 9014653088

n= 20

Means:

Y[bar]= ∑Y/n= 884971/20= 44248.55

X[bar]= ∑X/n= 256594/20= 12829.70

The estimated regression equation is:

Y= 68778.2406 - 1.9112X

Every time the college debt increases one dollar, the estimated mean of the current investments decreases 1.9112 dollars.

c)

The hypotheses to test if there is a linear regression between the two variables are two tailed:

H₀: β = 0

H₁: β ≠ 0

α: 0.01

To make this test you can use either a Student t or the Snedecor's F (ANOVA)

Using t=<u>  b - β  </u>=<u>  -1.91 - 0  </u>= -31.83

                 Sb         0.06

The critical region and the p-value for this test are two tailed.

The p-value is: 0.0001

The p-value is less than the level of signification, the decision is to reject the null hypothesis.

Using the

F= \frac{MSTr}{MSEr}= \frac{4472537017.96}{4400485.72} =1016.37

The rejection region using the ANOVA is one-tailed to the right, and so is the p-value.

The p-value is: 0.0001

Using this approach, the decision is also to reject the null hypothesis.

The conclusion is that at a 1% significance level, there is a linear regression between the current investment and the college debt.

The correct statement is:

There is a significant linear relationship between college debt and current investment because the P-value is less than 0.1.

d)

To predict what value will take Y to a given value of X you have to replace it in the estimated regression equation.

Y/X=$5000

Y= 68778.2406 - 1.9112*5000

Y= $59222.2406

The current investment of an individual that had a $5000 college debt is $59222.2406.

e)

To estimate the proportion of variation of the dependent variable that is explained/ given by the independent variable you have to calculate the coefficient of determination R².

R^2= \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{sumY^2-\frac{(sumY)^2}{n} }

R^2= \frac{-1.9112^2[4515520748-\frac{(256594)^2}{20} ]}{43710429303-\frac{(884971)^2}{20} }

R²= 0.9818

This means that 98.18% of the variability of the current investments are explained by the college debt at graduation under the estimated regression model: Y= 68778.2406 - 1.9112X

I hope it helps!

5 0
1 year ago
A toddler is allowed to dress himself on Mondays, Wednesdays, and Fridays. For each of his shirt, pants, and shoes, he is equall
avanturin [10]

Answer:

0.0286 = 2.86% probability that today is Monday.

Step-by-step explanation:

Conditional Probability

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

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

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

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Dressed correctly

Event B: Monday

Probability of being dressed correctly:

100% = 1 out of 4/7(mom dresses).

(0.5)^3 = 0.125 out of 3/7(toddler dresses himself). So

P(A) = 0.125\frac{3}{7} + \frac{4}{7} = \frac{0.125*3 + 4}{7} = \frac{4.375}{7} = 0.625

Probability of being dressed correctly and being Monday:

The toddler dresses himself on Monday, so (0.5)^3 = 0.125 probability of him being dressed correctly, 1/7 probability of being Monday, so:

P(A \cap B) = 0.125\frac{1}{7} = 0.0179

What is the probability that today is Monday?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0179}{0.625} = 0.0286

0.0286 = 2.86% probability that today is Monday.

4 0
1 year ago
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