Lena needs $50 to buy some notebooks. She had saved $2 and plans to work as a babysitter to earn $8 per hour.

Oliver works at a bookstore.He packed 20 identical paperbacks and 9 identical textbooks in a box. The total mass of the books wa

max2010maxim [7]

In the problem, the first equation should be represented like this base on the variable given in the problem:

20p + 9t = 44.4

with that equation, the second equation would is given by this formula, in response with the additional number of paperback and textbook

21p + 14t = 51

to get the system equation in getting the mass of each variable, you should subtract the two equation to simplified the formula.

20p + 9t = 44.4
- 21p + 14t = 51
-------------------------
p + 5t = 6.6 // this is the system equation that could be use in getting the mass of each object

8 0

3 years ago

Triangle s=71.5 R=71.5. What is the measure of T?

Dovator [93]

T is 37

37+71.5+71.5=180

6 0

3 years ago

Under average driving conditions, the life lengths of automobile tires of a certain brand are found to follow an exponential dis

wariber [46]

Answer:

a) $P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368$

b) $P(X>30000|X>15000)=P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}, x>0$

And 0 for other case. Let X the random variable that represent "life lengths of automobile tires of a certain brand" and we know that the distribution is given by:

$X \sim Exp(\lambda=\frac{1}{30000})$

The cumulative distribution function is given by:

$F(X) = 1- e^{-\frac{x}{\mu}}$

Part a

We want to find this probability:

$P(X>30000)$ and for this case we can use the cumulative distribution function to find it like this:

$P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368$

Part b

For this case w want to find this probability

$P(X>30000|X>15000)$

We have an important property on the exponential distribution called "Memoryless" property and says this:

$P(X>a+t| X>t)=P(X>a)$

On this case if we use this property we have this: $P(X>30000|X>15000)=P(X>15000+15000|X>15000)=P(X>15000)$

We can use the definition of the density function and find this probability:

$P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607$

7 0

2 years ago

Simplify the expression z^2-z(z+3)+3z

Alisiya [41]

z^2-z(z+3)+3z

z^2 - z^2 - 3z + 3z

0 + 0

0

z^2-z(z+3)+3z = 0

5 0

2 years ago

Read 2 more answers

A scale drawing of an automobile has a scale of 1 in.

Paul [167]

The correct question in the attached figure

we know that
the scale is 1 in (scale drawing)= (1/2) foot (actual)-----> 1/(1/2)=2 in/ft
[scale]=[scale drawing]/[actual]
then
[scale drawing]=[scale]*[actual]
therefore
for actual width of the car 8 ft
[scale drawing]=[2 in/ft]*[8 ft]=16 in

the answer is 16 in

5 0

2 years ago