Can you help me please??

Help me!! A water tank is 1/8 full. The tank is 3/4 full when 42 gallons of water is added to the tank. How much can the water t

chubhunter [2.5K]

Answer(s):

1.) 56 gallons

2.) 7 gallons

3.) 28 gallons

Step-by-step explanation:

Q1: How much water can the tank hold?

Based on the question, we know that 3/4 of the tank equals 42 gallons. Divide 42 by the numerator, 3.

42/3 = 14

Now, multiply 14 by the denominator, 4.

14 x 4 = 56.

So, the tank can hold 56 gallons. [If you'd like to double check this answer, take the fraction 42/56 and simplify it to get 3/4.

Q2: How much water was originally in the tank?

We know that the tank's maximum capacity is 56 gallons. So, if it was 1/8 originally full, multiply 56 by 1/8. [Remember that 56 as a fraction can be written as 56/1]

1/8(56/1) = 7

This means that the tank was originally 7 gallons full. [If you'd like to double check this answer, take the fraction 1/8 and multiply the numerator and denominator by 7. This gets you 7/56, or 7 gallons out of the total capacity, 56 gallons.]

Q3: How much water is in the tank when it is 1/2 full?

Earlier, we found out that the maximum capacity of the tank is 56 gallons. To find out how many gallons it holds when its half full, divide 56 by 2.

56/2 = 28

So, when the tank is 1/2 full, it contains 28 gallons. [If you want to double check this answer, take the fraction 28/56 and simplify it. When you do, you get 1/2.]

5 0

3 years ago

Read 2 more answers

A coin is thrown at random into the red triangle below.

Yakvenalex [24]

it is impossible

Step-by-step explanation:

8 0

2 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

$\mu_s=np=500*0.90=450$

The standard deviation will be:

$\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7$

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

$z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932$

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

$\mu=np=300*0.932=279.6$

The standard deviation will be:

$\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36$

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

$P(270\leq x\leq280)=P(269.5$