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Nadusha1986 [10]
3 years ago
13

Suppose a population of 50 crickets doubles in size every 3 months. How many crickets will there be after 2 years?

Mathematics
2 answers:
Verizon [17]3 years ago
8 0

This is a Geometric Progression problem.

We can use the formula

= a1 + q^{n-1}

= 50 x 2^(8-1)

50 is starting number

2 is doubled

8 is how many sets of 3 months in 2 years

= 50 x 2^(7)

= 50 x 128

= 6400

Rus_ich [418]3 years ago
5 0

Answer:

800 crickets.

Step-by-step explanation:

If the population starts at 50 crickets and doubles every 3 months that's 100 crickets every three months.

There are 12 months in a year.

So, we have to multiply 100 by 4.

That gets us 400.

So that is 400 crickets each year.

Since it is 2 years it would be 800 crickets.

Hope this helps!

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laila [671]

Answer:

if sinx=3/5=P/H

using pythogoras thoeren find B

B²= H²-P²

B²= 25-9

B=4

Tanx = P/B= 3/4

3.

n(S)= 6

n(O) = 3 {1,3,5)

P(O) = n(O)/n(s)= 3/6=1/2

Step-by-step explanation:

6 0
3 years ago
Jolinda scored 21 points during her basketball game. Her team scores a total of 28 points. What fraction of the team's points do
Natalka [10]

Answer:3/4


Step-by-step explanation:

if you divide 21 by 28 you get 3/4

3 0
3 years ago
[Algebra 2] Increasing/decreasing functions. Picture included.
Romashka [77]

Answer:

1.) Constant

2.) -7 < x < -3

3.) (-5,8)

4 0
3 years ago
Read 2 more answers
Farmers know that driving heavy equipment on wet soil compresses the soil and injures future crops. Here are data on the "penetr
Greeley [361]

Answer:

Step-by-step explanation:

Hello!

To see if driving heavy equipment on wet soil compresses it causing harm to future crops, the penetrability of two types of soil were measured:

Sample 1: Compressed soil

X₁: penetrability of a plot with compressed soil.

n₁= 20 plots

X[bar]₁= 2.90

S₁= 0.14

Sample 2: Intermediate soil

X₂: penetrability of a plot with intermediate soil.

n₂= 20 (with outlier) n₂= 19 plots (without outlier)

X[bar]₂= 3.34 (with outlier) X[bar]₂= 2.29 (without outlier)

S₂= 0.32 (with outlier) S₂= 0.24 (without outlier)

Outlier: 4.26

Assuming all conditions are met and ignoring the outlier in the second sample, you have to construct a 99% CI for the difference between the average penetration in the compressed soil and the intermediate soil. To do so, you have to use a t-statistic for two independent samples:

Parámeter of interest: μ₁-μ₂

Interval:

[(X[bar]₁-X[bar]₂)±t_{n_1+n_2-2;1-\alpha/2}*Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }]

t_{n_1+n_2-2;1-\alpha/2}= t_{20+19-2;1-(0.01/2)}= t_{37; 0.995}= 2.715

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{19*0.0196+18*0.0576}{20+19-2} }= 0.195= 0.20

[(2.90-2.29)±2.715*0.20\sqrt{\frac{1}{20} +\frac{1}{19} }]

[0.436; 0.784]

I hope this helps!

5 0
3 years ago
1. A tank is 3/5 full of water. After 330 litres of water is drawn out, it becomes 2/7 full. Find the capacity of the tank in li
Andreas93 [3]

Answer:

1050

Step-by-step explanation:

Let x = full capacity

\frac{3}{5} x=\frac{2}{7} x+330

Move the variable to the left side by subtracting both sides by \frac{2}{7} x

\frac{3}{5} x-\frac{2}{7}x=\frac{2}{7} x+330 -\frac{2}{7}x

\frac{3}{5} x-\frac{2}{7} x=330

Combine the like terms (don't forget about common denominator)

\frac{21}{35} x-\frac{10}{35} x=330

\frac{11}{35} x=330

Multiply both sides by \frac{35}{11} to isolate the x

(\frac{35}{11})\frac{11}{35} x=330(\frac{35}{11})

x = 1050

5 0
3 years ago
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