Find each rate and unit rate 420 miles in 7 hours

One year, the population of a city was 352,000. Several years later it was 337,920.

kvasek [131]

hi

here is the method :

you take Value 1 and Value 2

And you do : ( Value 2 - Value 1) / (Value 1 ) * 100

Here we have : Value 1 = 352 000

Value 2 = 337 920

So now we apply :

( 337 920 - 352 000 ) / ( 352 000) * 100 = -4

Population decreased by by 4 %

You can also make :

352 000 * X = 337 920

X = 337 920 / 352 000

X = 0.96

if the multiplier is 0.96 , it's mean that we lost 1 -0.96 = 0.04

which mean - 4%

8 0

1 year ago

Can someone pls find the slope and y-intercept ASAP!!

Semenov [28]

Answer:

4/1. with an intercept of 5

3 0

3 years ago

Can anyone PLEASE HELP ME with this question i really need help

choli [55]

The missing angle is 49

7 0

2 years ago

Read 2 more answers

Ben has 18 photos to put in a photo album. He can put 4 in each page if he fills 4 pages, how many photos are on the fifth page?

Oksana_A [137]

Answer:

since 4 * 4 = 16

we can subtract 18-16=2

so the answer is 2

6 0

2 years ago

The table shows the average annual cost of tuition at 4-year institutions from 2003 to 2010.

nata0808 [166]

Answer: 1) The best estimate for the average cost of tuition at a 4-year institution starting in 2020 =$ 31524.31

2) The slope of regression line b=937.97 represents the rate of change of average annual cost of tuition at 4-year institutions (y) from 2003 to 2010(x). Here,average annual cost of tuition at 4-year institutions is dependent on school years .

Step-by-step explanation:

1) For the given situation we need to find linear regression equation Y=a+bX for the given situation.

Let x be the number of years starting with 2003 to 2010.

i.e. n=8

and y be the average annual cost of tuition at 4-year institutions from 2003 to 2010.

With reference to table we get

$\sum x=36\\\sum y=150894\\\sum x^2=204\\\sum xy=718418$

By using above values find a and b for Y=a+bX, where b is the slope of regression line.

$a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}=\frac{150894(204)-(36)718418}{8(204)-(36)^2}=\frac{30782376-25863048}{1632-1296}=\frac{4919328}{336}\\\\=14640.85$

and

$b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}=\frac{8(718418)-(36)150894}{8(204)-(36)^2}=\frac{5747344-5432184}{1632-1296}=\frac{315160}{336}\\\\=937.97$

∴ To find average cost of tuition at a 4-year institution starting in 2020.(as n becomes 18 for year 2020 if starts from 2003 ⇒X=18)

So, Y= 14640.85 + 937.97×18 = 31524.31

∴The best estimate for the average cost of tuition at a 4-year institution starting in 2020 = $31524.31

4 0

3 years ago