Answer:
He needs 8.75 ft² polka dot paper to cover the board.
Step-by-step explanation:
How much polka dot paper she need to cover the board is the area of the board . The is a rectangle with length 3.5 ft long and width of 2.5 ft wide .
The area of the board(rectangle) = length × width
Area = LW
where
L= 3.5 ft
W = 2.5 ft
Area = 3.5 × 2.5
Area = 8.75 ft²
He needs 8.75 ft² polka dot paper to cover the board.
Answer:
Step-by-step explanation:
Given a function 
, we called the rate of change to the number that represents the increase or decrease that the function experiences when increasing the independent variable from one value "
" to another "
".
The rate of change of between and can be calculated as follows:
For:
Let's find 
and 
, where:
![[x_1,x_2]=[-4,3]](https://tex.z-dn.net/?f=%5Bx_1%2Cx_2%5D%3D%5B-4%2C3%5D)
So:
And for:
Let's find and , where:
So:
<em>Translation:</em>
Dada una función , llamábamos tasa de variación al número que representa el aumento o disminución que experimenta la función al aumentar la variable independiente de un valor "" a otro "".
La tasa de variación de entre y , puede ser calculada de la siguiente forma:
Para:
Encontremos y , donde:
Entonces:
Y para:
Encontremos y , donde:
Entonces:
Step-by-step explanation:
S = { 1, 2, 3, 4, 5, 6 7, 8 }
n ( S ) = 8
Let A be the event of getting 4,
A = { 4 }
n ( A ) = 1
P ( A )
= n ( A ) / n ( S )
= 1 / 8
Therefore, the probability of spinning a 4 is 1 / 8.
S = { A, B, A, C, A, B }
n ( S ) = 6
Let Y be the event of getting C,
Y = { C }
n ( Y ) = 1
P ( Y )
= n ( Y ) / n ( S )
= 1 / 6
Therefore, the probability of spinning a C is 1 / 6.
Answer:
H0: p ≤ 0.50
Ha: p > 0.50
Step-by-step explanation:
The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.
For the case above;
Let p represent the proportion of households in the city that gave a charitable donation in the past year
The null hypothesis is that the proportion of households in the city that gave a charitable donation in the past year is not more that 0.50
H0: p ≤ 0.50
The alternative hypothesis is that the proportion of households in the city that gave a charitable donation in the past year is more that 0.50
Ha: p > 0.50
x=10 because you add the variables to get 5x and you do 46-6 to get 40. Then you subtract 40 from 90 to get 50 and divide by 5 to get 10 as ur answer.
