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vfiekz [6]
2 years ago
7

2y+5x=4 please answer

Mathematics
1 answer:
marshall27 [118]2 years ago
5 0

Answer:x=x=

−2

5

y+

4

5

Step-by-step explanation:

Let's solve for x.

2y+5x=4

Step 1: Add -2y to both sides.

5x+2y+−2y=4+−2y

5x=−2y+4

Step 2: Divide both sides by 5.

5x

5

=

−2y+4

5

x=

−2

5

y+

4

5

Answer:

x=

−2

5

y+

4

5

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Write the standard form of the line that contains a slope -3/8 and passes through the point (5, -4).
Kipish [7]

Answer:

y+3/8x=-17/8

Step-by-step explanation:

y=mx+b

y=-3/8x+b

-4=-3/8(5)+b

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-32/8+15/8=b

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y=-3/8x-17/8

change to standard form where x and y are on one side of the equation

y+3/8x=-17/8

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Answer:

The formula for a quadrilateral depends on the shape.

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2 years ago
Over what interval is the function differentiable? Y=3/2(x-1)
jolli1 [7]

<u>Answer-</u>

<em>The interval where the function is diffrentiable is </em>

[-\infty,1)\ \bigcup\ (1,\infty]

<u>Solution-</u>

The given expression is,

y=\dfrac{3}{2(x-1)}

The function will be differentiable where it is continuous and it will not be differentiable, where the function is not continuous.

The function continuous everywhere except at x = 1, because

\Rightarrow 2(x-1)=0\\\\\Rightarrow x-1=0\\\\\Rightarrow x=1

at x = 1, its limit does not exist.

Therefore, apart from x=1, this function is differentiable everywhere. The interval will be

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3 0
2 years ago
A survey of 25 randomly selected customers found the ages shown (in years). The mean is 32.64 years and the standard deviation i
____ [38]

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Part A</u>

Assuming that conditions have been met for the interval, we use the formula \displaystyle CI=\bar{x}\pm t\frac{s}{\sqrt{n}} where \bar{x} represents the sample mean, t represents the critical value, s represents the sample standard deviation, and n is the sample size.

The critical value of t for an 80% confidence level with degrees of freedom df=n-1=25-1=24 is equivalent to t=1.317836

Thus, we can compute the confidence interval:

\displaystyle CI=\bar{x}\pm t\frac{s}{\sqrt{n}}\\\\CI=32.64\pm1.317836\biggr(\frac{9.39}{\sqrt{25}}\biggr)\\\\CI\approx\{30.17,35.11\}

Therefore, we are 80% confident that the true mean age of all customers is between 30.17 and 35.11 years.

<u>Part B</u>

The margin of error is \displaystyle t\frac{s}{\sqrt{n}}=1.317836\biggr(\frac{9.39}{\sqrt{25}}\biggr)\approx2.47

6 0
2 years ago
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