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3 years ago

Pretty ez ngl but need help extra points

Mathematics

2 answers:

VLD [36.1K]3 years ago

7 0

Answer:

Step-by-step explanation:

Amy spent 1/6 of the 90 dollars on school supplies, which means to multiply 1/6 by 90

1/6*90

6 and 90 are both factors of 3, so cancel

1/2*30

30/2

$15

Answer is D

PS. you can put the fraction 15/90 and see that it is equal to 1/6 if you're not sure/ if you want to double check :)

baherus [9]3 years ago

5 0

Answer:

Concept: Algebraic Manipulation

90* 1/6 = 15
Hence its 15 dollars
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In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponen

Mashcka [7]

Answer:

a) P' = P

$P(t) = 24e^{0.693t}$ where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

$\frac{dP}{dt} = P$

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

$P(t) = P_0e^{kt}$

where $P_0 = 24$ is the initial population

After 6 month (t = 1), the population is doubled to 48

$P(1) = 24e^k = 48$

$e^k = 2$

$k = ln(2) = 0.693$

Therefore $P(t) = 24e^{0.693t}$

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

$24e^{0.693t} = 1000000$

$e^{0.693t} = 1000000 / 24 = 41667$

$0.693t = ln(41667) = 10.64$

$t = 10.64 / 0.693 = 15.35$

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. $P' = P_0ke^{kt}$

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

$24e^{0.5k) = 48$

$0.5k = ln2 = 0.693$

$k = 1.386$

Therefore, $P' = 1.386*24e^{1.386t}$

At the mid of the 3rd year, where t = 2.5, we can calculate P'

$P' = 1.386*24e^{1.386*2.5} = 1064.67$ rabbits/year

4 0

3 years ago

Joseph claims that a scatterplot in which the y-values increase as the x-values increase must have a linear association. Amy cla

quester [9]

Answer:

Amy is correct because a nonlinear association could increase along the whole data set, while being steeper in some parts than others. The scatterplot could be linear or nonlinear.

Step-by-step explanation:

I just took the quiz, I hope this helps :).

7 0

3 years ago

Read 2 more answers

Suppose that y = k * (x - 1/3) ^ 2 is a parabola in the xy -plane that passes through the point (2/3, 1) :Find k and the length

Dennis_Churaev [7]

Answer:

k = 9

length of chord = 2/3

Step-by-step explanation:

Equation of parabola: $y=k (x-\frac13)^2$

Part 1

If the curve passes through point $(\frac23 ,1)$ , this means that when $x=\dfrac23$ , $y = 1$

Substitute these values into the equation and solve for $k$ :

$\implies 1=k \left(\dfrac23-\dfrac13\right)^2$

$\implies 1=k \left(\dfrac13 \right)^2$

Apply the exponent rule $\left(\dfrac{a}{b} \right)^c=\dfrac{a^c}{b^c}$ :

$\implies 1=k \left(\dfrac{1^2}{3^2} \right)$

$\implies 1=\dfrac{1}{9}k$

$\implies k=9$

Part 2

The chord of a parabola is a line segment whose endpoints are points on the parabola.

We are told that one end of the chord is at $(\frac23 ,1)$ and that the chord is horizontal. Therefore, the y-coordinate of the other end of the chord will also be 1. Substitute y = 1 into the equation for the parabola and solve for x:

$\implies 1=9 \left(x-\dfrac13 \right)^2$