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Luden [163]
2 years ago
15

The product of a number and 3 less than the number is 40. Find the number.

Mathematics
1 answer:
Nuetrik [128]2 years ago
4 0

By solving a quadratic equation we will see that the number can be:

8 or -5.

<h3>How to find the number?</h3>

The product of a number N, and 3 less than the number, is equal to 40.

This is written as:

N*(N - 3) = 40

Now we need to solve this for N.

N^2 - 3N = 40

N^2 - 3N - 40 = 0

Then we need to solve this quadratic equation:

N = \frac{-(-3) \pm \sqrt{(-3)^2 - 4*1*(-40)} }{2} \\\\N = \frac{3 \pm 13 }{2}

Then we have two solutions:

  • N = (3 - 13)/2 = -5
  • N = (3 + 13)/2 = 8

These are the two possible solutions.

If you want to learn more about quadratic equations:

brainly.com/question/1214333

#SPJ1

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The Question’s above.
Nastasia [14]

Answer:

What questions above?

Step-by-step explanation:

3 0
3 years ago
The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific ten-hour
Alexus [3.1K]

Answer:

a) the probability that the defective board was produced during the first hour of operation is \frac{1}{10} or 0.1000

b) the probability that the defective board was produced during the  last hour of operation is \frac{1}{10} or 0.1000

c) the required probability is 0.2000

Step-by-step explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) = \left \{ {{\frac{1}{b-a} }\\\ }} \right   _0;   ( a ≤ y ≤ b )_{elsewhere

= \left \{ {{\frac{1}{10-0} }\\\ }} \right   _0;   ( 0 ≤ y ≤ 10 )_{elsewhere

f(y) = \left \{ {{\frac{1}{10} }\\\ }} \right   _0;   ( 0 ≤ y ≤ 10 )_{elsewhere

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 )   =   \int\limits^1_0 {f(y)} \, dy

we substitute

=    \int\limits^1_0 {\frac{1}{10} } \, dy

= \frac{1}{10} [y]^1_0

= \frac{1}{10} [ 1 - 0 ]

= \frac{1}{10} or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is \frac{1}{10} or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) =    \int\limits^{10}_9 {f(y)} \, dy

we substitute

=    \int\limits^{10}_9 {\frac{1}{10} } \, dy

= \frac{1}{10} [y]^{10}_9

= \frac{1}{10} [ 10 - 9 ]

= \frac{1}{10} or 0.1000

Therefore, the probability that the defective board was produced during the  last hour of operation is \frac{1}{10} or 0.1000

c)

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

 = (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)

= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000

4 0
3 years ago
If repetition of digits is allowed, how many different ways 5-digit codes are possible with the condition that the first digit s
Nesterboy [21]

Answer:

b) 27,000

Step-by-step explanation:

3 × 10 × 10 × 10 × 9 = 27000

'3' because the first digit can only be 3,6, or 9

'9' because the last digit cannot be zero, leaving only 9 options

7 0
3 years ago
P(blue) someone plz help me
Anarel [89]
What is the question¿
5 0
3 years ago
Find the value of x to the nearest degree.<br> A. 58<br> B. 52<br> C. 38<br> D. 32
slega [8]

Answer:

52⁰

Step-by-step explanation:

Cos x⁰ = adjacent/hypotenus

Cos x⁰ = 8/13

Cos x⁰ = 0.615

x⁰ = cos^-1 (0.615)

x⁰ = 52⁰

3 0
3 years ago
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