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3 years ago

Please help me now plz I will mark you brainliest

Mathematics

1 answer:

Anestetic [448]3 years ago

6 0

Answer:

? =6.97

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

sin theta = opp/ hyp

sin 35 = 4 / ?

Multiply each side by ?

? sin 35 = 4

Divide each side by sin 35

? = 4/sin 35

? =6.973787182

To the nearest hundredth

? =6.97

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What is 5x4=? +3<br> Plz help I for got

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Answer is ) 5x4=20. 20+3=23

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3 years ago

Assuming that the sample mean carapace length is greater than 3.39 inches, what is the probability that the sample mean carapace

joja [24]

Answer:

The answer is "".

Step-by-step explanation:

Please find the complete question in the attached file.

We select a sample size n from the confidence interval with the mean $\mu$ and default $\sigma$ , then the mean take seriously given as the straight line with a z score given by the confidence interval

$\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\$

Using formula:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

$P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)$

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution

$=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z$

the required probability: $P(z>0.8349)=1-P(z$

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2 years ago

3. A rare species of aquatic insect was discovered in the Amazon rainforest. To protect the species, environmentalists declared

navik [9.2K]

The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.

Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).

(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get

P(t)=40

(50(1+0.05t))/(2+0.01t)=40

Cross multiply both sides, we get

50(1+0.05t)=40(2+0.01t)

Apply the distributive property a(b+c)=ab+ac, we get

50+2.5t=80+0.4t

Subtract 0.4t and 50 from both sides, we get

50+2.5t-0.4t-50=80+0.4t-0.4t-50

2.1t=30

Divide both sides with 2.1, we get

t=14.29 months

(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get

$\begin{aligned}\lim_{t \rightarrow \infty}P(t)&=\lim_{t \rightarrow \infty}\frac{50(1+0.05t)}{2+0.01t}\\ &=\lim_{t \rightarrow \infty}\frac{50(\frac{1}{t}+0.05)}{\frac{2}{t}+0.01}\\ &=50\times \frac{0.05}{0.01}\\ &=250\end$

(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.

Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.

Learn more about limiting factor from here brainly.com/question/18415071.

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