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Licemer1 [7]
3 years ago
6

What is 48 - (-31) Pre-Algebra

Mathematics
1 answer:
pav-90 [236]3 years ago
6 0

Answer:

it equals 79

Step-by-step explanation:

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I need help please!!
Semmy [17]

Answer:

Step-by-step explanation:

5.   <2=<10=102

<3=<9=78

<4=<10=102

<5=<7=38

<6=180-38=142

<8=142

<9=180-102=78

<11=<9=78

<12=<10=102

<13=<7=38

<14=<16=<6=<8=180-38=142

<15=<7=38

<16=142

5 0
3 years ago
Triangle is three centimeters longer than twice the length of th shortest side.the thrid side is twice as long as the shortest s
aliya0001 [1]

Answer:

17

Step-by-step explanation:

x + 2x + 3 + 2x = 88

5x = 85

x = 17

3 0
4 years ago
Solve the equation.<br> - 6y + 8 = - 4(2y + 1)
Alex_Xolod [135]

Answer:y=2/7

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
A box contains 20 lightbulbs, of which 5 are defective. If 4 lightbulbs are picked from the box randomly, the probability that a
madam [21]

Answer:

Option D. 938/969

Step-by-step explanation:

At most 2 defective means 2 or less than 2 bulbs are defective

So, we have 3 cases:

a) No defective bulb.       b) 1 defective bulb.      c) 2 defective bulbs

Case a) No defective bulb

Total number of bulbs = 20

Number of defective bulbs = 5

Number of non-defective bulbs = 15

Total number of ways to select 0 defective bulb = 15C4 = 1365

Case b) 1 defective bulb

Total number of ways to select 1 defective bulb = 15C3 x 5C1 = 2275

Case c) 2 defective bulbs

Total number of ways to select 2 defective bulbs = 15C2 x 5C2 = 1050

Therefore, total number of ways to select at most 2 defective bulbs = 1365 + 2275 + 1050 = 4690

Total number of ways to select 4 bulbs from 20 = 20C4 = 4845

Therefore, probability of selecting at most 2 defective bulbs = \frac{4690}{4845}=\frac{938}{969}

Therefore, option D gives the correct answer.

6 0
4 years ago
Is my answer correct or incorrect? True or false
olchik [2.2K]

Answer:

Youre correct

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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