All categories

3 years ago

Is a redwood tree an autograph or heterotroph?

1 answer:

7 0

autotrophs or heterotrophs:
(a) a giant redwood tree,
(b) a spider,
(c) a rose bush,
<span>(d) a mushroom,
(e) a blue whale.

</span>autotrophs=a)and (c)
<span>heterotrophs=b)d)and e)</span><span>

Hope This helped!</span>

You might be interested in

What expression is equivalent to (^3√2^5)^1/4

Nezavi [6.7K]

Soo first apply the exponent rule (2^5)^1/3*1/4
1/3*1/4= 1/12
so (2^5)^1/12
then you apply the exponent rule again 2^5*1/2
so 5*1/12= 5/12
So the answer would be 2^5/12

6 0

4 years ago

Read 2 more answers

What is the point-slope form of a line with a slope -3 that contains the point (10,-1)

Citrus2011 [14]

Here's the formula:

y - y1 = m(x - x1)

Substitute numbers accordingly:

y1: so 1 goes in the y1 spot (you switch the signs because it was already negative)

x1: and 10 goes to the x1 spot

m: -3 belongs in m

4 0

3 years ago

How to find BE and BH

ohaa [14]

Use the Pythagoras Theorem in the corresponding right angled triangles.

5 0

3 years ago

Read 2 more answers

Raiz quadrada de 2 ?

damaskus [11]

Tu respuesta: 1.41421

7 0

3 years ago

Suppose that a local TV station conducts a survey of a random sample of 120 registered voters in order to predict the winner of

forsale [732]

Answer:

a) The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b) The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

Step-by-step explanation:

Question a:

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

Sample of 120 registered voters in order to predict the winner of a local election. The Democrat candidate was favored by 62 of the respondents.

So 120 - 62 = 58 favored the Republican candidate, so:

$n = 120, \pi = \frac{58}{120} = 0.4833$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 - 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.3658$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 + 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.6001$

The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b. If a candidate needs a simple majority of the votes to win the election, can the Republican candidate be confident of victory? Justify your response with an appropriate statistical argument.

The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

8 0

3 years ago