What are the soulutions of x^2+6x-6=10?

Mrs.Anderson is giving a test in her third period class she has decided to record the amount of time that each student takes to

ICE Princess25 [194]

The graph Plot of the given results shown represents; A relation only

<h3>How to interpret a scatter plot?</h3>

For it to be a function, then no x-value can produce more than 2 y-values. However, we see that at x = 25, y has 2 values. Similarly, at x = 15, y has two values.

Thus, the graph plot can only be a relation as it is not a function as seen from the points plotted in the graph.

The complete question is;

Mrs. Anderton is giving a test in her third-period class. She has decided to record the amount of time that each student takes to finish the test (in minutes) and compare that to the grade each student receives on the test (out of 100). A plot of her results is below. Which of the following does this situation represent?

A. both a relation and a function

B. a function only

C. neither a relation nor a function

D. a relation only

Read more about Scatter Plot at; brainly.com/question/6592115

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8 0

2 years ago

15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.

Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = $30C5 = \frac{30 !}{(30-5)!5!} \\$

$30C5 = 142506$ ways

Number of ways of selecting 5 good bulbs from 26 bulbs = $26C5 = \frac{26 !}{(26-5)!5!} \\$

$26C5 = 65780$ ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = $\frac{26C4 * 4C1}{30C5}$

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) = $\frac{26C3 * 4C2}{30C5}$

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) = $\frac{26C2 * 4C3}{30C5}$

Pr(3 defective) = 0.009

Pr(4 defective) = $\frac{26C1 * 4C4}{30C5}$

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0

3 years ago

Help and get 6 points!

timofeeve [1]

10 hours was the full time she painted

7 0

3 years ago

Read 2 more answers

I'm not sure where to start and how to differentiate this.

Mazyrski [523]

The picture is not clear. let me assume y = (x^4)ln(x^3)

product rule :
d f(x)g(x) = f(x) dg(x) + g(x) df(x)

dy/dx = (x^4)d[ln(x^3)/dx] + d[(x^4)/dx] ln(x^3)
= (x^4)d[ln(x^3)/dx] + 4(x^3) ln(x^3)

look at d[ln(x^3)/dx]
d[ln(x^3)/dx]
= d[ln(x^3)/dx][d(x^3)/d(x^3)]
= d[ln(x^3)/d(x^3)][d(x^3)/dx]
= [1/(x^3)][3x^2] = 3/x
... chain rule (in detail)

end up with
dy/dx = (x^4)[3/x] + 4(x^3) ln(x^3)
= x^3[3 + 4ln(x^3)]

5 0

3 years ago

A triangle has angle measurements of 56°, 56°, and 68°. Is this triangle scalene?

slamgirl [31]

Answer:

No. It is an isoceles triangle. A scalene triangle would have NO equal angles.

Step-by-step explanation:

3 0

3 years ago