Ask question

Ask question

All categories

3 years ago

8

Jarell is grocery shopping for his BBQ cookout this weekend. He is grilling 13lb. burgers and expects about 20 people. How many

whole lbs. of ground beef should be purchase?

1 answer:

user100 [1]3 years ago

8 0

Jarrell should purchase 7 lbs of ground beef.

Number of attendees = 20

Fraction of burger per person = 1/3 lbs

Total fraction of burger required to feed 20 persons :

Number of attendees × fraction per person

20 × 1/3 = 6.667

The number of whole grouhd beef that should be purchased is 7 lbs

Learn more : brainly.com/question/2610134

You might be interested in

Julia charged $612 to her credit card. The cost of her next purchase caused her to go over her credit limit of $2000.

Fynjy0 [20]

She has already spent 612, and by adding C 612 has gone ABOVE 2000
So 612 + C MUST be Greater Than 2000
The answer is D

7 0

3 years ago

Read 2 more answers

kenny6666 [7]

The answer you are looking for is -5x^5-5x^4+6x^3-x^2-7x+20
-I hope this is the answer you are looking for, please feel free to post your future questions on Brainly.com.

4 0

2 years ago

(SUPER EASY) ILL GIVE BRAINLIEST TOO. <br><br><br> just answer all the questions you know.

wolverine [178]

Answer:

what is the question

Step-by-step explanation:

5 0

2 years ago

Find the area of the shaded region. Round the nearest hundredth if necessary. YZ=14.2m

andreyandreev [35.5K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

$A_1 = 67.58 \ in^2$

2

$A_2 =415.4 \ ft^2$

3

$A_3 = 8.48 \ cm^2$

4

$A_4 = 480.38 \ m^2$

Step-by-step explanation:

Generally the area of a sector is mathematically represented as

$A = \frac{\theta}{360} * \pi r^2$

Now at $r_1 = 11 in$ and $\theta_1 = 64^o$

$A_1 = \frac{64}{360} * 3.142 * 11^2$

$A_1 = 67.58 \ in^2$

Now at $r_2 = 20 ft$ in and $\theta_2 = 119 ^o$

$A_2 = \frac{119}{360} * 3.142 * 20^2$

$A_2 =415.4 \ ft^2$

Now at $r_3 = 6.5 cm$ and $\theta_3 = 23 ^o$

$A_3 = \frac{23}{360} * 3.142 * 6.5 ^2$

$A_3 = 8.48 \ cm^2$

Now at $r_4 = 14.2 m$ and $\theta_4 = 360 -87 = 273 ^o$

$A_4 = \frac{273}{360} * 3.142 * 14.2^2$

$A_4 = 480.38 \ m^2$

6 0

2 years ago

PLEASE HELP ME I BEG YOU !!!

boyakko [2]

Hi,

$f(x)=\log(x)$ FALSE

$f(x)=\log(x-4)+1$ TRUE

→ $f(x)=\log(x-2)$

$f(x)=\log(x+2)$ FALSE

$f(x)=2log(x)$ TRUE

4 0

3 years ago

Read 2 more answers