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2 years ago

If 10-2=33 , 37-23=83, 82-4=83, 55-7=32, then 76-45 ????

Mathematics

2 answers:

astra-53 [7]2 years ago

8 0

Answer:

✔ 76 – 45 = 35

*is the correct answer*

Step-by-step explanation:

"• 10 - 2 = 33 8 = 33 33 – 8 25 • 37-23=83 14 = 83 83 - 14 69 • 82-4= 43 78 = 43 78 – 43 35 • 55-7=32 48 = 32 16 Now on solving the above four answer further we will get, 69 – 25 = 44 35 – 16 = 19 Further on subtracting them, 44 – [ 19 = 35 Hence, the answer is 76 – 45 = 35

Alexxx [7]2 years ago

5 0

Answer:

i think value 35

Step-by-step explanation:

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Jane is taking two courses. The probablity she passes the first course is 0,7. The probablity she passes the second course is 0.

makkiz [27]

Answer:

a. 0.6

b. not independent

c. 0.1

d. 0.4

e. 0.3

Step-by-step explanation:

P(passing first course)=P(C1)=0.7

P(passing second course)=P(C2)=0.8

P(passing at least one course)=P(C1∪C2)=0.9

P( passes both courses)=P(C1∩C2)=?

We know that

P(A∪B)=P(A)+P(B)-P(A∩B)

P(A∩B)=P(A)+P(B)-P(A∪B)

So,

P( passes both courses)=P(C1∩C2)=P(C1)+P(C2)-P(C1∪C2)

P( passes both courses)=P(C1∩C2)=0.7+0.8-0.9

P( passes both courses)=P(C1∩C2)=0.6

Thus, the probability she passes both courses is 0.6.

The event of passing one course is independent of passing another course if

P(C1∩C2)=P(C1)*P(C2)

P(C1)*P(C2)=0.7*0.8=0.56

P(C1∩C2)=0.6

As,

0.6≠0.56

P(C1∩C2)≠P(C1)*P(C2),

So, the event of passing one course is dependent of passing another course.

P(not passing either course)=P(C1∪C2)'=1-P(C1∪C2)

P(not passing either course)=P(C1∪C2)'=1-0.9

P(not passing either course)=P(C1∪C2)'=0.1

Thus, the probability of not passing either course is 0.1.

P(not passing both courses)=P(C1∩C2)'=1-P(C1∩C2)

P(not passing both courses)=P(C1∩C2)'=1-0.6

P(not passing both courses)=P(C1∩C2)'=0.4

Thus, the probability of not passing both courses is 0.4.

P(passing exactly one course)=?

P(passing exactly course 1)=P(C1)-P(C1∩C2)=0.7-0.6=0.1

P(passing exactly course 2)=P(C2)-P(C1∩C2)=0.8-0.6=0.2

P(passing exactly one course)=P(passing exactly course 1)+P(passing exactly course 2)

P(passing exactly one course)=0.1+0.2

P(passing exactly one course)=0.3

Thus, the probability of passing exactly one course is 0.3.

3 0

2 years ago

Tags are placed to the left leg and right leg of a bear in a forest. Let A1 be the event that the left leg tag is lost and the e

Komok [63]

Answer:

0.75 = 75% probability that exactly one tag is lost, given that at least one tag is lost

Step-by-step explanation:

Independent events:

If two events, A and B, are independent, then:

$P(A \cap B) = P(A)*P(B)$

Conditional probability:

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

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: At least one tag is lost

Event B: Exactly one tag is lost.

Each tag has a 40% = 0.4 probability of being lost.

Probability of at least one tag is lost:

Either no tags are lost, or at least one is. The sum of the probabilities of these events is 1. Then

$p + P(A) = 1$

p is the probability none are lost. Each one has a 60% = 0.6 probability of not being lost, and they are independent. So

p = 0.6*0.6 = 0.36

Then

$P(A) = 1 - p = 1 - 0.36 = 0.64$

Intersection:

The intersection between at least one lost(A) and exactly one lost(B) is exactly one lost.

Then

Probability at least one lost:

First lost(0.4 probability) and second not lost(0.6 probability)

First not lost(0.6 probability) and second lost(0.4 probability)

$P(A \cap B) = 0.4*0.6 + 0.6*0.4 = 0.48$

Find the probability that exactly one tag is lost, given that at least one tag is lost (write it up to second decimal place).

$P(B|A) = \frac{0.48}{0.64} = 0.75$

0.75 = 75% probability that exactly one tag is lost, given that at least one tag is lost

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2 years ago

Solve y=4x+rx+6 for x.

MariettaO [177]

Answer:

x = y−6/r+4

Step-by-step explanation:

Let's solve for x.

y=4x+rx+6

Step 1: Flip the equation.

rx+4x+6=y

Step 2: Add -6 to both sides.

rx+4x+6+−6=y+−6

rx+4x=y−6

Step 3: Factor out variable x.

x(r+4)=y−6

Step 4: Divide both sides by r+4.

x(r+4)/r+4=y−6/r+4

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Otrada [13]

The answer is 17.6 miles per hour

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