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yarga [219]
3 years ago
8

It takes a computer 1.2 seconds to perform an operation. If a task contains eight operations how many seconds will it take for c

omputer to perform the task twice
Mathematics
2 answers:
Radda [10]3 years ago
8 0

Answer:

The answer is 9.6 seconds

serg [7]3 years ago
7 0

Answer:

19.2 seconds

Step-by-step explanation:

1.2 times 8 which equals 9.6

9.6 times 2

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Apply Euclid algorithm to find HCF of number 4052 and 420​
Sergeeva-Olga [200]

Answer:

ANSWER 4

Step-by-step explanation:

the hcf is 420 and then u say 2x2=4 simple

7 0
2 years ago
One solution of x^2-25=0 is 5 what is the other solution
zepelin [54]

Add 25 to both sides:


x^2 = 25


Square root both sides.


When you square root a perfect square, you will get a plus/minus result:


\sqrt{25} = \pm 5


x = \pm 5

\boxed{x = -5} \land \boxed{x = 5}


The other solution to this answer is x = -5.

5 0
4 years ago
Consider the Inequality <br> —2n +7 &gt; -11<br> What is the solution of the inequality
Marta_Voda [28]

Answer:

  • n < 9 or n = (-oo, 9)

Step-by-step explanation:

  • -2n +7 > -11
  • -2n > -11 - 7
  • -2n > -18
  • n < -18/-2
  • n < 9
  • n = (-oo, 9)
5 0
3 years ago
A commuter crosses one of three bridges, A, B, or C, to go home from work. The commuter crosses bridge A with probability 1/3, c
mihalych1998 [28]

Answer:

0.1333 = 13.33% probability that bridge B was used.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Arrives home by 6 pm

Event B: Bridge B used.

Probability of arriving home by 6 pm:

75% of 1/3(Bridge A)

60% of 1/6(Bridge B)

80% of 1/2(Bridge C)

So

P(A) = 0.75*\frac{1}{3} + 0.6*\frac{1}{6} + 0.8*\frac{1}{2} = 0.75

Probability of arriving home by 6 pm using Bridge B:

60% of 1/6. So

P(A \cap B) = 0.6*\frac{1}{6} = 0.1

Find the probability that bridge B was used.

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.75} = 0.1333

0.1333 = 13.33% probability that bridge B was used.

8 0
3 years ago
The theater arts committee estimated that it would sell 200 tickets for its upcoming play. The committee actually sold 224 ticke
MA_775_DIABLO [31]
12% error. To find out subtract the difference between the estimate and the actual amount sold and divide the difference by the estamite.
3 0
3 years ago
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