What is the purchase price of a 50-day T-bill with a maturity value of $1,801 that earns an annual interest rate of 2.528% (Assume 360 day/yr

Answer:

2/7 or about 29%

Step-by-step explanation:

2 are blue and there are 7 cards total so 2/7 and 2 divided by 7 is .2857 or .29 (29%).

Answer:

2/7

Step-by-step explanation:

There are 7 envelopes and 2 blue envelopes so the probability that the envelope will be blue is 2/7

Answer:

he ded

Step-by-step explanation:

\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \tohe no alive  because ⇆ω⇆π⊂∴∨α∈\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to

Final answer:

The standard form equation for this hyperbola, when vertices are (+-5,0) and one focus is (6,0), is x²/25 - y²/11 = 1.

Explanation:

In the question, we are given a hyperbola with vertices at (+-5,0) and one focus at (6,0). A hyperbola is defined by its distances from a given point to the two different foci, and its standard form equation along the x-axis can be written as

(x-h)²/a² - (y-k)²/b² = 1

, where (h, k) is the center of the hyperbola, a represents the distance from the center to each vertex, and b represents the distance from the center to each co-vertex. In this case,

h = 0

, since the center of the hyperbola is at the origin. The value of

a = 5

is the distance from the center to each vertex. Finally, the square of the distance c from the center to each focus is defined as

c² = a² + b²

, so we can find

b = sqrt(c² - a²)

. Here, c = 6, so b = sqrt(6² - 5²) = sqrt(11). Thus, the standard form equation of this hyperbola is

x²/25 - y²/11 = 1

.

Learn more about Hyperbola Standard form Equation here:

brainly.com/question/36025452

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3 9/10 because if you add all numerators it will equal 9 and you leave the denominator the same because in the equation all numbers have the sam denominator

Answer:

Undefined

Step-by-step explanation:

This is just a vertical line when mapped from the data points given, therefore it has an undefined slope.