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Gre4nikov [31]
3 years ago
10

The cost to mail a package is $5.00. Noah has postcard stamps that are worth $0.34 each and first-class stamps that are worth $0

.49 each
Mathematics
2 answers:
Kaylis [27]3 years ago
5 0

Answer:

md

dhjsdjxnsnsndqjdnks

Step-by-step explanation:

nssjndwjfnwjddnsjn

s2008m [1.1K]3 years ago
5 0
How much does he buy?
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6 0
3 years ago
Read 2 more answers
Can someone help me? ​
Oliga [24]
52-13=39
39/52x100=75
The percent of markdown: 75%
5 0
2 years ago
David is filling out orders for an online business and gets paid $1 for each order he fills out plus bonus of 25 cents per order
svp [43]

Answer:

David will make $481 (he earns the bonus)

Explanation:

<em>If he makes $1 for each order and he filled out 385 orders, then why can't we say he made $385?</em>

Because of this statement rights here:

"...and gets paid $1 for each order he fills out plus bonus of 25 cents per order if the average number of orders he completes per day within any of the given weeks exceeds 20."

So we need to find out if any of the 3 weeks has an average of 20+ orders per day.

<h2>David is filling out orders for an online business and gets paid $1 for each order he fills out</h2>

(x is the amount of orders he fills out)

profit = $1x

<h2>plus bonus of 25 cents per order if the average number of orders he completes per day within any of the given weeks exceeds 20. </h2>

if any average orders per day is > 20 in any week

bonus profit = $1.25x

<h2>The ratio of the number of orders he processed during the first week to the number of orders he processed during the second week is 3:2, </h2>

first week     second week

             3a : 2a

<h2>while the the ratio that compares the number of orders he filled out during the first and the third weeks is 4 to 5 respectively. </h2>

first week   third week

           4a : 5a

<h2>What amount of money will David make at the end of three weeks if the total number of orders he filled out was 385?</h2>

sum of all ratios of a = 385

So we have

3a : <u>2a</u> (first week to <u>second week</u>)

4a : <em>5a </em>(first week to <em>third week</em>)

Notice how the first two numbers are both from the first week. Let's use the Least Common Multiple to make them equal while still keeping ratios.

LCM of 3 and 4: 12 = 3 * 4

12a : <u>8a</u> ( times 4 )

12a : <em>15a</em> ( times 3 )

Now that we have the same value, we can create a big ratio

first week <u>second week</u> <em>third week</em>

   12a     :        <u>8a</u>          :      <em>15a</em>

we know that these ratios will all equal 385. Since ratios are equal no matter how big we make them, we can say that

12a + <u>8</u>a + <em>15</em>a = 385 (a is a variable to scale up the ratio)

which is the same as

(12 + <u>8</u> + <em>15</em>) * a = 385

(<em><u>35</u></em>) * a = 385

35a = 385

if we solve for a by dividing 35 on both sides we get

a = 11

This gives us how much to multiply the RATIO by to get the ACTUAL NUMBER of orders completed. Let's plug 11 for 'a' and see what happens.

12a + <u>8</u>a + <em>15</em>a = 385

12(11) + <u>8</u>(11) + <em>15</em>(11) = 385

132 + <u>88</u> + <em>165</em> = 385     (Check that out, the number of orders each week!)

<u>220</u> + <em>165</em> = 385

<em><u>385</u></em> = 385

Bingo! All the math works out. So, looking back at the verryyy top of this problem, the reason why it wasn't as easy as $385 was because of the bonus.

The bonus gives David $1.25 per order instead of $1 per order if any of the weeks have an average ORDER PER DAY of anything bigger than 20. If we know the real numbers of orders for every week (132, <u>88</u>, and <em>165</em>), then we can divide it by 7 to get the average order per day. Let's choose <em>165 </em>(the <em>third week</em>) because it is the biggest and has the greatest chance of meeting our goal.

165 orders / 7 days (7 days in a week) = 23.57 orders per day

Is this greater than 20 orders per day?

YES!

So now we can safely say that the bonus is there or not, and in this case, the bonus IS there because there is a week where David had more than 20 orders per day.

So instead of using

profit = $1x

We will use

bonus profit = $1.25x

(x is the amount of orders completed)

So if we know he completed 385 orders, and we know he earned the bonus, we plug in 385 for x for the bonus function

bonus profit = $1.25x

bonus profit = $1.25 * 385

bonus profit = $481.25

If necessary, round your answer to the nearest dollar.

So for the very end, all we have to do is round it to the nearest dollar.

$481.25 rounds to $481.

And we're done!

8 0
3 years ago
(a) The probability that an automobile being filled with gasoline will also need an oil change is 0.35; the probability that it
cricket20 [7]

Answer:

0.714 ; 0.556 ; (A1 n A2 n A3) = 0

Step-by-step explanation:

Given that :

P(oil) = 0.35

P(filter) = 0.45

P(oil n filter) = 0.25

1)

P(filter | Oil) = P(oil n filter) / P(oil)

P(filter | Oil) = 0.25 / 0.35 = 0.714

11)

P(Oil | filter) = P(oil n filter) / P(filter)

P(Oil | filter ) = 0.25 / 0.45 = 0.556

Number of red aces = 2

Queen or king = 4 + 4 = 8

Number on card greater than 3 and less than 8 = {4,5,6,7} = 4 card ; for 4 suits = 4 * 4 = 16

Total number of cards in deck = 52

Choosing without replacement :

Hence,

A1 = 2 / 52

A2 = 8 / 51

A3 = 16 / 50

(A1 n A2 n A3) ;this means card common to all three events.

However, (A1 n A2 n A3) = 0 because ;

1.) No red ace card is greater than 3 and less than 8.

11.) An ace card can neither be a king nor a queen

3 0
2 years ago
Professor Vinculum investigated the migration season of the Bulbul bird from their natural wetlands to a warmer climate. He foun
coldgirl [10]

The population of the Bulbul birds at the start of the migration season is 1800.

<h3>Exponential function</h3>

An exponential function is given by:

y = abˣ

Let y, x are variables, a is the initial value and b is the multiplication factor.

Let P be the population of the bird t days since the start of the migration season.

P=1350+400(1.25)^{-t}\\\\At\ the\ start\ of\ the\ season(t=0):\\\\P=1350+400(1.25)^{-0}\\\\P=1350+450=1800

The population of the Bulbul birds at the start of the migration season is 1800.

Find out more on Exponential function at: brainly.com/question/12940982

8 0
2 years ago
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