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3 years ago

What is 14/9k+13/9-2k+2/9+2/9k-6

Mathematics

1 answer:

Ierofanga [76]3 years ago

8 0

Answer:

both answers are pictures below

Step-by-step explanation:

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I know it is basic but I need help 11m+13=m+23

zloy xaker [14]

Answer:

m = 1

Step-by-step explanation:

11m + 13 = m + 23

Get m by itself by subtracting m from both sides.

10m + 13 = 23

Subtract 13 from both sides.

10m = 10

DIvide by 10.

m = 1

8 0

3 years ago

Read 2 more answers

Translate the word phrase into a variable expression. five more than the product of x and 6

Natasha_Volkova [10]

Answer:

6X+5

Step-by-step explanation:

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3 years ago

Helpppppp for 10points !!!

g100num [7]

Slope is x so it’s -1/5
pretty sure
haven’t done this in a year

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3 years ago

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Jefferson works part-time and earns $1,520 in four weeks how much does he earn each week

EleoNora [17]

Answer: $380 per week

Step-by-step explanation:

Assuming that Jefferson works the same amount each week, you need to divide the total that he earns by four weeks in order to determine how much he made each week.

$1,520 divided by 4 equals $380 per week.

5 0

3 years ago

A system consists of two components C1 and C2, each of which must be operative in order for the overall system to function. Let

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

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

So now the probability of the second system is given as

$P(W_1W_2)=1-P(W_1'W_2')\\P(W_1W_2)=1-0.02\\P(W_1W_2)=0.98$

So 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.

8 0

3 years ago