3 2/3 + 2 3/5 + x = 8

A snail can crawl 2 fiths of a meter in a minute. How many minutes will it take the snail to crawl 6 meters? Enter your answer i

Natasha2012 [34]

15 minutes because if you add 2/5 together 5 times you get 2 and you need 6 so 2+2+2=6 so you add 5+5+5=15 so your answer is 15.

6 0

3 years ago

I'm writing a short story and I'm stuck on what should happen next. could you give me some ideas?

GarryVolchara [31]

Answer:

Either they converse s little more like if Kate asks "You came just at the right time! When did you get here?" or Alyiah says something like "Now come with me, I'm taking you to somewhere safe"

Step-by-step explanation:

This story sounds super cool, keep it up !!!

4 0

3 years ago

Solve for x: (x+a)(x-b) = x^2 -1 How do I solve this ?

Sveta_85 [38]

You must factor.

$x^2-1=(x+1)(x-1)$

The factored form is equal to the assigned form.

$(x+1)(x-1)=(x+a)(x-b)$

Values a, b correspond to the factored 1, -1.

Value of x however remains unknown. But we can change that.

Given factored result,

$(x+1)(x-1)$

We can set expression equal to 0 then find such x that when added with number a or b results with 0.

$(x+1)(x-1)=0$

Hence x has 2 solutions.

$x+1=0\Rightarrow\boxed{x_1=-1} \\x-1=0\Rightarrow\boxed{x_2=1}$

Hope this helps.

6 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 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%.

4 0

3 years ago

What is the median of the data set?

weeeeeb [17]

Answer:

A) 67

Step-by-step explanation:

5 0

3 years ago

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