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3 years ago

Someone please help me. Im giving brainliest!!

Mathematics

2 answers:

Sindrei [870]3 years ago

7 0

Answer:

h=23

Step-by-step explanation:

if u plug 49 in the equation , you make h subject of formula by collecting like terms you'll have something like

h=49-26

h=23

lozanna [386]3 years ago

3 0

Answer:x=23

Step-by-step explanation:

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What is the answer to the question 2+2+x-7=21?

k0ka [10]

Lets get started :)

2 + 2 + x - 7 = 21
11 + x = 21

We have to subtract 11 on both sides to isolate x

11 ( - 11 ) + x = 21 ( - 11 )
11 and 11 cancels out

x = 10

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3 years ago

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What is the closest point on the line y = 2x - 5 to (4, 8)?

Gekata [30.6K]

Answer: I think it is (6.5, 8).

Step-by-step explanation:

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2 years ago

The answer to the problem.

bija089 [108]

The picture shows the answer I got!

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4 years ago

An average person in the United States throws away 4.5 × 10–1 kg of trash per day. In 2013, there were about 3.2 × 106 people in

kolbaska11 [484]

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3 years ago

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 genotype AABBcc, 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%.

Genotype AaBBcc:

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%

Genotype AaBbCc:

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%.

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3 years ago