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rewona [7]
3 years ago
6

Please help me out on this !

Mathematics
1 answer:
vichka [17]3 years ago
8 0
All to red 13:8
Green to all 5:13
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You are part of the research group that hypothesized that relative humidity played a role in the transpiration differences. What
aleksandrvk [35]

Answer:

<h2>A) Population A has a high water potential and loss due to transpiration because of the increased humidity.</h2>

Step-by-step explanation:

Notice that the hypothesis is that relative humidity affects the transpiration of the body. So, in order to prove such hypothesis, we need to find enough evidence about water loss due to higher humidity. Knowing the relative humidity as the relation between water vapor and the pressure of it.

Therefore, the right answer is A, because we need to prove that a high water potential and loss due to transpitation happens because of the increased humidity.

6 0
3 years ago
Help me please with the last 2 questions!
madam [21]

Answer:

what grade work is this 8th?

6 0
3 years ago
Find the average rate of change of each function over the interval [0, 2]. Match each representation with its respective average
WINSTONCH [101]

The average rate of change is 4.

<h3>What is function?</h3>

An expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable).

The rate of change in interval [0, 2]

r=[h(2)-h(0)]/2

h(2)=(2)²+2(2)-6

h(2)=4+4-6

h(2)=2

h(0)=(0)²+2(0)-6

h(0)=0+0-6

h(0)=-6

So, r=[h(2)-h(0)]/2

r=[2-(-6)]/2

r=(8)/2

r=4

Learn more about this concept here:

brainly.com/question/9645290

#SPJ1

6 0
2 years ago
At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro
Gnom [1K]

Answer:

A sample of 1068 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of \pi, so we use the worst case scenario, which is \pi = 0.5

Then

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}

n = 1067.11

Rounding up

A sample of 1068 is needed.

8 0
3 years ago
WRITE IN ALGEBRIC EXPRESSION<br> c) Subtract 3times of k from 25
Kobotan [32]
K=22
Hope this helps.
7 0
3 years ago
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