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2 years ago

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 mean temperature for the first 4 days in January was 1°C.

soldier1979 [14.2K]

-4 degrees I think.

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2 years ago

A scuba diver used the expression below to describe his position in relation to sea level.

Ipatiy [6.2K]

2 + (-20) + 8.....when sea level is 0.

starting on a platform 2 ft above sea level....so it is 2
dive down 20 ft....so this is below sea level...so it is -20
rise 8 ft...so it is 8

so ur answer is : 2nd answer choice

6 0

2 years ago

Read 2 more answers

You and a friend each randomly draw a card from a standard deck. What is the probability that at least one of you is holding a f

Andreyy89

Answer:

0.24

Step-by-step explanation:

Total cards = 52

Face cards = heart {K,Q,J} , Spade {K,Q,J} , Club {K,Q,J} , Diamond{K,Q,J} = 12

Now we are given that You and a friend each randomly draw a card from a standard deck.

Probability that no one getting face card:

Favorable events = total cards - face cards = 52 -12= 40

Total cards = 52

So, probability that no one is getting face card = $\frac{40}{52} =0.76$

Now we are supposed to find the probability that at least one of you is holding a face card

So, probability = 1 - probability that no one is getting face card

=1 - 0.76

= 0.24

So, probability that at least one of you is holding a face card is 0.24

8 0

2 years ago

Read 2 more answers

ana and christion collect stamps.together they have 500 stamps in their collection. ana has 150 fewer stamps then christian.how

mezya [45]

Lets represent Ana and Christion using the Letters A and C
a+c=500
c=a+150
now substitute c for a+150 back into the first equation
a+a+150=500
a+a=350
2a=350
Divide by two
a=175
Now that we know annie has 175 all we do is subtract that from the total (500)
C=325
Sure enough, 175 is 150 less than 325
Hope that helped, send me a message if you need clearing up :D

6 0

2 years ago

Suppose a certain airline uses passenger seats that are 16.2 inches wide. Assume that adult men have hip breadths that are norma

Pachacha [2.7K]

Answer:

Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.

There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

Step-by-step explanation:

Problems of normally distributed samples can be 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

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

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem

Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that $\mu = 14.4, \sigma = 1.1$ .

What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?

For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:

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

$Z = \frac{16.2 - 14.4}{1.1}$