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Airida [17]
3 years ago
13

You fitted a regression model on the height of a particular pine tree, after 20 years of growth, as a function of the elevation

the tree is growing in a coastal mountain. You found that linear regression was appropriate to study this question i.e., the relationship was linear. The regression model showed that tree height decreased 0.2 meters per meter gained in elevation and that tree height at sea level height (Om elevation) was predicted to be 110 m. The slope of this model was:______ (use the correct sign) while the intercept was:________ . The model predicts that a tree growing at an elevation of 25 m should be________ m tall. While the predicted elevation in the mountain that a 108 m tall tree will be found is________m.
Mathematics
1 answer:
tia_tia [17]3 years ago
3 0

Answer:

chjbbnnnnnffccccgg

gfggghhb

Step-by-step explanation:

cbbhvggffdgtxgiciuyyyyytybbbhhhhhbbbbhhhhhhhhhhhhbbhhhbhbhhhhhbjh

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Find the equation of the line that contains the point (4,-2) and is perpendicular to the line y=-2x+8.
snow_tiger [21]

Answer:

\frac{1}{2} x - 4

Step-by-step explanation:

Perpendicular lines will have slopes that are the multiplicative additive inverse of each other: a line with slope 3/4 is perpendicular to a line with slope -4/3, for example.  You need a line that is perpendicular to a line with slope -2 (reading that from the -2 in the -2x portion of the given equation, which is written in slope-intercept form), so your new line must have slope +1/2.

With the slope and a point, we can come up with an equation using this formula:

y - y-coordinate = slope (x - x-coordinate)

So we have y - (-2) = 1/2 (x - 4).

Simplify the equation:  y + 2 = 1/2 x - 2

Subtract 2 from both sides: y + 2 - 2 = 1/2 x - 2 - 2

Simplify: y = 1/2 x - 4.

4 0
2 years ago
Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

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

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

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

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

5 0
3 years ago
What is the value of 4 (3) - 17​
yawa3891 [41]

Answer:

4 times 3 is 12 and 12-17 is -5.

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
In a random sample of 8 people, the mean commute time to work was 36.5 minutes and the standard deviation was 7.4 minutes. A 98%
ANEK [815]

Answer:

Step-by-step explanation:

Given that

Sample size = n = 8: sample mean = 36.5 and s = sample std dev = 7.4 minutes

98% CI using t = (28.7, 44.3) with margin of error as 7.8

Now if sigma = population std dev is known we use Z critical value

Margin of error = 2.33*8.7/\sqrt 8 = 7.167

CI = (36.5 ±7.167)

This is not wider than that used using t distribution

The margin of error of u is ___7.167

98% confidence interval using the standard normal distribution is (29.333__,43.667__)

d)the confidence interval found using the standard normal distrubution is narrower than the confidence interval found using the students t-distribution

3 0
3 years ago
Find the vertex of y = x^2 – 2x – 15
svet-max [94.6K]
The vertex is ( 1, -16 ).
7 0
3 years ago
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