HELP ME! ITS WORTH 55 POINTS PLZ DONT LIE!

Bernice found that the slope of a trend line is –3 and a point on the trend line is (8, 10). She wanted to find the equation of

shusha [124]

Answer:b

Step-by-step explanation:

5 0

3 years ago

Help or sus lots of points

zimovet [89]

answer is steeper

Step-by-step explanation:

3 0

2 years ago

A person invest $1200 in an account that earns 2% interest compound quarterly. Find when the value of the investment reaches $15

Basile [38]

Answer:

After 11 years the value of the investment reaches $1500.00

.

Step-by-step explanation:

The formula used for finding time (when the value reaches certain amount) is:

$A= P(1+\frac{r}{n})^{nt}$

where A= Future VAlue

P= Principal Value

r= rate of interest (in decimal)

n= no of times investment is compounded

t= time

Putting the values given and finding Time t,

A= $1500

P= $1200

r= 2% or 0.02

n= 4 (compound quarterly)

$A= P(1+\frac{r}{n})^{nt}$

$1500= 1200(1+\frac{0.02}{4})^{4t}$

Dividing both sides by 1200 and solving 0.02/4 = 0.005

$1.25= (1+0.005)^{4t}$

$1.25= (1.005)^{4t}$

Since t is in power we take the logarithm ln on both sides.

The rule of logarithm says that the exponent can be multiplied with the base when taking log

$\ln1.25=ln( 1.005)^{4t}\\\ln1.25=4t * ln( 1.005)\\0.22 = 4t * 0.005\\Solving\,\,\\\frac{0.22}{4*0.005} = t\\=> t= 11\, years$

8 0

3 years ago

A dog chases a squirrel. The dog is originally 200 feet away from the squirrel. The dog's speed is 150 feet per minute. The squi

Dafna11 [192]

Answer:

15 seconds

Step-by-step explanation:

If you make a table of values for the dog and the squirrel using d = rt, then the rates are easy: the dog's rate is 150 and the squirrel's is 100. The t is what we are looking for, so that's our unknown, and the distance is a bit tricky, but let's look at what we know: the dog is 200 feet behind the squirrel, so when the dog catches up to the squirrel, he has run some distance d plus the 200 feet to catch up. Since we don't know what d is, we will just call it d! Now it seems as though we have 2 unknowns which is a problem. However, if we solve both equations (the one for the dog and the one for the squirrel) for t, we can set them equal to each other. Here's the dog's equation:

d = rt

d+200 = 150t

And the squirrel's:

d = 100t

If we solve both for t and set them equal to each other we have:

$\frac{150}{d+200} =\frac{100}{d}$

Now we can cross multiply to solve for d:

150d = 100d + 20,000 and

50d = 20,000

d = 400

But we're not looking for the distance the squirrel traveled before the dog caught it, we are looking for how long it took. So sub that d value back into one of the equations we have solved for t and do the math:

$t=\frac{100}{d}=\frac{100}{400} =\frac{1}{4}$