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Phantasy [73]
1 year ago
13

A scientist is trying to determine the relationship between clams, snails, and squid. When she generates a molecular clock, she

is surprised to see that clams and squid are more closely related than clams and snails. Which did the molecular clock data show?
Mathematics
2 answers:
Troyanec [42]1 year ago
5 0
<span>Molecular clocks use the rate of mutation for biomolecules to work out when two life forms diverged during their development. If the molecular clock shows clams are more closely related to squid than snails, this suggests that the divergence between clams and squid occurred more recently in prehistory than the divergence of clams and snails.</span>
Degger [83]1 year ago
4 0

Answer:

Clams and squid have been evolving separately for a shorter time than clams and snails.


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Which situation involves descriptive statistics?
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The experimental probability that Kevin will catch a fly ball is equal to 7/8. About what percent of the time will Kevin catch a
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Answer:

87.5% percent chance.

Trust me it's right.

6 0
1 year ago
You deposit $300 in a savings account that pays 6% interest compounded semiannually. How much will you have at the middle of the
Makovka662 [10]

Answer:

  • The total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 0.5 years is $ 309.00.

  • The total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 1 year is $ 318.27.

Step-by-step explanation:

a)  How much will you have at the middle of the first year?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

where

  • Principle = P
  • Annual rate = r
  • Compound = n
  • Time  = (t in years)
  • A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 0.5 years

To determine:

Total amount = A = ?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

substituting the values

A=300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(0.5\right)}

A=300\cdot \frac{2.06}{2}

A=\frac{618}{2}

A=309 $

Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 0.5 years is $ 309.00.

Part b) How much at the end of one year?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

where

  • Principle = P
  • Annual rate = r
  • Compound = n
  • Time  = (t in years)
  • A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 1 years

To determine:

Total amount = A = ?

so using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

so substituting the values

A\:=\:300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(1\right)}

A=300\cdot \frac{2.06^2}{2^2}

A=318.27 $

Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 1 year is $ 318.27.

3 0
1 year ago
(-2x²y)³ / (xy²z)² * (-yz)² Hello guys, im having trouble with this babe...
Masja [62]
\frac{(-2x^2y)^3}{(xy^2z)^2(-yz)^2}

On top: (-2)³ = -2 × -2 × -2 = -8
(x²)³ = x^6 (the exponents multiply)
and of course, (y)³ = y³

On the bottom: (xy²z)² = x² y^4 z²
(-yz)² = y²z²
Multiplying these together, the exponents add and we get x² y^6 z^4.

\frac{-8x^6y^3}{x^2y^6z^4}

So, your reasoning is correct for what you have so far.

Your next step would be cancelling shared factors from the top and bottom.
Just like with regular fractions, if the numerator and denominator are divisible by the same number, you can divide them by it to simplify. (ex: 4/6 = 2/3)

Well, x^6 and x^2 are both divisible by x^2, right?
We can also cancel the y^3.

It might help to visualise the factors like this:

\frac{-8xxxxxxyyy}{xxyyyyyyzzzz}

Once you've cancelled out x² and y³ from each, you're left with

\boxed{\frac{-8x^4}{y^3z^4}}
7 0
2 years ago
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