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2 years ago

Find the value of x to the nearest degree 1. 83 2. 53 3. 67 4.22

Mathematics

1 answer:

nydimaria [60]2 years ago

6 0

Answer:

x = 23

Step-by-step explanation:

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

sin theta = opp side/ hypotenuse

sin x = 3 / sqrt(58)

take the inverse sin of each side

sin ^-1 (sin x) = sin ^-1( 3 / sqrt(58))

x = sin ^-1( 3 / sqrt(58))

x=23.19859051

To the nearest degree

x = 23

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A parallelogram with sides of 6 and 10 has an area of 30. Find the measure of the small angle of the parallelogram.

dmitriy555 [2]

A = 2 * (0.5ab) + b (10 - a) = ab + 10b - ab = 10b

10b = 30√2; b = 3√2

sin α = 3√2 / 6; α = 45 degrees

Small angle: 45°; Large angle: 135°

4 0

2 years ago

Use the drop-down menus to complete the proof. By the unique line postulate, you can draw only one segment, AC BC CD. Using the

inn [45]

The first one is; BC
The second one is; Reflection
The third one is; A
The fourth one is; Length

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2 years ago

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If 10a+10b=35, what is the average (arithmetic mean) of a and b?

worty [1.4K]

Answer:

1.75

Step-by-step explanation:

If a and b are two numbers, then their arithmetic mean is

$\dfrac{a+b}{2}$

Given:

$10a+10b=35$

Divide this equation by 10:

$a+b=3.5$

Now, divide it by 2:

$\dfrac{a+b}{2}=\dfrac{3.5}{2}\\ \\\dfrac{a+b}{2}=1.75$

6 0

2 years ago

Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba

Maksim231197 [3]

Answer:

(a) Probability mass function

P(X=0) = 0.0602

P(X=1) = 0.0908

P(X=2) = 0.1704

P(X=3) = 0.2055

P(X=4) = 0.1285

P(X=5) = 0.1550

P(X=6) = 0.1427

P(X=7) = 0.0390

P(X=8) = 0.0147

NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.

(b) Cumulative distribution function of X

F(X=0) = 0.0602

F(X=1) = 0.1510

F(X=2) = 0.3214

F(X=3) = 0.5269

F(X=4) = 0.6554

F(X=5) = 0.8104

F(X=6) = 0.9531

F(X=7) = 0.9921

F(X=8) = 1.0068

Step-by-step explanation:

Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).

For X=0

This happens when every couple and the singles are on time (ot).

$P(X=0)=P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot)\\\\P(X=0)=(1-0.43)^{5}=0.57^5= 0.0602$

For X=1

This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:

$P(X=1) = P(\#4=late)+P(\#5=late)=2*P(\#4=late)\\\\P(X=1) = 2*P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=late)*P(\#5=ot)\\\\P(X=1) = 2*0.57*0.57*0.57*0.43*0.57\\\\P(X=1) = 2*0.57^4*0.43=2*0.0454=0.0908$

For X=2

This happens when

1) Only one of the three couples is late, and the others cooples and singles are on time.

2) When both singles are late , and the couples are on time.

$P(X=2)=3*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=2)=3*(0.43*0.57^4)+(0.43^2*0.57^3)=0.1362+0.0342=0.1704$

For X=3

This happens when

1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.

$P(X=3)=6*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=3)=6*(0.43^2*0.57^3)=6*0.342=0.2055$

For X=4

This happens when

1) Only two couples are late. There are 3 combinations of these.

2) Only one couple and both singles are late. Only one combination of these situation.

$P(X=4)=3*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=4)=3*(0.43^2*0.57^3)+(0.43^3*0.57^2)\\\\P(X=4)=3*0.0342+ 0.0258=0.1027+0.0258=0.1285$

For X=5

This happens when

1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.

$P(X=6)=6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=6*(0.43^3*0.57^2)=6*0.0258=0.1550$

For X=6

This happens when

1) Only the three couples are late (1 combination)

2) Only two couples (3 combinations) and one single (2 combinations) are late

$P(X=6)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=ot)*P(\#5=ot)+6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=(0.43^3*0.57^2)+6*(0.43^4*0.57)\\\\P(X=6)=0.0258+6*0.0195=0.0258+0.1169=0.1427$

For X=7

This happens when

1) Only one of the singles is on time (2 combinations)

$P(X=7)=2*P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=ot)\\\\P(X=7)=2*0.43^4*0.57=0.0390$

For X=8

This happens when everybody is late

$P(X=8)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=l)\\\\P(X=8) = 0.43^5=0.0147$

8 0

2 years ago

A group of 55 football players were polled about which synthetic grass they like to play on, Chemstrand or Chemgrass. The data f

cricket20 [7]

From the Venn diagram: 15 players like Chemstrand, 17 players like Chemgrass, 13 players like both Chemstrand and Chemgrass while 10 players like neither Chemstrand nor Chemgrass.

The missing values in the frequency table are x - representing the number of players that like both Chemstrand and Chemgrass, y - representing the number of players that like Chemgrass but do not like Chemstrand and z - representing the number of players likes Chemstrand but do not like Chemgrass.

The number of players that like both Chemstrand and Chemgrass is 13. The number of players that like Chemgrass but do not like Chemstrand is 17. The number of players likes Chemstrand but do not like Chemgrass is 15.

Therefore, x = 13, y = 17 and z = 15

6 0

2 years ago

Read 2 more answers