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Virty [35]
3 years ago
9

How is the graph of y = (x minus 1) squared minus 3 transformed to produce the graph of y = one-half (x + 4) squared?

Mathematics
2 answers:
Valentin [98]3 years ago
6 0

Answer:

A

Step-by-step explanation:

edgenuity 2020

hope this helps!

san4es73 [151]3 years ago
3 0

Step-by-step explanation:

y = (x − 1)² − 3 → y = ½ (x + 4)²

The graph is shifted 3 units up, 5 units left, and scaled vertically by ½.

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Ghella [55]

Answer:

Step-by-step explanation:

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2 years ago
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This data set represents the number of cups of coffee sold in a caf_ between 8 a.m. and 10 a.m. every day for 14 days.
Elis [28]
Put numbers in order: 4,5,6,6,7,8,9,9,10,10,12,12,14,15

Cut the list into two equal parts: 4,5,6,6,7,8,9 and 9,10,10,12,12,14,15

Find middle numbers: 4,5,6,6,7,8,9 and 9,10,10,12,12,14,15

Lower quartile: 6
Upper quartile: 12

The difference of the values of the first and third quartiles of the data set is 12 - 6 = 6
5 0
3 years ago
Write the first five terms of the sequence defined by the explicit formula
denis-greek [22]

<u>Answer:</u>

The correct answer option is: 19, 13, 3, -11, -29.

<u>Step-by-step explanation:</u>

We are given the following explicit formula for an arithmetic sequence:

a_n=21-2n^2

So to find the first five terms of this sequence, we will substitute the values of n (num here in the given formula:

a_1=21-2(1)^2\\\\a_1=19


a_2=21-2(2)^2\\\\a_2=13


a_3=21-2(3)^2\\\\a_3=3


a_4=21-2(4)^2\\\\a_4=-11


a_5=21-2(5)^2\\\\a_5=-29

Therefore, the first five terms of the sequence defined by the explicit formula a_n=21-2n^2 are 19, 13, 3, -11, -29.


5 0
3 years ago
Read 2 more answers
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MrRa [10]

Answer:

a) 0.984

b) 20 people

Step-by-step explanation:

a)

If The probability that a person carries the gene is 0.1, then in a sample of 20 people, 2 should carry the gene.

Now, we want to know how many samples there are with this property.

Since we have 20 elements where 18 are alike (do not carry the gene) and 2 are alike (carry the gene), we have to compute the number of permutations of 20 elements in which 18 are alike and 2 are alike. This number is

\frac{20!}{18!2!}=190

In this 190 20-tuples there are only 3 where the 2 carriers of the gene are in the first 3 places, namely

(1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

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So there are 190-3 = 187 elements in which the first 3 elements have no 2 carriers, hence the probability that 4 or more people will have to be tested until 2 of them with the gene are detected is 187/190 = 0.984 (98.4%) rounded to three decimal places.

b)

Given that the probability that a person carries the gene is 0.1, then in a sample of 20 people, 20*0.1 = 2 should carry the gene.

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3 years ago
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