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suter [353]
3 years ago
14

Y=3/8x−4 write in standard form

Mathematics
2 answers:
Naddik [55]3 years ago
5 0
Answer:

y=3/8x-4
Move the expression to the left
y-3/8x=-4
Multiply both sides by 8
8y-3x=-32
Reorder the terms
-3x+8y=-32
Change the signs and standard form is...
3x-8y=32


andriy [413]3 years ago
3 0

Answer:

3x-8y=32

Step-by-step explanation:

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ELEN [110]
Maybe use photo math i guess
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What is the common denominator for ¾ and 9/10?
Contact [7]

Answer:

20 so the fractions would be 3/4=15/20 and 9/10=18/20

Step-by-step explanation:

because 4 x 5 = 20 and 10 x 2 = 20

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URGENT!!!!!! Find the value of q in the following system so that the solution to the system is——
valkas [14]
<h3>Answer:  Q = 8</h3>

====================================================

Explanation:

The left hand side of the first equation is x-3y

The left hand side of the second equation is 2x-6y = 2(x-3y). Note how it's simply double of the first expression x-3y

If we multiply both sides of the first equation by 2, we get

x-3y = 4

2(x-3y) = 2*4

2x-6y = 8

Meaning that 2x-6y = 8 is equivalent to x-3y = 4. Both produce the same line leading to infinitely many solutions. Each solution will lay along the line x-3y = 4.

We can say each solution is in the set {(x,y): x-3y = 4}

Which is the same as saying each solution is of the form (3y+4,y)

4 0
3 years ago
We are interested in the amount that students study per week. Suppose you collected the following data in hours {4.4, 5.2, 6.4,
olga nikolaevna [1]

Answer:

Step-by-step explanation:

Hello!

The objective is to estimate the average time a student studies per week.

A sample of 8 students was taken and the time they spent studying in one week was recorded.

4.4, 5.2, 6.4, 6.8, 7.1, 7.3, 8.3, 8.4

n= 8

X[bar]= ∑X/n= 53.9/8= 6.7375 ≅ 6.74

S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[376.75-(53.9²)/8]= 1.94

S= 1.39

Assuming that the variable "weekly time a student spends studying" has a normal distribution, since the sample is small, the statistic to use to perform the estimation is the student's t, the formula for the interval is:

X[bar] ± t_{n-1;1-\alpha /2}* (S/√n)

t_{n-1;1-\alpha /2}= t_{7;0.975}= 2.365

6.74 ± 2.365 * (1.36/√8)

[5.6;7.88]

Using a confidence level of 95% you'd expect that the average time a student spends studying per week is contained by the interval [5.6;7.88]

I hope this helps!

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