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Dvinal [7]
3 years ago
12

Please help! Must get done ASAP!!! ^^

Mathematics
2 answers:
pentagon [3]3 years ago
3 0

Answer:

No

No

Yes

Yes

Step-by-step explanation:

Try it.

lara [203]3 years ago
3 0

Answer: 1st: No, 2nd: No, 3rd: Yes, 4th: Yes

Step-by-step explanation: You only flip the inequality when you divide a negative number on both sides. The only reason why you would divide would be to get the variable by itself -  in this case the coefficient would need to be negative. That only applies to the third and 4th equations.

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What is the surface area of a right circular cylindrical oil can, if the radius of its base is 4 inches and its height is 11 inc
Alla [95]

Step-by-step explanation:

Given

Radius (r) = 4 inches

height (h) = 11 inches

Total surface area of cylindrical oil can is

= 2πr ( r + h)

= 2 * 22 / 7 * 4 ( 4 + 11)

= 2 * 22/7 * 4 * 15

= 377.14 inches ²

4 0
3 years ago
Read 2 more answers
A side view of the top of a building is shown in the diagram below. The measure of angle MNT is 72 degrees. The measure of angle
bulgar [2K]

Answer:

142°

Step-by-step explanation:

Make a line parallel to RT to create alternate internal angles

7 0
3 years ago
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A bicycle lock has a four-digit code. The possible digits, 0 through 9, cannot be repeated. What is the probability that the loc
kozerog [31]
First let's find the number of the elements in the sample space, that is the total number of codes that can be produced.

The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one. 
and so on...

so in total there are 10*9*8*7 = 5040 codes.

a. What is the probability that the lock code will begin with 5?

Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.

Thus, there are 1*9*8*7=504 codes which start with 5.

so 

P(first digit is five)=\frac{n(first -digit- is- 5)}{n(all-codes)}= \frac{1*9*8*7 }{10*9*8*7 }= \frac{1}{10}=0.1

b. <span>What is the probability that the lock code will not contain the number 0? 

from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}

from which we can form in total 9*8*7*6 codes which do not contain 0.

P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6

Answer:

0.1 ; 0.6</span>
4 0
3 years ago
Read 2 more answers
Help ima mark BRAINLIST
maw [93]

Answer:

40%

Step-by-step explanation:

There are 5 seniors (2+3)

2 of them are males

P ( male given a senior)= male/ senior

                                       2/5

                                     .40

                                      40%

3 0
3 years ago
Math Test Helpppppp!!!!!!!!
soldier1979 [14.2K]
Simple problem:

When simplifying 3(x-6) they subtracted 3 instead of multiplied by 3.
3(x-6) should be 3x-18.
Now when you simplify:
3x-18+4x+12-6x
Which is x -6 =0
Or x =6
3 0
3 years ago
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