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Elodia [21]
3 years ago
14

Seed are often treated with fungicides to protect them in poor draining, wet environments. A small-scale trial, involving five t

reated and five untreated seeds, was conducted prior to a large-scale experiment to explore how much fungicide to apply. The seeds were planted in wet soil, and the number of emerging plants were counted. If the solution was not effective and four plants actually sprouted, what is the probability that a b c all four plants emerged from treated seeds
Mathematics
1 answer:
Natalija [7]3 years ago
3 0

Answer:

2.38%

Step-by-step explanation:

Given that:

The population size involving numbers of treated and untreated seeds N  = 10

the number of seeds that are treated which are observed = 5

the number of seeds that sprouted 'n' = 4

Now, use the method of hypergeometric probability, we have the formula:

p(y) = \dfrac{(^r_y)(^{N-r}_{n-y})}{(^N_n)}

∴

p(4) = \dfrac{(^5_4)(^{5}_{0})}{(^{10}_4)}

p(4) = \dfrac{\dfrac{5!}{4!(5-4)!}* \dfrac{5!}{0!(5-0)!}  }{\dfrac{10!}{4!(10-4)!} }

p(4) = \dfrac{\dfrac{5!}{4!}* \dfrac{5!}{(5)!}  }{\dfrac{10!}{4!(6)!} }

p(4) = \dfrac{5*1  }{210}

p(4) = 0.0238

p(4) = 2.38%

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4) 0= 4 +n/5 what is n
WARRIOR [948]

Answer:

n=-20

Step-by-step explanation:

First subtract four from both sides then you'll get

-4=n/5

Second multiply 5 to both sides then you'll get

n=-20

7 0
3 years ago
If prizes for a carnival are priced at 5 for 80 cents. how much would 250 prices cost?
sergij07 [2.7K]
The answer is 4,000 is the cost

3 0
3 years ago
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A 5×5×5 wooden cube was painted and then sawed into 1×1×1 cubes. How many 1×1×1 cubes are there? HELPPPPP!!!!!!!!!!!!!!!!!!!!!!
Shalnov [3]

Answer:

125

Step-by-step explanation:

1 x 1  x 1 = 1

5 x 5 x 5 = 125

125 / 1 = 125

125 is your answer

6 0
3 years ago
Pam has 90 m of fencing to enclose an area in a petting zoo with two dividers to separate three types of young animals. The thre
andrew-mc [135]

Answer:

The area function is

A=\frac{135}{2}x-\frac{9}{2}x^2.

The domain and range of A is (0,15m) and (0, 253.125 m^2].

Step-by-step explanation:

The given length of fencing is 90 m.

Let the length and width of each pen be x and y respectively as shown in the figure.

As there are 3 pens, so, the total area,

A= 3 xy \;\cdots (i)

From the figure the total length of fencing is 6x+4y.

Here, for a significant area for the animals, x>0 as well as y>0 as x and y are the sides of ben.

From the given value:

6x+4y=90\;\cdots (ii)

\Rightarrow  y=\frac {45}{2}-\frac{3x}{2}

Now, from equation (i)

A=3x\left(\frac {45}{2}-\frac{3x}{2}\right)

\Rightarrow A=\frac{135}{2}x-\frac{9}{2}x^2\;\cdots (iii)

This is the required area function in the terms of variable x.

For the domain of area function, from equation (ii)

x=15-\frac{2y}{3}

\Rightarrow x [as y>0]

So, the domain of area function is (0,15m).

For the range of area function:

As x \rightarrow 0 or y\rightarrow 0, then A\rightarrow 0 [from equation (i)]

\Rightarrow A>0

Now, differentiate the area function with respect to x .

\frac {dA}{dx}=\frac{135}{2}-9x

Equate \frac {dA}{dx}  to zero to get the extremum point.

\frac {dA}{dx}=0

\Rightarrow \frac{135}{2}-9x=0

\Rightarrow x=\frac{15}{2}

Check this point by double differentiation

\frac {d^2A}{dx^2}=-9

As,  \frac {d^2A}{dx^2}, so, point x=\frac{15}{2} is corresponding to maxima.

Put this value back to equation (iii) to get the maximum value of area function. We have

A=\frac{135}{2}\times \frac {15}{2}-\frac{9}{2}\times \left(\frac {15}{2}\right)^2

\Rightarrow A=253.125 m^2

Hence, the range of area function is (0, 253.125 m^2].

4 0
3 years ago
A local phone company charges a monthly fee of $34.99 plus $.05 for each minute of long distance calls. Parts of minutes are rou
stepladder [879]

53.24-34.99=18.25

$1 will buy 20 minutes of long distance calling

x$18.25 = 365 minutes @ $34.99

$132-$34.99=$97.01

20x97=1,940 minutes

minimum=365 minutes

maximum=1,940 minutes

Hope this helps.

3 0
4 years ago
Read 2 more answers
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