All categories

3 years ago

The table shows the number of touchdowns Luis scored during his football games.

1 answer:

4 0

What is the probability of luis scoring 2 touchdowns?
C) what is the probability of luis scoring more than 1 touchdown?
D) what is the estimated value of the number of touchdowns luis scores? Show

You might be interested in

Ordered pairs that satisfy y=x^2+1

-BARSIC- [3]

Using an ordered pair is a way of graphing a linear equation on a coordinate plane.
{(-1, 2), (0, 1), (1, 2), (4, 5)}.

5 0

4 years ago

Simplify. 10 – 5.3 • (–4 + –6) a. 63 b. 47 c. –43 d. –47

Ulleksa [173]

$10-5.3*(-4+-6)$
$10-5.3*(-10)$
$10-(-53)=63$

4 0

3 years ago

Can someone please help me with this?

ss7ja [257]

Answer:

3/4

Step-by-step explanation:

There are 12 marbles in the bag

9 marbles are not blue

P(not blue) = marbles that are not blue / total marbles

= 9/12 = 3/4

7 0

3 years ago

Condense logarithmic equation PLEASE!

skelet666 [1.2K]

This is the answer: log_5(d^4 * k^12)

5 0

3 years ago

A certain bridge arch is in the shape of half an ellipse 106 feet wide and 33.9 feet high. At what horizontal distance from the

nata0808 [166]

Answer:

The horizontal distance from the center is 49.3883 feet

Step-by-step explanation:

The equation of an ellipse is equal to:

$\frac{x^2}{a^{2} } +\frac{y^2}{b^{2} } =1$

Where a is the half of the wide, b is the high of the ellipse, x is the horizontal distance from the center and y is the height of the ellipse at that distance.

Then, replacing a by 106/2 and b by 33.9, we get:

$\frac{x^2}{53^{2} } +\frac{y^2}{33.9^{2} } =1\\\frac{x^2}{2809} +\frac{y^2}{1149.21} =1$

Therefore, the horizontal distances from the center of the arch where the height is equal to 12.3 feet is calculated replacing y by 12.3 and solving for x as:

$\frac{x^2}{2809} +\frac{y^2}{1149.21} =1\\\frac{x^2}{2809} +\frac{12.3^2}{1149.21} =1\\\\\frac{x^2}{2809}=1-\frac{12.3^2}{1149.21}\\\\x^{2} =2809(0.8684)\\x=\sqrt{2809(0.8684)}\\x=49.3883$

So, the horizontal distance from the center is 49.3883 feet

8 0

4 years ago