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Fofino [41]
2 years ago
7

Use the graph to find the

Mathematics
1 answer:
Dmitry [639]2 years ago
7 0
The y intercept is 2, because that is the value of y when it crosses the y axis.

Since the line is pointing down, it has a negative slope, and since the y value goes down 2 for every 3 x values, the slope is -2/3. You should always right the slope in rise/run form, or change in y/change in x.

Finally, write the equation in the form of y=mx+b, where m is the slope and b is the y intercept. This means the equation of the line is y=-2/3x+2.
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A newspaper provided a​ "snapshot" illustrating poll results from 1910 professionals who interview job applicants. The illustrat
Darya [45]

Answer:

confidence level is missing

Step-by-step explanation:

<em>1.confidence level  </em>

The results can be given only in a predetermined confidence level

<em>2. point estimate</em>

The illustration states the estimate 26% of the professionals who interview job applicants said the biggest interview turnoff is that the applicant did not make an effort to learn about the job or the company.

<em>3.sample size</em>

Sample size is given as 1910 people

<em>4.confidence interval </em>

Confidence interval is given ±3 around the point of estimate

3 0
3 years ago
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
amm1812
3^2 + 4^2 = 5^2
9 + 16 = 25
25 = 25

5^2 + 12^2 = 13^2
25 + 144 = 169
169 = 169

9^2 + 12^2 = 15^2
81 + 144 = 225
225 = 225

<span>a^2 + b^2 is = to c^2</span>
8 0
3 years ago
Simplify<br> 21^6y^5/14x^2y^9
nevsk [136]

1 2 2 5 2 3 0 3 ^1 4 ^2/2

6 0
3 years ago
Consider the differential equation y'' − y' − 30y = 0. Verify that the functions e−5x and e6x form a fundamental set of solution
Alex Ar [27]

Answer:

Step-by-step explanation:

We have to take the derivatives for both functions and replace in the differential equation. Hence

for y=e^{-5x}:

y(x)=e^{-5x}\\y'(x)=-5e^{-5x}\\y''(x)=25e^{-5x}\\

for y=e^{6x}:

y(x)=e^{6x}\\y'(x)=6e^{6x}\\y''(x)=36e^{6x}\\

Now we replace in the differential equation  y'' − y' − 30y = 0

for y=e^{-5x}:

25e^{-5x}+5e^{-5x}-30e^{-5x}=0\\25+5-30=0

for y=e^{6x}:

36e^{6x}-6e^{6x}-30=0\\36-6+30=0

Now, to know if both function are linearly independent we calculate the Wronskian

W(f,g)=fg'-f'g

W(e^{-5x},e^{6x})=(e^{-5x})(6e^{6x})-(-5e^{-5x})(e^{6x})\neq 0

I hope this is useful for you

Best regard

7 0
4 years ago
Please help !!!!!!!!!!!!
klasskru [66]
This is the answer/Hope this helps

5 0
3 years ago
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