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4 years ago

What is the following sum? 4√5+2√5 a. 6√10 b. 8√10 c. 6√5 d. 8√5

2 answers:

6 0

Since the radicands (the number under the radical) are the same, add the coefficients and keep the radicand.
4√5 + 2√5 = 6√5
LETTER C

Ber [7]4 years ago

5 0

For this case we have the following expression:

$4 \sqrt {5} +2 \sqrt {5}$

We can rewrite the expression by taking a common factor.

We have then:

$\sqrt {5} (4 + 2)$

From here, add the expression within the parenthesis we have:

$6 \sqrt {5}$

Answer:

The equivalent expression for $4 \sqrt {5} +2 \sqrt {5}$ is given by:

$6 \sqrt {5}$

Option C

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hram777 [196]

Answer:

The reliability of the first system to work is 0.72 whereas the reliability of the second system to work is 0.98.As the reliability of the second system is more than the first one so the second system is more reliable.

Step-by-step explanation:

For first system as given in the attached diagram gives,

$P(W_1W_2)=P(W_1) \times P(W_2)$

As the systems are independent.

The given data indicates that

$P(W_1)$ is given as 0.9
$P(W_2)$ is given as 0.8

Now the probability of the system is given as

$P(W_1W_2)=P(W_1) \times P(W_2)\\P(W_1W_2)=0.9 \times 0.8\\P(W_1W_2)=0.72$

So the reliability of the first system to work is 0.72.

For the second system is given as

$P(W_1W_2)=1-P(W_1'W_2')$

Where

$P(W_1'W_2')$ is the probability where both of the systems does not work. this is calculated as

$P(W_1'W_2')=P(W_1') \times P(W_2')\\P(W_1'W_2')=(1-P(W_1)) \times (1-P(W_2))\\P(W_1'W_2')=(1-0.9) \times (1-0.8)\\P(W_1'W_2')=(0.1) \times (0.2)\\P(W_1'W_2')=0.02$