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3 years ago

What is her change in elevation after 5 Hours

Mathematics

1 answer:

Iteru [2.4K]3 years ago

5 0

Can you add additional information about the problem?

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An adult short-eared owl weighed 1/2 pound. A baby short-eared owl weighed 2/9 pound. What was the total weight of the two owls?

bezimeni [28]

Total weight of the two owls are 2/3

3 0

3 years ago

PLEASE HELP!!!!! Find the volume of a cone that has a diameter 30ft and a height of 12ft.

Leviafan [203]

Answer:

the answer is 2827.4

Step-by-step explanation:

$\pi \: {r}^{2} \frac{h}{3}$

5 0

3 years ago

I need help with this pls

Alexxx [7]

Hello!

What is an x-intercept:

⇒ value of x when the value of y equals '0'

$8x + 5y=25\\8x+5(0)=25\\8x=25\\\\x=\dfrac{25}{8}$

x-intercept is 25/8

What is a y-intercept:

⇒ value of y when the value of x equals '0'

$8x+5y=25\\8(0)+5y=25\\5y=25\\y=5$

y-intercept is 5

Hope that helps!

5 0

1 year ago

The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth

Art [367]

Answer:

a) $N_0=250\; k=0.15$

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies $t=0$ . And $N(0)=N_0$ . The process is:

$N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250$

b) After 2 days means $t=48$ . So, we just replace and operate:

$N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}$

c) $N(t_1)=4000; \;t_1=?$

$N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\ \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h$

d) $t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)$

$N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h$

6 0

3 years ago

A halloween assorted bag of candy contains 4 Snickers, 2 packages of M&M's, & 3 Butterfingers. What is the probability o

a_sh-v [17]

Answer:

2/27

Step-by-step explanation:

In the question we have:

4 snickers

2 packages of M&M's

3 Butterfingers.

Total number of candies in the Halloween bag = 9

Probability of selecting some M&M's and putting them back = 2/9

Probability of selecting a Butter finger = 3/9

Therefore, the probability of selecting some M&M's, putting them back, and then selecting a Butterfinger =

Probability of selecting some M&M's and putting them back × Probability of selecting a Butter finger

2/9 × 3/9

= 6/ 81

= 2/27

3 0

3 years ago