Ask question

Ask question

All categories

3 years ago

11

Janice lost a total of 75 pounds over the course of 12 months. How many pounds on average did Janice lose each moth? Answer with

a mixed number.

1 answer:

Licemer1 [7]3 years ago

4 0

Answer:

The awnser is 6.25 pounds

Step-by-step explanation: 75/12 = 6.25 which means every month she lost a total of 6.25 pounds and on a 12 month average she lost 75 pounds

You might be interested in

What is a good estimate for this angle measurement?

ycow [4]

C. 75 degreesssssss:)

7 0

2 years ago

Read 2 more answers

1. Which figure comes next in the pattern below?

LenaWriter [7]

The awnser is D
Purrr

6 0

3 years ago

The distance from Earth to Mars is about 35,000,000 miles. What is the order of magnitude of this number?. . . . . . . . A.. . 6

Musya8 [376]

The scientific notation of the given, 35,000,000 is 3.5 x 10^7. The order of magnitude of a scientific notation is the exponent of its multiplier, 10. From the derived scientific notation above, the exponent of 10 is 7. Thus, the answer is letter D.

7 0

3 years ago

In a certain region, 6% of a city’s population moves to the surrounding suburbs each year, and 4% of the suburban population mov

Charra [1.4K]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Following are the differential equation:

$\to P(C,S)t+1 = P(C,S)t+(- 0.06Ct +0.04St, 0.06Ct - 0.04St ) \ \ \ \\\\or \ \ \ \ \ \ \ \\\\ \to (0.94Ct + 0.04St, 0.96St + 0.06Ct)$

In equation:

$\to t = 0, \ as \ 2019\\\\\to Ct = 10,000,000 \\\\\to St = 800,000,000\\$

$\to P(C, S)1 = (0.94Ct + 0.04St, 0.96St + 0.06Ct) \\\\$

$= (0.94(10000000) + 0.04(800000), 0.96(800000) + 0.06(10000000))\\\\ = (9,400,000 + 32,000), (768,000 +6000000))\\\\ = (9432000, 1368000)$

$\to P(C, S)2 = (0.94(9432000) + 0.04(1368000), 0.96(1368000) + 0.06(9432000))$

$=(8,866,080 + 54,720), (1,313,280 + 565,920)\\\\=(8,920,800, 1,879,200)$

3 0

3 years ago

HELPPPPP!!!! MEEE!!! PLEASE!!! ASAPPP!!!!

mario62 [17]

Umm, what grade are you in . i’ll see if i can help

6 0

3 years ago