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4 years ago

Select each inequality that 0 is a solution to. Q No. 2

Mathematics

2 answers:

PtichkaEL [24]4 years ago

3 0

Answer:

Can you reformat?

So far, 0 is a solution to none of these.

Darina [25.2K]4 years ago

3 0

Answer:

1. nope

2. not sure

3. nope

Step-by-step explanation:

-3(y - 1) > 3

-3y + 3 > 3

-3y > 0

y > 0

0 is not a solution since 0 is not greater than 0

I'm not sure what this question means, is there a $\geq or \leq$ between the 4 and the 0?

y + 8 < 2y + 8

8 < 2y - y + 8

8 - 8 < y

0 < y

0 is not a solution because 0 is not greater than 0

im confused- I prolly got something wrong.

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Which equation is not a linear function?

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Answer:

the second answer Is the right one

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3 years ago

Which of the following expressions represents "twelve diminished by six times a number"?

ahrayia [7]

Diminished means "made smaller", so in other words, subtracted. This makes the phrase "twelve subtracted by six times a number", or more simply "twelve minus six times a number".

Answer:
A. 12 - 6n

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3 years ago

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Need help with this math question

anastassius [24]

Answer:

The vertex is: $(6, 8)$

Step-by-step explanation:

First solve the equation for the variable y

$x^2-4y-12x+68=0$

Add 4y on both sides of the equation

$4y=x^2-4y+4y-12x+68$

$4y=x^2-12x+68$

Notice that now the equation has the general form of a parabola

$ax^2 +bx +c$

In this case

$a=1\\b=-12\\c=68$

Add $(\frac{b}{2}) ^ 2$ and subtract $(\frac{b}{2}) ^ 2$ on the right side of the equation

$(\frac{b}{2}) ^ 2=(\frac{-12}{2}) ^ 2\\\\(\frac{b}{2}) ^ 2=(-6) ^ 2\\\\(\frac{b}{2}) ^ 2=36$

$4y=(x^2-12x+36)-36+68$

Factor the expression that is inside the parentheses

$4y=(x-6)^2+32$

Divide both sides of the equality between 4

$\frac{4}{4}y=\frac{1}{4}(x-6)^2+\frac{32}{4}$

$y=\frac{1}{4}(x-6)^2+8$

For an equation of the form

$y=a(x-h)^2 +k$

the vertex is: (h, k)

In this case

$h=6\\k =8$

the vertex is: $(6, 8)$

5 0

3 years ago

Simplicity the expression (0.4^3)

julia-pushkina [17]

0.4^3 = 0.4 * 0.4 * 0.4 = 0.064

5 0

4 years ago

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A company establishes a fund of 120 from which it wants to pay an amount,C, to any of its 20 employees who achieve a high-perfor

Leno4ka [110]

Answer:

$C=120/2=60$

Step by step Explanation'

To solve this problem, we will need to apply trial-and-error calculation with the binomial distribution, even though it appears like Central Limit Theorem but it's not.

For us to know the value of C , we will look for a minimum integer such that having 'n' number of high performance level of employee has the probability below 0.01.

Determine the maximum value of C, then the maximum value that C can have is 120/n

Let us represent X as the number of employees with high performance with a binomial distribution of

P =0.02( since the percentage of chance of achieving a high performance level is 2%)

n = 20 ( number of employees who achieve a high performance level)

The probability of X= 0 can be calculated

P( X= 0) = 0.98^n

$P(X=0)=0.98^20$

$P(X=0)=0.668$

$P(X=1)=0.02*20*0.98^19$

$P(X=1)=0.272$

$P(X=2)=0.02^2*20*0.98^18$

$P(X=2)=0.053$

Summation of P( X= 0)+ P( X= 1)+P( X= 2) will give us the value of 0.993 which is greater than 0.99( 1% that the fund will be inadequate to cover all payments for high performance.)

BUT the summation of P( X= 0)+ P( X= 1) will give the value of 0.94 which doesn't exceed the 0.99 value,

Therefore, the minimum value of integer in such a way that P(X >2) is less than 0.01 have n= 2

then the maximum value that C can have is 120/n

$C=120/2=60$

7 0

4 years ago