HELP !!!!!!MEEEEEEEEEEEE!!!!!!!!!!

What's the equation of the line slope 1/4 y-intercept: 3

kvasek [131]

In y = mx + b form, the m is ur slope and the b is ur y int.

y = mx + b
y = 1/4x + 3 <=== slope intercept form

or...
y = 1/4x + 3
-1/4x + y = 3
x - 4y = -12 <=== standard form

8 0

3 years ago

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Confusing stuff! Giving out 50 big points!

UkoKoshka [18]

If the gum costs 75 cents today, which is 3 cents less than three times what the pack cost 20 years ago, we just need to arrange what we know into an equation.
We know that the price of gum today is 75 cents, so we can leave that on one side of the equation as a constant. Also, since we are dealing only with cents and not dollars, there is no need to shift the decimal point to the left.

? = 75

Now, we can work the equation.

The problem tells us that 75 is 3 cents less than three times the old cost. So, if x is the old cost, we can work out the equation based on that.
It says that it's 3 cents less than 3 times x. The question is basically the equation in word form.

3x - 3 = 75

'3x' being the "three times the old cost" part, '- 3' being the "3 cents less" part, and '75' of course being the current price of a pack of gum.
Option C is your correct answer.

BTW if we were to solve this, we would see that the old price of a pack of gum is:

3x - 3 = 75
+ 3 + 3

3x = 78
/3 /3

x = 26

20 years ago, a 75-cent pack of gum cost 26 cents.
Hope that helped! =)

7 0

3 years ago

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A study was conducted to estimate the difference in the mean salaries of elementary school teachers from two neighboring states.

umka21 [38]

Answer:

$(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70$

$(28900-30300) +2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = 501.698$

The confidence interval would be $-3301.70 \leq \mu \leq 501.698$ and since the confidence interval contains the value of 0 we don't have enough evidence to conclude that the difference between the two states for the salary of teachers are significantly different.

Step-by-step explanation:

We have the following info given by the problem

$\bar X_1 = 28900$ the sample mean for the salaries of teachers in Indiana

$s_1 = 2300$ the sample deviation for the salary of teachers in Indiana

$n_1 =10$ the sample size from Indiana

$\bar X_2 = 30300$ the sample mean for the salaries of teachers in Michigan

$s_2 = 2100$ the sample deviation for the salary of teachers in Michigan

$n_2 =14$ the sample size from Michigan

We want to find a confidence interval for the difference in the two means and the formula for this case is given by;

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom for this case are:

$df = n_1 +n_2 -2= 10+14-2 =22$

The confidence is 95%so then the significance is $\alpha=0.05$ and the $\alpha/2 =0.025$ , we need to find a critical value in the t distribution with 22 degrees of freedom who accumulates 0.025 of the area on each tail and we got:

$t_{\alpha/2}= 2.07$

And now replacing in the formula for the confidence interval we got:

$(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70$

$(28900-30300) +2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = 501.698$

The confidence interval would be $-3301.70 \leq \mu \leq 501.698$ and since the confidence interval contains the value of 0 we don't have enough evidence to conclude that the difference between the two states for the salary of teachers are significantly different.

6 0

3 years ago

Write 0.003 as the product of a whole number and a power of 10

Arte-miy333 [17]

3 × $10^{-3}$

expressing as a whole number, using the digits available, we obtain

3

Now consider the position of the decimal point in the original 0.003

To get it back to where it should be we require to move the decimal point 3 places to the left , denoted by a negative exponent

0.003 = 3 × $10^{-3}$

8 0

3 years ago

Which point is a solution to y ≤ 3x- 4? O A. (0,4) O B. (-2, 0) OC. (0, 0) OD. (3,1)

Nina [5.8K]

Answer:

(D). (3, 1)

Step-by-step explanation:

y ≤ 3x - 4

(A). (0, 4)

4 ≤ 3(0) - 4

4 ≤ - 4 (False statement)

(B). (-2, 0)

0 ≤ 3(- 2) - 4

0 ≤ - 10 (False statement)

(C). (0, 0)

0 ≤ 3(0) - 4

0 ≤ - 4 (False statement)

(D). (3, 1)

1 ≤ 3(3) - 4

1 ≤ 5 (True statement) 

4 0

3 years ago

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