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3 years ago

400 is 1/10 of what number

Mathematics

1 answer:

den301095 [7]3 years ago

4 0

400 is 1/10 of 4000.

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Help me please<br><br> x/3+12=2

victus00 [196]

Answer:

Step-by-step explanation:

minus 12 on both sides

updated equation: x/3=-10

multiply 3 on both sides

updated equation

x = -30

3 0

2 years ago

Please help I'm trying to study and don't know answer

Alexxx [7]

Since the total is 42 divide by two since they are the same length so it would be 21

8 0

3 years ago

For questions 2 and 3, simplify the polynomial.

Dafna11 [192]

3. B
Hope this helps ..........................

4 0

3 years ago

HELP ASAP!!!!!!

WARRIOR [948]

Answer:

x=26

C = 91

Step-by-step explanation:

The sum of the angles of a triangle add to 180 degrees

A + B+ C = 180

35+ 52+ 3(x+5) = 180

Distribute

35+52+3x+15 = 180

Combine like terms

102 +3x = 180

Subtract 102 from each side

102-102 +3x=180-102

3x= 78

Divide by 3

3x/3 = 78/3

x=26

Now we need to find angle C

c = 3(x+5)

C = 3(26+5)

3(31)

6 0

3 years ago

If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calcul

Wittaler [7]

Answer:

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

3 failures every twenty weeks

This means that for 1 week, $\mu = \frac{3}{20} = 0.15$

Calculate the probability that there will not be more than one failure during a particular week.

Probability of at most one failure, so:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

Then

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.15}*0.15^{0}}{(0)!} = 0.8607$

$P(X = 1) = \frac{e^{-0.15}*0.15^{1}}{(1)!} = 0.1291$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.8607 + 0.1291 = 0.9898$

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

6 0

2 years ago