I need to know the answer please

PLEASE HELPPO I BEFGG

faltersainse [42]

Answer:

-1.75, -1 1/2, -3/4, 1/4, 0.75

Step-by-step explanation:

I hope this helped and if it did I would appreciate it if you marked me Brainliest. thank you and have a nice day!

7 0

3 years ago

What is an equation in point-slope form for the line that passes through the points (-3,5) and (2, -3)?

zloy xaker [14]

Answer:

y - 5 = -8/7(x + 3)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate
y₁ - y coordinate
m - slope

Step-by-step explanation:

Step 1: Define

Point (-3, 5)

Point (2, -3)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m

Substitute [SF]: $m=\frac{-3-5}{2+5}$
Subtract/Add: $m=\frac{-8}{7}$

Step 3: Write Equation

Plug in variables into the general form.

y - 5 = -8/7(x + 3)

y + 3 = -8/7(x - 2)

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

Please can you simplify -x+2z+x+2y

slavikrds [6]

Answer:

2y + 2z

Step-by-step explanation:

-x + 2z + x + 2y

= (-1 + 1)x + 2y + 2z

= 2y + 2z

Hope this helped!

3 0

3 years ago

John consistently saves

bagirrra123 [75]

Answer:

7789 i think this is the answer

Step-by-step explanation:

6 0

3 years ago