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vazorg [7]
2 years ago
10

Can somebody plz help?

Mathematics
2 answers:
Natalija [7]2 years ago
6 0

Answer:

what do you need I'm not the smartest person but I may be able to help :)

babymother [125]2 years ago
6 0

Answer:

help with what?

Step-by-step explanation:

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What’s the slope of a line that passes through the points (-3,7) and (-3, 4)?
Dmitry_Shevchenko [17]

Answer: The slope should be infinity.

Step-by-step explanation: I hope this helps you!

4 0
3 years ago
Read 2 more answers
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
3 years ago
A commuter crosses one of three bridges, A, B, or C, to go home from work. The commuter crosses bridge A with probability 1/3, c
mihalych1998 [28]

Answer:

0.1333 = 13.33% probability that bridge B was used.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Arrives home by 6 pm

Event B: Bridge B used.

Probability of arriving home by 6 pm:

75% of 1/3(Bridge A)

60% of 1/6(Bridge B)

80% of 1/2(Bridge C)

So

P(A) = 0.75*\frac{1}{3} + 0.6*\frac{1}{6} + 0.8*\frac{1}{2} = 0.75

Probability of arriving home by 6 pm using Bridge B:

60% of 1/6. So

P(A \cap B) = 0.6*\frac{1}{6} = 0.1

Find the probability that bridge B was used.

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.75} = 0.1333

0.1333 = 13.33% probability that bridge B was used.

8 0
3 years ago
A patrolman spends three times as many hours patrolling as she does completing paper work. If she works 36 hours per week, how m
fomenos

Answer:

27 hours

Step-by-step explanation:

Let her work "x" hours for paperwork

Since, patrolling amount of hours is THREE TIMES that,

Patrolling = 3x hours

Total Hours = 36

That is including paperhour work (x) and patrolling (3x). So we can setup an equation in x and solve for x first:

x + 3x = 36

4x = 36

x = 36/4

x = 9 hours

Amount of Hours for Patarolling = 3x hours = 3(9) = 27 hours

So, she spends 27 hours patrolling

4 0
3 years ago
Explain how you compared the fractions in Exercise 9
natima [27]

that question doesnt make sense!

6 0
3 years ago
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