10 liters = 100,000 millimeters True or False?

A glass is

Marrrta [24]

total volume of the glass is $}x =200cm^3$ .

Step-by-step explanation:

Here we have , A glass is 1/4 full. Then 70cm3 of orange juice is poured in.

The glass is now 3/5 full. We need to find that What is the total volume of the glass . Let's find out :

Suppose that volume of glass is x ,so according to question we have the following equation

$\frac{x}{4}+70 = \frac{3x}{5}$ , Now, solve this to find value of x

⇒ $\frac{x}{4}+70 = \frac{3x}{5}$

⇒ $70 = \frac{3x}{5}- \frac{x}{4}$

⇒ $70 = \frac{3x((4)-x(5)}{5(4)}$

⇒ $70 = \frac{12x-5x}{20}$

⇒ $70(20) = 12x-5x$

⇒ $7x = 1400$

⇒ $\frac{7}{7}x = \frac{1400}{7}$

⇒ $}x =200cm^3$

Therefore, total volume of the glass is $}x =200cm^3$ .

5 0

3 years ago

Identify the coefficient in the expression x + 4. A x B 1 C 0 D 4

Hunter-Best [27]

Answer:

1

Step-by-step explanation:

The coefficient is the number in front of the variable. Since there is not a number written, there is an implied 1

x+4

1x+4

The coefficient is 1

5 0

3 years ago

-4.7 divided by 1\4 ????????????????

Nady [450]

Answer:

-18.8

Step-by-step explanation:

1/4 as a decimal is 0.25 so then if you divide -4.7 by 0.25 the answer is -18.8 :)

I hope this helped :)

3 0

3 years ago

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The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection.

This means that,

Probability that at least one of them needs replacement at the first inspection =

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago

I need this question done by March 4,2019 What is the answer to a + 2 = 5 + 4a ?

Ronch [10]

A = -1

Hope this is right

3 0

3 years ago

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