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3 years ago

in a bag the coins of 50 Paisa 25 paisa and 10 paisa are in the ratio of 2:3: 5 and its total value is rupees 90

Mathematics

1 answer:

IgorLugansk [536]3 years ago

5 0

Answer:

It is given 50 paise, 25 paise and 10 paise are in the ratio 2:3:5. =4:3:2.

Step-by-step explanation:

Hope this helps

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Solve for each variable.

dem82 [27]

Answer:

a = 38⁰

b = 54⁰

c = 54⁰

Step-by-step explanation:

a + 142⁰ = 180⁰ because a straight line = 180⁰

b + 38⁰ + 88⁰ = 180⁰ because the angles of a triangle equal 180⁰, and the missing measurement = 38⁰ because it is a vertical angle to angle a

c + 88⁰ = 142⁰ because the angles are corresponding angles

5 0

3 years ago

If the filling equipment is functioning properly what is the probability that the volume of oil in a randomly selected barrel wi

Sophie [7]

Answer:

$P(X>55.4)=P(\frac{X-\mu}{\sigma}>\frac{55.4-\mu}{\sigma})=P(Z>\frac{55.4-55}{0.5})=P(Z>0.8)$

$P(Z>0.8)=1-P(Z\leq 0.8)$

$P(Z>0.8)=1-0.788=0.212$

Step-by-step explanation:

Assuming this previous info : "Trucks carry barrels of crude oil from a port in Texas to a distributer in New Mexico on long trailers. Filling equipment is used to fill the barrels with the oil. When functioning properly, the actual volume of oil loaded into each barrel by the filling equipment at the port is approximately normally distributed with a mean of 55 gallons and a standard deviation of 0.5 gallons. If the mean is greater than 55.4 gallons, the filling mechanism is overfilling."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the amount filling of a population, and for this case we know the distribution for X is given by:

$X \sim N(55,0.5)$

Where $\mu=55$ and $\sigma=0.5$

We are interested on this probability

$P(X>55.4)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>55.4)=P(\frac{X-\mu}{\sigma}>\frac{55.4-\mu}{\sigma})=P(Z>\frac{55.4-55}{0.5})=P(Z>0.8)$

And we can find this probability using the complement rule:

$P(Z>0.8)=1-P(Z\leq 0.8)$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(Z>0.8)=1-0.788=0.212$

8 0

2 years ago

Simplify (x^3/8)^3/4

Viktor [21]

When an exponent is raised to another exponent, you can multiply the powers to simplify them.

3/8*3/4

*Multiply numerators*

3*3=9

*Multiply denominators*

8*4=32

x^9/32

Hope this helps!!

4 0

2 years ago

Read 2 more answers

Base area: 12.5 m2; height: 1.2 m

Novay_Z [31]

Answer:

Volume = 150 m^2

Step-by-step explanation:

Using Volume formula of prism, we get L*W*H, but since we know the area of the base is L*W, which is 12.5, we multiply 12.5 by 12 to get 150 m^2.

7 0

3 years ago

[2/(n+2)] - (3/n+2) = 5 Solve by u-substitution

IgorLugansk [536]

let n+2=u so, the equation became= [2/u]-[3/u]=5 => [(2+3)/u]=5 => 5/u=5 => u=5/5=1 thus, u=1 we know u=n+2 so, n+2=1 => n=1-2=-1 so, n=-1

7 0

2 years ago

in a bag the coins of 50 Paisa 25 paisa and 10 paisa are in the ratio of 2:3: 5 and its total value is rupees 90​

in a bag the coins of 50 Paisa 25 paisa and 10 paisa are in the ratio of 2:3: 5 and its total value is rupees 90