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2 years ago

Help me please: answer as an improper fraction

Mathematics

1 answer:

rewona [7]2 years ago

7 0

1) $\frac{7}{24}$
2) $\frac{2}{7}$
3) $\frac{1}{5}$
4) $\frac{3}{14}$
5) $\frac{1}{3}$
6) $\frac{7}{20}$
7) $\frac{6}{10}$
8) $\frac{1}{12}$
9) $\frac{1}{3}$
10) $\frac{1}{90}$
11) $\frac{1}{40}$
12) $\frac{1}{50}$
13) $\frac{20}{21}$
14) $\frac{1}{4}$
15) $\frac{5}{6}$
16) $\frac{9}{35}$
17) $\frac{35}{12}$ ( $2 \frac{11}{12}$ )
18) $\frac{9}{1}$ (9)

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A geneticist needs to grow a stock of fruit flies for her experiments.

antoniya [11.8K]

Answer :B

Step-by-step explanation:

200/1.8

\ $\sqrt{x}$

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2 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
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$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

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2 years ago

A rectangle has a perimeter of 26 cm and one of its sides has length of 5 cm

otez555 [7]

Answer:5 by 8 by 5 by 8

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3 years ago

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3 years ago

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Here, angles 4 and 5 are supplementary. No need to know their exact measurements. Together, their sum is 180°

Angles 1 and 2 are also supplementary, another 180°

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