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Montano1993 [528]
2 years ago
6

Drag the labels to the appropriate locations on this diagram. Use labels of Group 1 to identify the structures involved in plant

terrestrial adaptations. Then identify the function of each structure on the target below it. Some targets have already been filled in.
SAT
1 answer:
Masja [62]2 years ago
7 0

The matchup is that Mycorrhiza is the association between roots of higher plant and fungus.

What is  Mycorrhizae?

This is known as an element that has or is one where there is a symbiotic association between plant roots and fungi.

Note that Their major role is to bring about a stable and better nutrient and water uptake by the host plant and thus The matchup is that Mycorrhiza is the association between roots of higher plant and fungus.

Learn more about terrestrial adaptations from

brainly.com/question/12406061

#SPJ1

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4 years ago
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1) What percent of 440 is 369?
zimovet [89]

Answer:

119.24

Explanation:

Step 1: We make the assumption that 369 is 100% since it is our output value.

Step 2: We next represent the value we seek with x​.

Step 3: From step 1, it follows that 100%=36​.

Step 4: In the same vein, x% = 440​.

Step 5: This gives us a pair of simple equations:100%=369(1) 440(2) = x%

Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS(left hand side) of both equations have the same unit (%); we have 100%/x% = 369/440

Step 7: Taking the inverse (or reciprocal) of both sides yields x%/100% = 440/369

Therefore, 440​ is 119.24% of 369.

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3 years ago
Antique Accents tracks their daily profits and has found that the distribution of profits is approximately normal with a mean of
mezya [45]

Answer;

a) 0.434

b) 0.983

c) 0.367

Explanation:

The exact question with the given parameters wasn't obtained online, but the same question, albeit with different parameters is then obtained. Hopefully, this Helps to solve the complete question with the required parameters.

Antique Accents tracks their daily profits and has found that the distribution of profis is approximately normal with a mean of $17,700.00 and a standard deviation of about $900.00. Using this information, answer the following questions For full marks your answer should be accurate to at least three decimal places. Compute the probability that tomorrow's profit will be

a) less than $16,791 or greater than $18,231

b) greater than $15,783

c) between $17,997 and $20,130

Solution

This is a normal distribution problem with

Mean = μ = $17,700

Standard deviation = σ = $900

a) less than $16,791 or greater than $18,231. P(x < 16,791) or P(X > 18,231) = P(X < 16,791) + P(x > 18,231)

We first standardize 16,791 and 18,231

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 16791

z = (x - μ)/σ = (16791 - 17700)/900 = - 1.01

For 18231

z = (x - μ)/σ = (18231 - 17700)/900 = 0.59

To determine the required probability

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

We'll use data from the normal probability table for these probabilities

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

P(z < -1.01) = 0.15625

P(z > 0.59) = 1 - (z ≤ 0.59) = 1 - 0.7224 = 0.2776

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59) = 0.15625 + 0.2776 = 0.43385 = 0.434 to 3 d.p

b) greater than $15,783. P(x > 15783)

We standardize 15783

z = (x - μ)/σ = (15783 - 17700)/900 = -2.13

To determine the required probability

P(x > 15783) = P(z > -2.13)

We'll use data from the normal probability table for this probability

P(x > 15783) = P(z > -2.13) = 1 - P(z ≤ - 2.13)

= 1 - 0.01659 = 0.98341 = 0.983 to 3 d.p.

c) between $17,997 and $20,130.

P(17,997 < x < 20,130)

We first standardize 17,997 and 20,130

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 17,997

z = (x - μ)/σ = (17,997 - 17700)/900 = 0.33

For 20,130

z = (x - μ)/σ = (20,130 - 17700)/900 = 2.70

To determine the required probability

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

We'll use data from the normal probability table for these probabilities

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

= P(z < 2.70) - P(z < 0.33)

= 0.99653 - 0.62930

= 0.36723 = 0.367 to 3 d.p.

Hope this Helps!!!

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marusya05 [52]

Answer:

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Explanation:

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