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xz_007 [3.2K]
1 year ago
14

Determine whether the dimensions of each rectangle described of Taylor's first rectangle.

Mathematics
1 answer:
Svetllana [295]1 year ago
8 0

Step-by-step explanation:

yes

no

yes

yes

that is the correct answer

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A set of 3 consecutive integers has a sum of 27. which integers are they
Scilla [17]
They are 8, 9 and 10. 
5 0
3 years ago
Determine f^-1(2) given the following: f(1 = -3f(2) = -2f(-1) = 2f(3) = -1
lana [24]

We assume that the function is as follows:

f^{-1_{}}(2)=?

Then, we have that:

f^{-1}(x)=\frac{1}{f(x)}

Thus, if we have that:

f(2)=-2\Rightarrow f^{-1^{}_{}}(x)=\frac{1}{f(x)}\Rightarrow f^{-1}(2)=\frac{1}{f(2)}=\frac{1}{-2}\Rightarrow f^{-1}(2)=-\frac{1}{2}

Therefore, the value for the function is f^-1(2) = -1/2 or

f^{-1}(2)=-\frac{1}{2}

8 0
11 months ago
Which one is it pleaseee helppppp!
ladessa [460]

Answer:

-2/1

Step-by-step explanation:

i hope this helps :)

5 0
3 years ago
Read 2 more answers
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
Consider the equation 5x-2y=3. If possible, find a second linear equation to create a system of equations that has: Exactly one
nekit [7.7K]
A) One solution: 5x +2y = 0 . . . . (any line with a different slope)

b) Two solutions: not possible

c) No solutions: 5x -2y = 0 . . . . (any different line with the same slope)

d) Infinitely many solutions: 10x -4y = 6 . . . . (any other equation for the same line)
4 0
3 years ago
Read 2 more answers
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