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Westkost [7]
3 years ago
12

A basketball player recorded her points for 11 consecutive games. Her points are listed below. 13, 20, 15, 22, 24, 21, 18, 20, 2

5, 17, 27 How many points would she need to score in her 12th game in order to decrease the interquartile range by 1 point?
Mathematics
2 answers:
yaroslaw [1]3 years ago
8 0

Answer:

answer is C ( 23)

Step-by-step explanation:

answered on edu

Jlenok [28]3 years ago
4 0

Answer:

the answer is 23 on edguniity

Step-by-step explanation:


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What’s A= a+b divided by 2
pentagon [3]

Step-by-step explanation:

A=a+b/2

2×A=a+b

2A=a+b

2a-a=b

a=b

hope it helps

3 0
3 years ago
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Todd takes 3 hours to weed the garden. Sophie takes 3.5 hours to weed the garden. How long would they take to weed the garden wo
nignag [31]

Answer:

I think the answer would be 6.5

5 0
4 years ago
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What is the area of a rectangle with vertices (-8, -2), (-3,-2), (-3,-6), and (-8. -6)?
noname [10]

Answer:

31 square units

8 0
3 years ago
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AaBbCc x AaBbCc (use for all 3 questions) 1. What is the probability that this individual will: AABBcc 2. What is the probabilit
tankabanditka [31]

Answer: 1. P = 1/64

             2. P = 1/32

             3. P = 1/8

Step-by-step explanation:

In genetics, an ofspring inherits one copy of an allele of each parent, which can be described in the tables below called Punnett Square:

For crossing Aa x Aa:

       A      a

A    AA   Aa

a    Aa    aa

For crossing Bb x Bb:

        B      b

B     BB    Bb

b     Bb    bb

For crossing Cc x Cc:

       C      c

C    CC    Cc

c     Cc     cc

We can separate them because they are assorted independently.

For offspring with <u>genotype</u> <u>AABBcc</u>, probability will be:

P(AA) = 1/4

P(BB) = 1/4

P(cc) = 1/4

As all three probabilities has to happen at the same time, it is a "E" rule:

P(AABBcc) = (\frac{1}{4}) (\frac{1}{4}) (\frac{1}{4})

P(AABBcc) = 1/64

Probability for the individual to be AABBcc is 1/64 or 1.56%.

<u>Genotype</u> <u>AaBBcc</u>:

P(Aa) = 2/4 = 1/2

P(BB) = 1/4

P(cc) = 1/4

P(AaBBcc) = (\frac{1}{2}) (\frac{1}{4}) (\frac{1}{4})

P(AaBBcc) = 1/32

Probability for the individual to be AaBBcc is 1/32 or 3.12%

<u>Genotype</u> <u>AaBbCc</u>:

P(Aa) = 1/2

P(Bb) = 1/2

P(Cc) = 1/2

P(AaBbCc) = (\frac{1}{2}) (\frac{1}{2}) (\frac{1}{2})

P(AaBbCc) = 1/8

Probability for the individual to be AaBbCc is 1/8 or 12.5%.

4 0
3 years ago
Can someone please help me with this question? I have been stuck on it for a while now​
frez [133]

Answer:

<em>w = 12 cm  </em>and <em> l = 20 cm</em>

Step-by-step explanation:

3 0
3 years ago
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