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ohaa [14]
3 years ago
8

A farmer planted corn in two different fields. He planted 500 seeds of regular corn in Field A and 500 seeds of experimental cor

n in Field B. At
harvest time, more ears of corn had grown in Field A than in Field B. The farmer concluded that more corn grew in Field A because of the type

of corn planted.

Choose two other possible variables that could have caused more corn to grow in Field A.
Mathematics
1 answer:
slamgirl [31]3 years ago
8 0

Answer:

1. More pollination process in the regular corn planted in field A than that of field B.

2. Low pest and insect attack on corn planted in field A compared to that of B.

Step-by-step explanation:

1. Pollination is the process in which a plant becomes fertilized, so as to produced seeds. The process requires some agent which could be; air, human, wind etc.

Therefore more pollination of the corn planted in field A than those in B would lead to more yield (ears of corn harvested) than that of B.

2. Pests and insects are agents which could reduce the yield of the corn after harvest. Comparing the two fields A and B, if the corn planted in field A were not affected by pests or insects, while those planted in B were affected, then more ears of corn would be harvested in field A.

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A particular sound wave can be graphed using the function y = 1 sin 2x . Find the amplitude of the function.
tiny-mole [99]
Answer: 1

--------------------------------------------------------------

The equation y = 1*sin(2x) is in the form y = a*sin(bx-c)+d
where
a = 1
b = 2
c = 0
d = 0
The value of 'a' determines the amplitude
amplitude = |a| = |1| = 1
8 0
3 years ago
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Pending Review Question 7
vitfil [10]

Answer:

$ 2,200 for 8%, $ 1,300 for 12%

Step-by-step explanation:

I=PRT

Let Bank 1 be <em>x</em>

332= x (8%) + (3,500-x) 12%

332= x (0.08) + (3,500-x) 0.12   change the percent to decimal

332= 0.08x + 420 - 0.12x

0.04x= 420-332

x= \frac{88}{0.04}

x= 2,200 (Bank 1)

Bank 2= 3,500-Bank 1

Bank 2= 3,500- 2,200

Bank 2= 1,300

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3 years ago
What is 20:50 (Equivalent Ratios)
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Step-by-step explanation:

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Anyone want to help me with my math assignments ill post another question with the assignments on it
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whats the question?

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7 0
3 years ago
Consider two medical tests, A and B for a virus. Test A is 90% effective at recognizing the virus when it is present, but has a
Licemer1 [7]

1

P(V|A) is not 0.95. It is opposite:

P(A|V)=0.95

From the text we can also conclude, that

P(A|∼V)=0.1

P(B|V)=0.9

P(B|∼V)=0.05

P(V)=0.01

P(∼V)=0.99

What you need to calculate and compare is P(V|A) and P(V|B)

P(V∩A)=P(A)⋅P(V|A)⇒P(V|A)=P(V∩A)P(A)

P(V∩A) means, that Joe has a virus and it is detected, so

P(V∩A)=P(V)⋅P(A|V)=0.01⋅0.95=0.0095

P(A) is sum of two options: "Joe has virus and it is detected" and "Joe has no virus, but it was mistakenly detected", therefore:

P(A)=P(V)⋅P(A|V)+P(∼V)⋅P(A|∼V)=0.01⋅0.95+0.99⋅0.1=0.1085

7 0
3 years ago
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