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Rina8888 [55]
3 years ago
6

Solve for x please. :)

Mathematics
1 answer:
solmaris [256]3 years ago
6 0

Answer:

180-130=50

Step-by-step explanation:

done hope you like it

i use the 180 because its a straight line from the x to the 130

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The graph of a line and an exponential can intersect twice, once or not at all. Describe the possible number of intersections fo
Katyanochek1 [597]

Answer:

  c) parabola and circle: 0, 1, 2, 3, 4 times

  d) parabola and hyperbola: 1, 2, 3 times

Step-by-step explanation:

c. A parabola can miss a circle, be tangent to it in 1 or 2 places, intersect it 2 places and be tangent at a 3rd, or intersect in 4 places.

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d. A parabola must intersect a hyperbola in at least one place, but cannot intersect in more than 3 places. If the parabola is tangent to the hyperbola, the number of intersections will be 2.

If the parabola or the hyperbola are "off-axis", then the number of intersections may be 0 or 4 as well. Those cases seem to be excluded in this problem statement.

7 0
3 years ago
Which of the following is not a real number?
givi [52]

Answer:

im pretty sure its the -3 one

Step-by-step explanation:

7 0
3 years ago
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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
What is the ratio for each one?
blagie [28]
This is complicated because I’m typing on a phone, but
24:30 simplified is 4:5
30:54 simplified is 5:9

10:5 simplified is 2:1
5:15 simplified is 1:3

32:72 simplified is 4:9
72:104 simplified is 9:13

56:7 simplified is 8:1
7:63 simplified is 1:9
8 0
3 years ago
What is the value of x?<br> 16<br> 34
lord [1]

Answer:30

Step-by-step explanation:

7 0
3 years ago
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