All categories

2 years ago

A gallon of paint covers 400 square feet. How many square feet will 2 3/8 gallons of paint cover?

1 answer:

8 0

1 gal = 400 ft²
2 3/8 gal = ? ft²

to solve, use similar ratios with units gal/ft² and compare

(1)/(400) = (2 3/8)/(x)

*multiply both sides by (x) and divide both sides by (1)/(400), you get:

x = (400)*(2 3/8)/(1) = 400*(2 3/8) = 400*2 + 400*3/8 = 800 + 150 = 950 ft²

<u><em>answer is 950 ft²</em></u>

You might be interested in

Every weekend, jaylon downloads interesting music on his phone from an app. He downloads 15 songs at $1.99 and 7 songs at $2.50.

butalik [34]

Answer:

15*1.99

7*2.50

Step-by-step explanation:

I think you would multiply them (im not sure im not very good with math if its wrong im really sorry hope you get it though good luck :) ! )

6 0

2 years ago

Find the point estimate of the proportion of people who wear hearing aids if, in a random sample of 493 people, 40 people had he

Anna71 [15]

Point estimate proportion is the average of lower and upper interval limit.

Point estimate proportion )p) = x/n
Where x = observations, n=sample size

Substituting;

p= 40/493 = 0.0811

8 0

2 years ago

Read 2 more answers

A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$