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3 years ago

What is the expression x - 4 represents a negative integer. What integers could X represent?

2 answers:

3 0

The expression x-4 will come out to be a negative integer. That means X <em>has to be </em>something that when 4 is taken away from it, the answer is still negative.

Let's try 5.
5-4 is 1. Is 1 negative? No, so that doesn't work.
4-4 is 0. 0 isn't negative either, so that doesn't work.
3-4 is -1. Is -1 negative? YES! Let's try it for 2 as well, just to be safe.
2-4 is -2. Still negative.

X can be anything less than 4.

Kisachek [45]3 years ago

3 0

Answer:

Integers between (-∞, 4)

Step-by-step explanation:

Negative Integer is integers whose value is less than 0 i.e. -1, -2, -3, -4,.... so on.

The expression x - 4 will be negative if x is less than 4.

For Example, if we take integers greater than 4 or equal to 4 as the value of x then,

5 - 4 = +1 which is not a negative integer

4 - 4 = 0 which is again not a negative integer.

But, if we take x as less than 4 then we get,

2 - 4 = -2 i.e. negative integer

-3 - 4 = -7 i.e. also negative integer.

So, for all value of x which is less than 4, we will get a negative integer.

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Solve for y<br>7y + 9 - 10y = -18<br>What is y simplify your answer as much as possible.

Tpy6a [65]

Y = 9

7y - 10y = -3y
bring down the 9 = -18 so

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4 years ago

From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one def

jasenka [17]

Answer:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X 0 1 2 3

P(X) 0.92 0.03 0.03 0.02

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

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4 years ago

Which of the following best defines 3 to the power of 2 over 3?

Anika [276]

Answer:

Option B is correct that is cube root of 9.

Step-by-step explanation:

We have been given the expression:

$3^{\frac{2}{3}}$ after solving square power

It can be rewritten as $9^{\frac{1}{3}}$

Which is the cube root of 9

Hence, Option B is correct That is cube root of 9

Option A is discarded because square gets solve then only we get 9

Option C and D are discarded because square will be solve we will get 9.

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