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3 years ago

PLEASE HELP ME ILL MARK WITH BRAINLY

Mathematics

2 answers:

AlladinOne [14]3 years ago

8 0

Answer:

15(4)(3), 12(15),

Step-by-step explanation:

nika2105 [10]3 years ago

4 0

You can mark her/him as the brainiest now :}

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Write out the sample space for the given experiment. Use the following letters to indicate each choice: M for mushrooms, O for o

pshichka [43]

Answer: {(MBV), (MBH) , (MEV) , (MEH) , (OBV), (OBH) , (OEV) , (OEH)}

Step-by-step explanation:

Sample space is a set that contains all the possible outcomes for an event.

For salad:

Choices for extra toppings (mushrooms, olives) =2

Choices for meats (bacon, eggs) =2

Choices for dressings (vinaigrette, honey mustard.)= 2

By Fundamental counting principal .

The number of different salads can be made= (Choices for extra toppings) x (Choices for meats ) x (Choices for dressings)

= 2 x 2 x 2 = 8

All the possible outcomes = (MBV), (MBH) , (MEV) , (MEH) , (OBV), (OBH) , (OEV) , (OEH)

Required Sample space would be : {(MBV), (MBH) , (MEV) , (MEH) , (OBV), (OBH) , (OEV) , (OEH)}

5 0

3 years ago

A the fraction one half kilogram of flour is packed equally into 8 bags. How much flour is in each bag

marissa [1.9K]

Answer:

0.0625 kg per bag

or 1/16th of a kilogram per bag

Step-by-step explanation:

You'd need to divide the total amount the the number of sections.

In this case, it'd be 0.5 kg/8 bags, resulting in 0.0625 kg of flour per bag.

I hope this helps!

6 0

3 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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