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Fynjy0 [20]
4 years ago
10

140 apples fall on the ground in 35 Seconds how many apples fall in 1 second

Mathematics
2 answers:
snow_tiger [21]4 years ago
8 0
140a/35s=4a/s

.......................
Lubov Fominskaja [6]4 years ago
5 0
140 divided by 35, 4 fall per second.
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If the maximum of a data set is 50 and the minimum of the set is 10, what is the range?
Nonamiya [84]

Answer:

40

explanation:

The range is the difference between the smallest and highest numbers in a list or set. To find the range, first put all the numbers in order. Then subtract the lowest number from the highest. The answer gives you the range of the list.

5 0
3 years ago
HELP WITH THIS QUESTION NEED HELP ASAP GIVING BRAINLIST!!!!
ipn [44]

Answer:

1. he can go up to 4 rides

2. she can buy 3 flowers

Step-by-step explanation:

1. 19 >= 5+ 3x

19 - 5 >= 3x

14 >= 3x

If you solve this you can get 4 and balance 2

2. 40 >= 2 + 11x

38 >= 11x

If you solve this you can get 3 and balance 5

6 0
3 years ago
Read 2 more answers
Explain why the following form linearly dependent sets of vectors. (solve this problem by inspection.)
Hunter-Best [27]

A set of vectors {v1,v2,...,vk} is linearly independent if the vector equation x1v1 + x2v2 + .......... + xkvk = 0 has only the trivial solution

x1 = x2 = .... = xk =0. Then the set  {v1,v2,...,vk} is linearly dependent otherwise.

So putting in the formula we get

x1u1 + x2u2 + x3u3 = 0

x1u1 + x2u2 = -x3u3

au1 + bu2 = u3               ∵ (-x1/x3 = a & -x2/x3 = b)

On putting in the values

a(3,-1) + b(4,5) = (-4,7)

(3a+4b,-a+5b) = (-4,7)

On comparing we get

3a + 4b = -4 -(1)

-a + 5b = 7  -(2)

on solving these equations we get

b = 17/19 and a= -48/19

which is non trivial.

Thus the following form linearly dependent sets of vectors.

Learn more about linear dependence of vectors here :

brainly.com/question/13776214

#SPJ4

7 0
2 years ago
Suppose X and Y are random variables with joint density function. f(x, y) = 0.1e−(0.5x + 0.2y) if x ≥ 0, y ≥ 0 0 otherwise (a) I
Hatshy [7]

a. f_{X,Y} is a joint density function if its integral over the given support is 1:

\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac1{10}\int_0^\infty\int_0^\infty e^{-x/2-y/5}\,\mathrm dx\,\mathrm dy

=\displaystyle\frac1{10}\left(\int_0^\infty e^{-x/2}\,\mathrm dx\right)\left(\int_0^\infty e^{-y/5}\,\mathrm dy\right)=\frac1{10}\cdot2\cdot5=1

so the answer is yes.

b. We should first find the density of the marginal distribution, f_Y(y):

f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\frac1{10}\int_0^\infty e^{-x/2-y/5}\,\mathrm dy

f_Y(y)=\begin{cases}\dfrac15e^{-y/5}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}

Then

P(Y\ge8)=\displaystyle\int_8^\infty f_Y(y)\,\mathrm dy=e^{-8/5}

or about 0.2019.

For the other probability, we can use the joint PDF directly:

P(X\le5,Y\le8)=\displaystyle\int_0^5\int_0^8f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=1+e^{-41/10}-e^{-5/2}-e^{-8/5}

which is about 0.7326.

c. We already know the PDF for Y, so we just integrate:

E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\frac15\int_0^\infty ye^{-y/5}\,\mathrm dy=\boxed5

5 0
3 years ago
Eddie deposited $1200 into an account that earns 3% interest compounded 2 times per year. How much money will Eddie have in his
earnstyle [38]
A=p(1+i/m)^mn
A=1,200×(1+0.03÷2)^(2×7)
A=1,478.11
5 0
4 years ago
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