Abdullah earns $10 an hour delivering newspaper on his bicycle. He started with $20 in his banks account.

It is now 3:15 pm. Is it possible to drive 135 miles and arrive before 5:00 pm if ypu drove 55 mph? Explain your answer

Troyanec [42]

Well, how long will it take you to drive 135 miles at 55mph?

at 55 mph, you're doing 55 miles every hour, so we can simply get the quotient of 135/55 and that's how many hours it'll take you to drive 135 miles at that speed.

$\bf \cfrac{135}{55}\implies \cfrac{27}{11}\implies 2\frac{5}{11}~hours\implies \textit{about 2hrs, 27 minutes}$

so, it takes you that long, however, from 3:15 to 5:00pm there are only 45mins + 60mins or 1hr and 45 minutes, namely 1¾ hr.

so 2hrs and 27 minutes is much later than 1¾ hr, so, no dice, you can't arrive at 5pm, actually you'll arrive around 5:42pm.

6 0

2 years ago

Read 2 more answers

Determine what type of model best fits the given situation:

lyudmila [28]

Let value intially be = P

Then it is decreased by 20 %.

So 20% of P = $\frac{20}{100} \times P = 0.2P$

So after 1 year value is decreased by 0.2P

so value after 1 year will be = P - 0.2P (as its decreased so we will subtract 0.2P from original value P) = 0.8P-------------------------------------(1)

Similarly for 2nd year, this value 0.8P will again be decreased by 20 %

so 20% of 0.8P = $\frac{20}{100} \times 0.8P = (0.2)(0.8P)$

So after 2 years value is decreased by (0.2)(0.8P)

so value after 2 years will be = 0.8P - 0.2(0.8P)

taking 0.8P common out we get 0.8P(1-0.2)

= 0.8P(0.8)

$=P(0.8)^{2}$ -------------------------(2)

Similarly after 3 years, this value $P(0.8)^{2}$ will again be decreased by 20 %

so 20% of $P(0.8)^{2} \frac{20}{100} \times P(0.8)^{2} = (0.2)P(0.8)^{2}$

So after 3 years value is decreased by $(0.2)P(0.8)^{2}$

so value after 3 years will be = $P(0.8)^{2} - (0.2)P(0.8)^{2}$

taking $P(0.8)^{2}$ common out we get $P(0.8)^{2}(1-0.2)$

$P(0.8)^{2}(0.8)$

$P(0.8)^{3}$ -----------------------(3)

so from (1), (2), (3) we can see the following pattern

value after 1 year is P(0.8) or $P(0.8)^{1}$

value after 2 years is $P(0.8)^{2}$

value after 3 years is $P(0.8)^{3}$

so value after x years will be $P(0.8)^{x}$ ( whatever is the year, that is raised to power on 0.8)

So function is best described by exponential model

$y = P(0.8)^{x}$ where y is the value after x years

so thats the final answer

3 0

3 years ago

I can’t figure this one out! Someone please help

Zinaida [17]

Answer:

g= -2x+1

Step-by-step explanation:

-2x because it goes down two boxes for every one box to the right.

+1 because that is the y intercept

6 0

2 years ago

Read 2 more answers

The petronas towers in Kuala Lumpur, Malaysia, are 452 meters tall. A woman who is 1.75 meters tall stands 120 meters from the b

bazaltina [42]

Answer:

$75.1^{\circ}$

Step-by-step explanation:

In this problem, we have:

H = 452 m is the height of the Petronas tower

h = 1.75 m is the height of the woman

d = 120 m is the distance between the woman and the base of the tower

First of all, we notice that we want to find the angle of elevation between the woman's hat the top of the tower; this means that we have consider the difference between the height of the tower and the height of the woman, so

$H' = H-h = 452-1.75=450.25 m$

Now we notice that $d$ and $H'$ are the two sides of a right triangle, in which the angle of elevation is $\theta$ . Therefore, we can write the following relationship: