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Bogdan [553]
3 years ago
10

Cora is getting balloons for her uncle,s birthday party she wants each balloon string to be 6 feet long

Mathematics
2 answers:
storchak [24]3 years ago
7 0

Answer:

not enough information

Step-by-step explanation:

missing information and question so it's just a random paragraph

statuscvo [17]3 years ago
3 0

how many balloons is she getting?? i can not give you an answer with not enogh info sorry :(

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L
Harlamova29_29 [7]

Answer:

(Z + 2•)

Step-by-step explanation:

5 0
3 years ago
Which statements about the hyperbola are true? Check all that apply.
max2010maxim [7]

Answer:

1.) There is a focus at (0,4)

4.) There is an asymptote with a slope of 4/3

These are the only 2 correct.

Step-by-step explanation:

Just took the quiz od edge!

4 0
3 years ago
Read 2 more answers
Solve 0= 2t^3-21t^2+ 40t
sukhopar [10]
0=2t^3-21^2+40t
switch sides
2t^3-21t^2+40t=0
solve factoring 
t(t-8)(2t-5)=0
using the zero factor principle
t=0

solve
t-8=0:t=8
solve
2t-5=0:t= \frac{5}{2}
t=0,t=8,t= \frac{5}{2}
Hope this helps

4 0
3 years ago
1. What is the radius<br> (1 Point)<br> 11 cm
mr_godi [17]

Given that we need to determine the radius of the circle.

<u>Radius:</u>

By definition of radius of circle, the radius is the  length of the line which is drawn from the center of the circle to any point on the circle.

From the figure, we need to determine the radius of the circle.

As, we can see that, the distance from the center of the circle to the point on the circle is 11 cm.

Since, we know that, the radius is the distance from the center of the circle to the point on the circle then, the radius of the given circle is 11 cm

Thus, radius of the circle is 11 cm.

3 0
3 years ago
One hundred items are simultaneously put on a life test. Suppose the lifetimes
romanna [79]

Answer:

a) \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b) \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( i-1 )st and the ith failures. Then, the T_{i} are independent with \mathrm{T}_{\mathrm{i}} being exponential with rate \frac{(101-i)}{200} .

Therefore,

a) E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]

=\sum_{i=1}^{5} \frac{200}{101-i}

\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b)

The variance is given by, \mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]

\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

7 0
3 years ago
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