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AnnZ [28]
4 years ago
11

Which equation represents a line that passes through (4, 1/3) and has a slope of 3/4

Mathematics
2 answers:
Alex4 years ago
8 0

Answer:  9x - 12 y =32

Step-by-step explanation:

Since, the equation of a line passes through one point (x_1,y_1) and having the slope m is,

y-y_1 = m(x-x_1)

Here, x_1 = 4, y_1 = 1/3 and m= 3/4

Therefore, equation of the line,

y-1/3 = 3/4(x-4)

⇒ 3y -1 = 9/4 (x-4)

⇒ 12 y - 4 = 9x - 36

⇒ 9x - 12y = - 4 + 36

⇒ 9x +12y = 32

Which is required equation of the line.


Ipatiy [6.2K]4 years ago
5 0
Y = mx + b
m = slope of the line
b= y-intercept

Since you are given the point (4,1/3), the y-value is 1/3.
So plug in the numbers and solve for b:
(1/3)= (3/4)*(4) + b

After you solve the equation for b, you should get -2 2/3 or -8/3

No rewrite the equation with the y-intercept, b.

y = (3/4 x) - (8/3)
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4x^2-35x+49=0 solve by factoring
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Answer: x=7/4 or x=7

Step-by-step explanation:

step 1: Factor left side of equation.

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Step 2: Set factors equal to 0.

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3 0
3 years ago
A professor claims that his students' average score on the first exam of the semester is different than the average score on the
Marta_Voda [28]

Answer:

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=0.704

p_v = 2*P(t_{7} >0.704)=0.504

So the p value is higher the significance level given 0.1, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before score is equal 0. So then we don't have enough evidence to say that the score for the second exam is different than the score for the first exam.

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value before (first exam) , y = test value after (second exam)

The system of hypothesis for this case are:

Null hypothesis: \mu_y -\mu_x =0

Alternative hypothesis: \mu_y -\mu_x \neq 0

The first step is calculate the difference d_i=y_i-x_i

The statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=0.704

The next step is calculate the degrees of freedom given by:

df=n-1=8-1=7

Now we can calculate the p value, since we have a left tailed test the p value is given by:

p_v = 2*P(t_{7} >0.704)=0.504

So the p value is higher the significance level given 0.1, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before score is equal 0. So then we don't have enough evidence to say that the score for the second exam is different than the score for the first exam.

5 0
3 years ago
What is the answer to 18.7 and 18.6 rounding to the nearest tenth
aivan3 [116]
187=190 186=190 because 18.7 and 18.6 are 187 and 186 so 8 is the tenth place. So they both equal 190
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