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2 years ago

14

Which partial quotients could be added to find 396 ÷ 12?

2 answers:

NikAS [45]2 years ago

6 0

The partial quotients will be 33

Travka [436]2 years ago

3 0

Answer:

33 plz braniest i think i help you

Step-by-step explanation:

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2 years ago

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Paula's Flowers sells 39 arrangements per week. The flower shop down the street sells

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50:39

Step-by-step explanation:

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3 years ago

Solve the inequality 3/5x+30 is less than or equal to 150

FrozenT [24]

Answer: less than

Step-by-step explanation:

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2 years ago

Solve using the elimination method:<br> X-3y=7<br> 3x + 3y=9

sergij07 [2.7K]

Answer:

$x=4\\\\y=-1$

Step-by-step explanation:

Given Equation:

$x-3y=7$ Equation:1

$3x + 3y=9$ Equation:2

Dividing Equation:2 by '3' both the sides:

$x+y=3$

or $x=3-y$ Equation:3

Putting the vale of 'x' in Equation:1

$x-3y=7$

$(3-y)-3y=7\\\\3-y-3y=7\\\\3-4y=7\\\\$

Subtracting '3' both sides

$-4y=7-3\\\\-4y=4$

$y=\frac{4}{-4}$

$y=-1$

Putting value of 'y' in Equation:3

$x=3-y$

$x=3-(-1)\\\\x=3+1\\\\x=4$

The solution of the equations is :

$x=4\\\\y=-1$

7 0

3 years ago

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t

Angelina_Jolie [31]

Answer:

$P(X = 0) = 0.027$

$P(X = 1) = 0.189$

$P(X = 2) = 0.441$

$P(X = 3)= 0.343$

Step-by-step explanation:

The probability mass function P(X = x) is the probability that X happens x times.

When n trials happen, for each $x \leq n$ , the probability mass function is given by:

$P(X = x) = pe_{n,x}.p^{x}.(1-p)^{n-x}$

In which p is the probability that the event happens.

$pe_{n,x},$ is the permutation of n elements with x repetitions(when there are multiple events happening(like one passes and two not passing)). It can be calculated by the following formula:

$pe_{n,x} = \frac{n!}{x!}$

The sum of all P(X=x) must be 1.

In this problem

We have 3 trials, so $n = 3$

The probability that a wafer pass a test is 0.7, so $p = 0.7$

Determine the probability mass function of the number of wafers from a lot that pass the test.

$P(X = 0) = (0.7)^{0}.(0.3)^{3} = 0.027$

$P(X = 1) = pe_{3,1}.(0.7).(0.3)^{2} = 0.189$

$P(X = 2) = pe_{3,2}.(0.7)^{2}.(0.3) = 0.441$

$P(X = 3) = (0.7)^{3} = 0.343$

5 0

3 years ago