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Contact [7]
2 years ago
10

The graph shows a fir tree's diameter growth over a 50 year period.

Mathematics
1 answer:
Brilliant_brown [7]2 years ago
6 0

Answer:

The answer is B.) The diameter of the fir tree when planted and 20 inches

Step-by-step explanation:

The y-intercept is the initial diameter of the fir tree (10 inches). At the end of 50 years, the tree's diameter is 30 inches. Therefore, 30 − 10 = 20 inches of growth occurred over the 50 year period. The diameter of the fir tree when planted. The fir tree's diameter was 10 inches when it was planted.

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The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.
Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If Z \leq -2 or Z \geq 2, the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{19 - 22}{1}

Z = -3

Z = -3 \leq -2, so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0
1 year ago
Segment YB is x+3 units long and segment BW is 2x-9 units long. The diagonal YW is ___ units long.
Gnom [1K]

Answer:

3x - 6

Step-by-step explanation:

Segment YB = x + 3

Segment BW = 2x - 9

Add the two segments.

x + 3 + 2x - 9

3x - 6

The diagonal YW is 3x - 6 units long.

6 0
1 year ago
Read 2 more answers
Harriet would like to frame her graduation picture. The picture is 8 inches wide and 11 inches long. She wants the total area of
LekaFEV [45]

Answer: i explained

Step-by-step explanation: Just do

8 times 11 = *your answer*

then

*your answer* times/subtract/divided by  108

8 0
2 years ago
Let’s assume the following statements are true: Historically, 75% of the five-star football recruits in the nation go to univers
marishachu [46]
<span>Given:
75% of the five-star football recruits in the nation go to universities in the three most competitive athletic conferences. </span>→ 25% goes to other schools.
<span>
five-star recruits get full football scholarships 93% of the time, regardless of which conference they go to. </span>→ 7% of the 5-star recruits don't get full football scholarships.<span>

a. The probability that a randomly selected five-star recruit who chooses one of the best three conferences will be offered a full football scholarship? 
75% * 93% = 69.75%

b. What are the odds a randomly selected five-star recruit will not select a university from one of the three best conferences?
25% of selected five-star recruit will not select a university from one of the three best conferences. I got the number based on the given data. Since, 75% will go, the remaining percent won't go. Total percentage should be 100% of the population. 

c. Explain whether these are independent or dependent events. Are they Inclusive or exclusive? 
These are independent events. One can still go to different school and still be legible for the full football scholarship. 

For question 2, pls. see attachment.</span>

4 0
2 years ago
A store uses the expression –2p + 50 to model the number of backpacks it sells per day, where the price, p, can be anywhere from
Grace [21]

Your question is  store uses the expression –2p + 50 to model the number of backpacks it sells per day, where the price, p, can be anywhere from $9 to $15. Which price gives the store the maximum amount of revenue, and what is the maximum revenue?

The answer is C. $12.50 per backpack gives the maximum revenue; the maximum revenue is $312.50.

7 0
1 year ago
Read 2 more answers
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