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musickatia [10]

3 years ago

12

These graphs show production rates for a non-specialized lamp factory and a specialized lamp factory. Which is the GREATEST diff

erence between these factories?

1 answer:

denpristay [2]3 years ago

3 0

The specialized factory had the largest gains in production. Is your answer

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Which equation represents fertilization? haploid diploid = diploid diploid haploid = haploid haploid haploid = diploid diploid d

Crazy boy [7]

Answer: Haploid + Haploid = Diploid :)

6 0

2 years ago

How do you solve this kind of problem?

katen-ka-za [31]

First, we want to solve for b.

In order to solve for b, we plug in x=6 and f(x)=7 into the equation, since we know f(6)=7

7 = (3/2)(6) + b

7 = 9 + b

Subtract both sides by 9

b = -2

Now, let's insert this value into the equation

f(x) = (3/2)x - 2

Now, plug in x = -2 into the equation to calculate f(-2)

f(-2) = (3/2)(-2) - 2

= -3 -2

= -5

Thus, your answer is A.

Have an awesome day! :)

4 0

3 years ago

Ms. Sanchez’s students were asked how many hours they spend participating in organized sports each week. Each row of the table r

slava [35]

The mean of a dataset is the average of the dataset.

The mean of the sample in the third row is 2.2

The dataset is given as:

5 5 3 0 6 2 2 4 5 2 3 4 3 0 1 4 0 4 3 6

From the complete question, the samples in the third row are:

Sample: 3 4 3 0 1

The mean of a dataset is calculated using:

$\bar x = \frac{\sum x}{n}$

Where:

n = 5 i.e. the number of samples

So, we have:

$\bar x = \frac{3 + 4 + 3 + 0 + 1}{5}$

Add the expressions in the numerator

$\bar x = \frac{11}{5}$

Divide 11 by 5

$\bar x = 2.2$

Hence, the mean of the sample in the third row is 2.2

Read more about mean at:

brainly.com/question/11887128

7 0

2 years ago

Read 2 more answers

Once you have your heap structure created, next you must use it as a backing structure to a priority queue. Develop a PriorityQu

saveliy_v [14]

Using the knowledge in computational language in python it is possible to write a code that Develop a PriorityQueue data structure in a file.

<h3>Writting the code in python:</h3>

<em>import sys</em>

<em>class Node:</em>

<em>def __init__(self,key):</em>

<em>self.left = None</em>

<em>self.right = None</em>

<em>self.val = key</em>

<em>self.parent = None</em>

<em>class MinHeap:</em>

<em>def __init__(self, maxsize):</em>

<em>self.maxsize = maxsize</em>

<em>self.size = 0</em>

<em>self.Heap = Node(-1 * sys.maxsize )</em>

<em>self.FRONT = self.Heap</em>

<em># A utility function to do inorder tree traversal</em>

<em>def isLeaf(self, node):</em>

<em>if node.left == None and node.right == Node:</em>

<em>return True</em>

<em>return False</em>

<em># Function to swap two nodes of the heap</em>

<em>def swap(self, node1, node2):</em>

<em>node1.data, node2.data = node2.data, node1.data</em>

<em># Function to heapify the node at pos</em>

<em>def minHeapify(self, node):</em>

<em># If the node is a non-leaf node and greater</em>

<em># than any of its child</em>

<em>if not self.isLeaf(node):</em>

<em>if (node.data > node.left.data or</em>

<em>node.data > node.right.data):</em>

<em># Swap with the left child and heapify</em>

<em># the left child</em>

<em>if node.left.data < node.right.data:</em>

<em>self.swap(node, node.left)</em>

<em>self.minHeapify(node.left)</em>

<em># Swap with the right child and heapify</em>

<em># the right child</em>

<em>else:</em>

<em>self.swap(node, node.right)</em>

<em>self.minHeapify(node.right)</em>

<em># Function to insert a node into the heap</em>

<em>def insert(self, element):</em>

<em>if self.size >= self.maxsize :</em>

<em>return</em>

<em>self.size+= 1</em>

<em>self.bst_insert(FRONT, element)</em>

<em>current = FRONT</em>

<em>while current.parent != None and current.data < current.parent.data:</em>

<em>self.swap(current, current.parent)</em>

<em>current = current.parent</em>

<em># Function to print the contents of the heap</em>

<em>def Print(self):</em>

<em>self.inorder()</em>

<em># Function to build the min heap using</em>

<em># the minHeapify function</em>

<em>def inorder(self, root):</em>

<em>if root:</em>

<em>inorder(root.left)</em>

<em>print(root.val)</em>

<em>inorder(root.right)</em>

<em>def bst_insert(self, root, node):</em>

<em>if root is None:</em>

<em>root = node</em>

<em>else:</em>

<em>root.next = node</em>

<em>self.FRONT = node</em>

<em># Driver Code</em>

<em>if __name__ == "__main__":</em>

<em>r = Node(50)</em>

<em>bst_insert(r,Node(30))</em>

<em>bst_insert(r,Node(20))</em>

<em>bst_insert(r,Node(40))</em>

<em>bst_insert(r,Node(70))</em>

<em>bst_insert(r,Node(60))</em>

<em>bst_insert(r,Node(80))</em>

<em># Print inoder traversal of the BST</em>

<em>inorder(r)</em>

<em>print('The minHeap is ')</em>

<em>minHeap = MinHeap(15)</em>

<em>minHeap.insert(5)</em>

<em>minHeap.insert(3)</em>

<em>minHeap.insert(17)</em>

<em>minHeap.insert(10)</em>

<em>minHeap.insert(84)</em>

<em>minHeap.insert(19)</em>

<em>minHeap.insert(6)</em>

<em>minHeap.insert(22)</em>

<em>minHeap.insert(9)</em>

<em>minHeap.minHeap()</em>

<em>minHeap.Print()</em>

<em>print("The Min val is " + str(minHeap.remove()))</em>

See more about python at brainly.com/question/13437928

#SPJ1

6 0

1 year ago

The trait of “freckles” is dominant and represented by A. No freckles is recessive and represented by a. A researcher crossed tw

Ivahew [28]

Answer:

It's wrong.

Explanation:

In the Punnett square, we will use F to represent the dominant freckles gene and f to represent the recessive lack-of-freckles gene. The first individual would have FF, because homozygous means that both alleles are the same. The second individual would have Ff, because heterozygous means that it has 2 different alleles. If the first individual passes down one of its "F" alleles and the second individual passes down its "F" allele, the offspring would have 2 dominant alleles, and the researcher's conclusion is disproven.

5 0

3 years ago

Read 2 more answers