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4 years ago

Arnoldo needs to write this system in slope-intercept form. Which shows how he could do that?

Mathematics

2 answers:

likoan [24]4 years ago

6 0

Answer:

y=3/2x-3

y=1/8x-3/4

Step-by-step explanation:

I plugged these equations into an algebra calculator, the result was this. I also did the question on edge. TRUST ME LOL

Jobisdone [24]4 years ago

5 0

Answer:

$y = \frac{3}{2}x - 3$

$y = \frac{1}{8} x - \frac{7}{8}$

Step-by-step explanation:

The given system is

3x-2y=6

0.4(20y+15)=x

The slope intercept form is y=mx+c.

For the first equation, 3x-2y=6, we add -3x to both side -2y=-3x+6

$y = \frac{3}{2}x - 3$

For 0.4(20y+15)=x, we expand to get:

$8y + 7 = x$

$8y= x - 7$

Divide through by 8

$y = \frac{1}{8}x - \frac{7}{8}$

Therefore the system in slope-intercept form is:

$y = \frac{3}{2}x - 3$

$y = \frac{1}{8}x - \frac{7}{8}$

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According to records from the last 50 years, there is a 2/3 chance of June temperatures being above 80 F. Azul needs to find the

dangina [55]

Here we shall use the binomial distribution formula to evaluate the probability:
let the probability that the temp will be above 80 be P(F)=2/3;
Probability that the temp will be below 80 be P(F')=1/3
thus,
The probability that the temp will be above 80 for 20 days for the month of June will be:
P(F)=n!/((n-x)!x!)p^xq^(n-x)
p=2/3 and q=1/3, n=30, x=20
plugging the values in the equation we obtain:
P(F)=30!/(10!*20!)*(2/3)^20*(1/3)^(30-20)
P(F)=0.153

6 0

4 years ago

Read 2 more answers

Could someone help me in this problem? <br><br><br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%20-%203%7D%7B14%7D%20%20%3D%

dalvyx [7]

Answer:

The answer is $x=3$ .

Step-by-step explanation:

Given:

$\frac{x-3}{14}=0$

Multiply by 14 on both the side, it gives

$\frac{x-3}{14}\times 14=0\times 14\\ x-3=0$ .

Now we add 3 on both sides.

$x-3+3=0+3\\x=3$

Therefore, the answer is $x=3$ .

6 0

4 years ago

Help<br> Help help help help

Shkiper50 [21]

Answer:

Step-by-step explanation:

$(2^{3})^{3} = 2^{3*3}=2^{9}\\\\\\(\dfrac{4^{2}}{5^{3}})^{3} = \dfrac{(4^{2})^{3}}{(5^{3})^{3}}=\dfrac{4^{6}}{5^{9}}\\\\\dfrac{3^{6}}{3^{3}}=3^{6-3}=3^{3}\\\\\\a^{m}*a^{n}=a^{m+n}\\\\(a^{m})^{n}=a^{m*n}\\\\\dfrac{a^{m}}{a^{n}}=a^{m-n}$