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3 years ago

What value of x makes this true 1.5x=120?

Mathematics

2 answers:

Debora [2.8K]3 years ago

5 0

Answer: 80

Step-by-step explanation:

x.3/2 = 120

x = 120.2/3

x=80

navik [9.2K]3 years ago

3 0

80

120/1.5=x
Hopefully this helps

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A stockbroker has money in three accounts. The interest rates on the three accounts are 7%, 8%, 9%. If she has twice as much mon

Sveta_85 [38]

Answer:

x=amount invested at 7%

2x=amount invested at 8%

3x=amount invested at 9%

interest=principal*rate*time (time=1 year)

$150=0.07x+0.08*2x+0.09*3x

$150=0.07x+0.16x+0.27x

$150=0.50x

$1500=5x

x=$300 invested at 7%

2x=$600 invested at 8%

3x=$900 invested at 9%

Step-by-step explanation:

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2 years ago

Which 2 numbers have a mean average of 10 and a range of 4

Aloiza [94]

Simply the answer is 12 and 8.

Mark brainliest please

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2 years ago

A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated tha

8_murik_8 [283]

To model this situation, we are going to use the decay formula: $A=Pe^{rt}$
where
$A$ is the final pupolation
$P$ is the initial population
$e$ is the Euler's constant
$r$ is the decay rate
$t$ is the time in years

A. We know for our problem that the initial population is 1,250, so $P=1250$ ; we also know that after a year the population is 1000, so $A=1000$ and $t=1$ . Lets replace those values in our formula to find $r$ :
$A=Pe^{rt}$
$1000=1250e^{r}$
$e^{r}= \frac{1000}{1250}$
$e^{r}= \frac{4}{5}$
$ln(e^{r})=ln( \frac{4}{5} )$
$r=ln( \frac{4}{5} )$
$r=-02231$

Now that we have $r$ , we can write a function to model this scenario:
$A(t)=1250e^{-0.2231t}$ .

B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.

C.
- The function is decreasing
- The function doe snot have a x-intercept
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0:
$A(0)=1250e^{(-0.2231)(0)$
$A_{0}=1250e^{0}$
$A_{0}=1250$
- Over the interval [0,10], the function will have a minimum at t=10:
$A(10)=1250e^{(-0.2231)(10)$
$A_{10}=134.28$

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula: $m= \frac{A(0)-A(10)}{10-0}$
where
$m$ is the rate of change
$A(10)$ is the function evaluated at 10
$A(0)$ is the function evaluated at 0
We know from previous calculations that $A(10)=134.28$ and $A(0)=1250$ , so lets replace those values in our formula to find $m$ :
$m= \frac{134.28-1250}{10-0}$
$m= \frac{-1115.72}{10}$
$m=-111.572$
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.