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3 years ago

The centimorgan (cm) is a unit named in honor of thomas hunt morgan. to what is it equal?

Mathematics

1 answer:

grandymaker [24]3 years ago

8 0

$1 \ cm = 0.01 \ m.$

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Which option lists am expression that is not equivalent to 4^2/3

Vesnalui [34]

remember that

$x^{\frac{a}{b}}=\sqrt[b]{x^a}$

also $x^{-a}=\frac{1}{x^a}$

and $(a^b)^c=a^{bc}$

$4^{\frac{2}{3}}$

first one, that one is clearly not equal since 0.25≠4

2nd one, 0.25=1/4, so $0.25^{\frac{-2}{3}}=(\frac{1}{4})^{\frac{-2}{3}=$ $\frac{1}{(\frac{1}{4})^{\frac{2}{3}}}=4^{\frac{2}{3}}$ , which matches

3rd one, $\sqrt[3]{16}=\sqrt[3]{4^2}=4^{\frac{3}{2}}$ , which matches

4th one $(\sqrt[3]{4})^2=(4^{\frac{1}{3}})^2=4^{\frac{2}{3}}$ which matches

answer is $0.25^{\frac{2}{3}}$

4 0

3 years ago

Addison's Cafe offers two kinds of espresso: single-shot and double-shot. Yesterday afternoon, the cafe sold 70 espressos in all

Valentin [98]

Answer:

The number of single-shot espressos sold by the cafe was 49.

Step-by-step explanation:

It is provided that Addison's Cafe offers two kinds of espresso: single-shot and double-shot.

Total number of espressos sold by the cafe yesterday afternoon was,

N = 70

The proportion of single-shot espresso sold was, p = 0.70.

Let X = number of single-shot espressos sold.

Compute the number of single-shot espressos sold by the cafe as follows:

$E(X)=N\times p$

$=70\times 0.70\\=49$

Thus, the number of single-shot espressos sold by the cafe was 49.

6 0

3 years ago

Am I doing something wrong????

ser-zykov [4K]

Same thing is happening to me!

7 0

2 years ago

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BC=10, AC=x, and AB=2x-21. Solve for x ,and find the length of AB.

KonstantinChe [14]

Answer:

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2 years ago

Calculate the value for each determinant:

Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

$\begin{vmatrix}3 & -1\\ -3 & 4\end{vmatrix}=(3\times 4)-(-1)\times (-3)$

= 12 - 3

= 9

$\begin{vmatrix}8 & -1\\ -4 & -1\end{vmatrix}=8\times(-1)-(-1)\times(-4)$

= -8 - 4

= -12

$\begin{vmatrix}-12 & -1\\ 12 & 5\end{vmatrix}=(-12)\times(5)-(-1)\times(12)$

= -60 + 12

= -48

$\begin{vmatrix}2 & -1\\ 2 & -1\end{vmatrix}=2\times(-1)-(2)\times(-1)$

= -2 + 2

= 0

$\begin{vmatrix}0 & -1\\ 17 & 4\end{vmatrix}=0\times(4)-(-1)\times(17)$

= 17

$\begin{vmatrix}1 & 0\\ 0 & 1\end{vmatrix}=1\times(1)-(0)\times(0)$

= 1

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3 years ago

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