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2 years ago

How do I get the answer to this

Mathematics

2 answers:

Oxana [17]2 years ago

8 0

5x7=35. Voila. You're the 35th caller.

stepladder [879]2 years ago

4 0

You do 5 X 7 which is 35 because that means that you are both one in 7 callers and one in 5 callers. Hope this helps, have a nice day :)

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An evolutionary biologist examined the relative fitness of Escherichia coli bacteria grown for 2000 generations, about 300 days,

Varvara68 [4.7K]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: fitness of a line of E. coli grown on an acidic environment.

n= 6 E. coli lines

Recorded fitness for each line: 1.24, 1.22, 1.23, 1.24, 1.18, 1.09

The relative fitness of 1 indicates that both bacteria types are equally fit.

A relative fitness larger than 1 indicates that the acid-evolved line is more fit than the parental line kept at neutral pH when both are grown in acidic conditions.

Meaning that if the average fitness of the E. coli lines grown on an acidic environment is greater than 1 then they are better adjusted to live in acidic conditions, symbolically: μ > 1

The statistic hypotheses are:

H₀: μ ≤ 1

H₁: μ > 1

α: 0.05

Assuming that the variable has a normal distribution you have to apply a one-sample t-test:

$t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~~t_{n-1}$

X[bar]= 1.20

S= 0.06

$t_{H_0}= \frac{1.20-1}{\frac{0.06}{\sqrt{6} } } = 8.40$

The p-value for this test is 0.0002

Since the p-value= 0.0002 is less than α:0.05 the decision is to reject the null hypothesis.

Then at a 5% significance level, there is significant evidence to conclude that the bacteria evolved in acidic pH are better adapted to acidic conditions.

I hope you have a SUPER day!

7 0

2 years ago

Help please!!! | DON'T GUESS, please show work

Aleksandr-060686 [28]

For one of the triangular sides, 2.6 x 4 = 10.4 divided by 2 = 5.2
Other triangle; 2.6 x 3 = 7.8 /2 = 3.9
Rectangular side in front; 6.5 x 3 = 19.5
Back; 6.5 x 4 = 26
Base; 6.5 x 2.6 = 16.9
Answer: For the surface area it is 74.6 (SUPER SORRY IF THIS ISNT RIGHT OMG)

5 0

3 years ago

HELP, DUE TOMORROW!!!<br> How do you solve real world problems using expressions?

lutik1710 [3]

Expressions are problems with no answer. I would use it to represent inequalities and/or a basic formula

6 0

3 years ago

Read 2 more answers

Pots and Pans cost $195.50 and it is 20% off, what is the cost?

erica [24]

It cost $156.40 because 20% of 195.50 is $39.10 so you subtract that from the 195.50.

8 0

2 years ago

Read 2 more answers

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose

Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

In which

x is the number of sucesses

$
e = 2.71828$ is the Euler number

$\mu$ is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson mean:

$\mu = 0.2$

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

$P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819$

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017$

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so $p = 0.181$

We want to find P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671$

0.671 = 67.1% probability that neither contains a missing pulse

8 0

2 years ago