3.4×5

Kai has been saving up money to buy a new backpack that is originally priced at $88.95. The store is having a 40% off end of sea

Fiesta28 [93]

Answer:

yes, he can but the backpack.

Step-by-step explanation:

8 0

3 years ago

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deff fn [24]

true
Typically, the price charged must cover more than labor costs figured at an hourly rate. There may be costs related to rent, utilities, administration, equipment use and/or depreciation. These costs are often called overhead.

6 0

3 years ago

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5. Guests are given a random goodie bag at a party. There are less than 15 bags, but more than 8. Part A: How many bags were pre

Artemon [7]

Answer:

12 bags

Step-by-step explanation:

Number of the bags = between 8 and 15 and

1/2 of bags contain money
1/6 of bags contain candy
1/3 of bags contain a toy

As per above the number is divisible by 2, 3 and 6. LCM of the three numbers is 6.

The number between 8 and 15 and divisible by 6:

This number is 6*2 = 12

So the number of bags is 12

4 0

3 years ago

The sides of a rhombus with angle of 60° are 6 inches. Find the area of the rhombus.

ehidna [41]

1. Check the drawing of the rhombus ABCD in the picture attached.

2. m(CDA)=60°, and AC and BD be the diagonals and let their intersection point be O.

3. The diagonals:
i) bisect the angles so m(ODC)=60°/2=30°

ii) are perpendicular to each other, so m(DOC)=90°

4. In a right triangle, the length of the side opposite to the 30° angle is half of the hypothenuse, so OC=3 in.

5. By the pythagorean theorem, $DO= \sqrt{ DC^{2}- OC^{2} }= \sqrt{ 6^{2}- 3^{2} }=\sqrt{ 36- 9 }=\sqrt{ 27 }= \sqrt{9*3}=$
$=\sqrt{9}* \sqrt{3} =3\sqrt{3} (in)$

6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4* $\frac{DO*OC}{2}=4 \frac{3 \sqrt{3} *3}{2}$ = $=2*9 \sqrt{3}=18 \sqrt{3}$ ( $in^{2}$ )

6 0

3 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:

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

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 : 2a (first week to second week)

4a : 5a (first week to third week)

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 : 8a ( times 4 )

12a : 15a ( times 3 )

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

first week second week third week

12a : 8a : 15a

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 + 8a + 15a = 385 (a is a variable to scale up the ratio)

which is the same as

(12 + 8 + 15) * a = 385

(35) * 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 + 8a + 15a = 385

12(11) + 8(11) + 15(11) = 385

132 + 88 + 165 = 385 (Check that out, the number of orders each week!)

220 + 165 = 385

385 = 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, 88, and 165), then we can divide it by 7 to get the average order per day. Let's choose 165 (the third week) 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