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rodikova [14]
2 years ago
13

You need 12 gallons of paint to paint the exterior of your house. At one store the paint sells for $19.95 a gallon. At a second

store the same brand of paint sells for $37.98 for two gallons. Which store is less expensive?
Mathematics
2 answers:
Ronch [10]2 years ago
8 0
(12*19.95) : (37.98*(12/2))
$239.4 : $227.88
The second store is less expensive
FromTheMoon [43]2 years ago
3 0
The second store is cheaper. It is only 18.99 per gallon. Just divide the costs by 2. 
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PLEASE HELPPPP , WILL GIVE BRAINLIESTTTTT
insens350 [35]

Answer:

80%

Step-by-step explanation:

What I did was just play with the numbers. But first, you'd multiply 6.00 time 3 because it's for the three people. Then just multiply 18 with .8 and bingo! The answer you get is 14.40, so the discount would be 80%

Hope this helps!

5 0
1 year ago
Read 2 more answers
An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t
Neko [114]

Answer:

35.2% probability that the sample mean will be 246 pages or more

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 245 \sigma = 15, n = 33, s = \frac{15}{\sqrt{33}} = 2.61

What the probability that the sample mean will be 246 pages or more?

This is 1 subtracted by the pvalue of Z when X = 246. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{246 - 245}{2.61}

Z = 0.38

Z = 0.38 has a pvalue of 0.6480.

1 - 0.6480 = 0.3520

35.2% probability that the sample mean will be 246 pages or more

4 0
2 years ago
An urn contains six chips, numbered 1 through 6. two are chosen at random and their numbers are added together. what is the prob
vitfil [10]
<span>2/15 if drawn without replacement. 1/9 if drawn with replacement. Assuming that the chips are drawn without replacement, there are 6 * 5 different possibilities. And that's a low enough number to exhaustively enumerate them. So they are: 1,2 : 1,3 : 1,4 : 1,5 : 1,6 2,1 : 2,3 : 2,4 : 2,5 : 2,6 3,1 : 3,2 : 3.4 : 3,5 : 3,6 4,1 : 4,2 : 4.3 : 4,5 : 4,6 5,1 : 5,2 : 5.3 : 5,4 : 5,6 6,1 : 6,2 : 6.3 : 6,4 : 6,5 Of the above 30 possible draws, there are 4 that add up to 5. So the probability is 4/30 = 2/15 If the draw is done with replacement, then there are 36 possible draws. Once again, small enough to exhaustively list, they are: 1,1 : 1,2 : 1,3 : 1,4 : 1,5 : 1,6 2,1 : 2,2 : 2,3 : 2,4 : 2,5 : 2,6 3,1 : 3,2 : 3,3 : 3.4 : 3,5 : 3,6 4,1 : 4,2 : 4.3 : 4,4 : 4,5 : 4,6 5,1 : 5,2 : 5.3 : 5,4 : 5,5 : 5,6 6,1 : 6,2 : 6.3 : 6,4 : 6,5 : 6,6 And of the above 36 possibilities, exactly 4 add up to 5. So you have 4/36 = 1/9</span>
7 0
2 years ago
What is the Value of the underlined digit 646,950
Jobisdone [24]
I'm sorry, but what digit is underlined?
4 0
1 year ago
Read 2 more answers
Consider a single spin of the spinner.
alex41 [277]

Answer:

"landing on a shaded portion and landing on a 3"

"landing on an unshaded portion and landing on a number less than 2 "

Step-by-step explanation:

Mutually exclusive means the events will have no intersection.

Let's look at your first choice:

"landing on a shaded portion and landing on an even number"

Landing on a shaded portion would be 1 or 4.

Landing on an even number would be 2 or 4.

There is an intersection (they contain a common element), the 4.

These events are not mutually exclusive.

Let's look at your second choice:

"landing on a shaded portion and landing on a number greater than 3"

Landing on a shaded portion would be 1 or 4.

Landing on a number greater than 3 would be just 4.

There is an intersection; they both contain 4.

These events are not mutually exclusive.

Let's look at your third choice:

"landing on a shaded portion and landing on a 3"

Landing on a shaded portion would be 1 or 4.

Landing on 3 would just be 3.

There is no common elements in the lists listed.  These events have no intersection.

These events are mutually exclusive.

Let's look at your fourth choice:

"landing on an unshaded portion and landing on an odd number"

Landing on a unshaded portion would be 2 or 3.

Landing on an odd number would be 1 or 3.

There is an intersection; they both have 3 in common.

These events are not mutually exclusive.

Let's look at your fifth choice:

"landing on an unshaded portion and landing on a number less than 2 "

Landing on an unshaded portion would 2 or 3.

Landing on a number less than 2 would be 1.

There is no intersection.

These events are mutually exclusive.

6 0
2 years ago
Read 2 more answers
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