Answer:
960 ft^2
Step-by-step explanation:
There are 9 surfaces forming the solid.
5 of them are congruent squares and 4 of them are congruent triangles.
area of square = (side)^2
area of triangle = (base)(height)/2
total surface area = 5 * area of square + 4 * area of triangle
A = 5 * (12 ft)^2 + 4 * (12 ft * 10 ft)/2
A = 5 * 144 ft^2 + 4 * 60 ft^2
A = 720 ft^2 + 240 ft^2
A = 960 ft^2
Answer:
Solve for the first variable in one of the equations, then substitute the result into the other equation.
Point Form:
(
−
3
,
−
8
)
Equation Form:
x
=-
3
,
y
=
−
8
Step-by-step explanation:
9+10=19 it’s an addition problem and the sum is 19
Answer:
68% of jazz CDs play between 45 and 59 minutes.
Step-by-step explanation:
<u>The correct question is:</u> The playing time X of jazz CDs has the normal distribution with mean 52 and standard deviation 7; N(52, 7).
According to the 68-95-99.7 rule, what percentage of jazz CDs play between 45 and 59 minutes?
Let X = <u>playing time of jazz CDs</u>
SO, X ~ Normal(
)
The z-score probability distribution for the normal distribution is given by;
Z =
~ N(0,1)
Now, according to the 68-95-99.7 rule, it is stated that;
- 68% of the data values lie within one standard deviation points from the mean.
- 95% of the data values lie within two standard deviation points from the mean.
- 99.7% of the data values lie within three standard deviation points from the mean.
Here, we have to find the percentage of jazz CDs play between 45 and 59 minutes;
For 45 minutes, z-score is =
= -1
For 59 minutes, z-score is =
= 1
This means that our data values lie within 1 standard deviation points, so it is stated that 68% of jazz CDs play between 45 and 59 minutes.
Answer:
This is an exponential decay
Because the base of the exponent is 1/4.4 which is less than 1
Step-by-step explanation:
What is exponential growth?
when the base of our exponential is bigger than 1, which means those numbers get bigger.
What is exponential decay?
when the base of our exponential is in between 1 and 0 and those numbers get smaller.