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2 years ago

Two consecutive integers are such that 3 times the larger exceeds twice the smaller by 34. find the integers.

Mathematics

2 answers:

julsineya [31]2 years ago

7 0

x - a smaller number

(x+1) - a larger number

3(x+1) - 2x = 34

3x + 3 - 2x = 34

x= 31

(x+1) = 32

rodikova [14]2 years ago

5 0

Answer: larger = 32 and smaller = 31

Step-by-step explanation:

let's call the smaller x. The larger must be x+1, because they are consecutives.

3 times the larger exceeds twice the smaller by 34, so:

3(x+1) = 2x+34

3x+3=2x+34

3x-2x=34-3

x=31

x+1 = 32

Checking:

3.32 = 92

3.31+34 = 96 ok

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dangina [55]

Answer:

8% or 0.08

Step-by-step explanation:

Probability of missing the first pass = 40% = 0.40

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We have to find the probability that he misses both the passes. Since the two passes are independent of each other, the probability that he misses two passes will be:

Probability of missing 1st pass x Probability of missing 2nd pass

i.e.

Probability of missing two passes in a row = 0.40 x 0.20 = 0.08 = 8%

Thus, there is 8% probability that he misses two passes in a row.

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2 years ago

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Semenov [28]

Answer: We have two solutions:

1000 - 998 = 2

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Step-by-step explanation:

So we have the problem:

****-*** = 2

where each star is a different digit, so in this case, we have a 4 digit number minus a 3 digit number, and the difference is 2.

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So we could write this as:

1000 - 998 = 2

now, if we add one to each number, the difference will be the same, and the number of digits in each number will remain equal:

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2 years ago

An attempt to establish a video call via some social media app may fail with probability 0.1. If connection is established and i

xxMikexx [17]

Answer:

(1). y = x ~ Exp (1/3).

(2). Check attachment.

(3). EY = 3(1 - e^-2).

(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.

Step-by-step explanation:

Kindly check the attachment to aid in understanding the solution to the question.

So, from the question, we given the following parameters or information or data;

(A). The probability in which attempt to establish a video call via some social media app may fail with = 0.1.

(B). " If connection is established and if no connection failure occurs thereafter, then the duration of a typical video call in minutes is an exponential random variable X with E[X] = 3. "

(C). "due to an unfortunate bug in the app all calls are disconnected after 6 minutes. Let random variable Y denote the overall call duration (i.e., Y = 0 in case of failure to connect, Y = 6 when a call gets disconnected due to the bug, and Y = X otherwise.)."

(1). Hence, for FY(y) = y = x ~ Exp (1/3) for the condition that zero is equal to y = x < 6.

(2). Check attachment.

(3). EY = 3(1 - e^-2).

(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.

The condition to follow in order to solve this question is that y = 0 if x ≤ 0, y = x if 0 ≤ x ≤ 6 and y = 6 if x ≥ 6.