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3 years ago

45-29it is hard I get comefused

Mathematics

1 answer:

Korvikt [17]3 years ago

6 0

The answer is 16 for 45-29

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Represent A Quadratic Function Using:

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Sub the Xs in the eqn

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2 years ago

Lori wants to trade more than 3/10 but less than 4/5 of her stickers.

mote1985 [20]

Answer:

7/10, 6/10, or 5/10

Step-by-step explanation:

So, first, we change 4/5 into tenths. That equals 8/10. 3/10 and 8/10.

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3 years ago

Calista boxes items for a packaging company. She can use only cube-shaped boxes, as

Shtirlitz [24]

Answer:

0.295in^3

Step-by-step explanation:

I divide the 56 by 8 by 27 and got 295 reaped and i keep the exponent and the inces

6 0

3 years ago

Step by step how to solve -2.3x=.46

Igoryamba

<u>-2.3 x = 0.46</u>
Step 1:
Divide each side by -2.3 : x = - 0.46 / 2.3

Step 2:
To simplify the fraction,
divide -0.46 by 2.3 : <em>x = - 0.2
</em>

7 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