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2 years ago

What is the scale if: 3.2 cm on the map correspond to 4 km in the actual distance?

Mathematics

1 answer:

aliya0001 [1]2 years ago

5 0

Scaling on map is defined as the ratio of a distance on the map to the corresponding distance on the ground.

Here, 3.2 centimeters on the map is equivalent to 4 kilometers.

We can use unitary method to find, how many kilometer/kilometers each centimeter corresponds to.

Using the unitary method:

$3.2 cm=4 km$

$1 cm=\frac{4}{3.2} =1.25km$

Therefore, the scale of map represents 1.25 kilometers for each centimeter.

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The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find

fenix001 [56]

Answer:

y2 = (6x + 7)/36 + (Dx + E)e^x

Step-by-step explanation:

The method of reduction of order is applicable for second-order differential equations.

For a known solution y1 of a 2nd order differential equation, this method assumes a second solution in the form Uy1 which satisfies the said differential equation. It then assumes a reduced order for U'' (w' = U'').

The differential equation becomes easy to solve, and all that is left are integration and substitutions.

Check attachments for the solution to this problem.

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1 year ago

Exams are approaching and Helen is allocating time to studying for exams. She feels that with the appropriate amount of studying

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Answer: 0.05

Step-by-step explanation:

Let M = Event of getting an A in Marketing class.

S = Event of getting an A in Spanish class,

i.e. P(M) = 0.80 , P(S) = 0.60 and P(M∩S)=0.45

Required probability = P(neither M nor S)

= P(M'∩S')

= P(M∪S)' [∵P(A'∩B')=P(A∪B)']

=1- P(M∪S) [∵P(A')=1-P(A)]

= 1- (P(M)+P(S)- P(M∩S)) [∵P(A∪B)=P(A)+P(B)-P(A∩B)]

= 1- (0.80+0.60-0.45)

= 1- 0.95

= 0.05

hence, the probability that Helen does not get an A in either class= 0.05

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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

1 year ago

a direct variation function contains the points ( -8, -6 ) and (12,9 ) . Which equation represents the function ?

9966 [12]

Answer:

y = $\frac{3}{4}$ x

Step-by-step explanation:

Given that the quantities vary directly then the equation relating them is

y = kx ← k is the constant of variation

To find k use either of the 2 given points

Using (12, 9), that is x = 12 when y = 9, then

k = $\frac{y}{x}$ = $\frac{9}{12}$ = $\frac{3}{4}$

y = $\frac{3}{4}$ x ← equation of variation

8 0

1 year ago

An acute triangle has two sides measuring 8 cm and 10 cm . What is the best representation of the possible range of values for t

Mekhanik [1.2K]

Answer:

the right answer is 2 < s< 18

Step-by-step explanation:

The Third side of the triangle must be greater than the difference of the other two sides which is 10 -8= 2 and smaller than the sum of the other two sides of the triangle which is 10+8= 18.

so third side should be greater than 2 and less than 18.

So right answer is 2 < s< 18

5 0

1 year ago