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3 years ago

7.

Mathematics

2 answers:

Tema [17]3 years ago

6 0

Answer:

The corect answer is $77,06.

Step-by-step explanation:

First let's start by analyzing the information we have:

This is what Yvette spent for each product:

A 26 pound turkey for $1.49 per pound: 26 x 1,49 = 38,74

Two bags of potatoes for $5.75 each: 5,75 x 2 =11,50

6 pounds of stuffing for $2.47 per pound: 2,47x6 =14,82

8 cans of pumpkin for $1.50 each: 8x1,50=12

Once we have these results we just have to add them in order to get the final result of all the money Yvette spent:

38.74 + 11.50 + 14.82 + 12 = 77.06

The correct answer is 77.06.

maria [59]3 years ago

4 0

Answer:

77.06

Step-by-step explanation:

Multiply each item by price and amount and add them all up.

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American Vending (AV) supplies vended food to a large university. Because students often kick the machines out of anger and frus

aev [14]

Answer:

case 2 with two workers is the optimal decision.

Step-by-step explanation:

Case 1—One worker:A= 3/hour Poisson, ¡x =5/hour exponential The average number of machines in the system isL = - 3. = 4 = lJr machines' ix-A 5 - 3 2 2Downtime cost is $25 X 1.5 = $37.50 per hour; repair cost is $4.00 per hour; and total cost per hour for 1worker is $37.50 + $4.00

= $41.50.Downtime (1.5 X $25) = $37.50 Labor (1 worker X $4) = 4.00

$41.50

Case 2—Two workers: K = 3, pl= 7L= r= = 0.75 machine1 p. -A 7 - 3Downtime (0.75 X $25) = S J 8.75Labor (2 workers X S4.00) = 8.00S26.75Case III—Three workers:A= 3, p= 8L= ——r = 5- ^= § = 0.60 machinepi -A 8 - 3 5Downtime (0.60 X $25) = $15.00 Labor (3 workers X $4) = 12.00 $27.00

Comparing the costs for one, two, three workers, we see that case 2 with two workers is the optimal decision.

3 0

3 years ago

I need help... on my homework it says to write each expression in radical form, or write each radical and exponential form . For

pantera1 [17]

$\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt{13}\implies \sqrt[2]{13^1}\implies 13^{\frac{1}{2}}$

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3 years ago

FIRST PROPER ANSWER GETS BRAINLIEST ON TEST PLEASE HELP!!!

Irina-Kira [14]

Area of a semicircle = $\pi$ r² / 2
r= 5/2 = 2.5

So: A= 3.14 x 2.5² / 2 A = 9.8125 so rounding to the nearest hundredth is A=9.81

8 0

3 years ago

Read 2 more answers

If I had 12 marbles and then gave two away and now I have ten out of twelve marbles left what fraction of the marbles did I give

Ostrovityanka [42]

Okay so you have 10/12 marbles (ten out of twelve). Both 10 and 12 can be divided by 2, right? So divide both of those by 2 and you get 5/6. That’s your answer. Hope this helps!

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3 years ago

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Ifteen bales of hale were loaded into a truck. Some of the bales weight 150 pounds, and some weighed 100 punds. Hpw many bales o

horrorfan [7]

Answer:

There are 7 bales of 100 pounds and 8 bales of 150 pounds hale

Step-by-step explanation:

Let the amount of bale of 100 pounds be x and the bale of hale of 150pounds be y

If fifteen bales of hale were loaded into a truck, then x+y = 15 ... 1

Also if some of the bales weight 150 pounds, and some weighed 100 pounds with total weight of pounds to be 1900 pounds, then mathematically;

150y + 100x = 1900

Dividing through by 100;

1.5y+x = 19 ..... 2

Solving equation 1 and 2 simultaneously;

x+y = 15 ... 1

1.5y+x = 19 .... 2

Subtracting 1 from 2

y-1.5y = 15-19

-0.5y = -4

Divide both sides by 0.5

-0.5y/-0.5 = -4/-0.5

y = 8

Substituting y = 8 into equation 1, we will have;

x+y = 15

x+8 = 15

x = 15-8

x = 7

Hence there are 7 bales of 100 pounds and 8 bales of 150 pounds hale

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3 years ago