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leva [86]
3 years ago
14

Please help with this question.

Mathematics
1 answer:
Gnesinka [82]3 years ago
4 0

Answer:

teqawg veqwargth4ve fqhvreh

Step-by-step explanation:

aferd thgaReghaergheargh

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A scout troop is planting trees at the state park. The troop planted 7 trees in 3 hours. At that rate, how many trees will they
Novay_Z [31]

Answer:

I believe 57

Step-by-step explanation:

7 trees per 3 hours

6 hours = 14 trees

plus the 2 hours which I'm assuming 5 to 4 trees

so you multiply 19 trees per day

19 x 3 = 57 if I'm wrong please correct me :)

7 0
3 years ago
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klemol [59]

Answer:

The equation is equashon

Step-by-step explanation:

:D

7 0
3 years ago
Can somebody help me? anybody know the answer
tigry1 [53]

Answer:

  25 = c

Step-by-step explanation:

You had a good start on filling in the values in the Pythagorean Theorem. The work needed to be finished.

  a^2 +b^2 = c^2\\\\20^2 +15^2=c^2\\\\400+225=c^2\\\\625=c^2\\\\\sqrt{625}=\sqrt{c^2}\\\\\boxed{25=c}

7 0
3 years ago
Please help, I have no idea how to do this
Sever21 [200]

Answer:

the surface area increases by 37.38

Step-by-step explanation:

This is the equation for future reference.

A=2(wl+hl+hw)

3 0
3 years ago
Read 2 more answers
The number of customers waiting for gift-wrap service at a department store is an rv X with possible values 0, 1, 2, 3, 4 and co
FrozenT [24]

Let Y_i denote the number of packages brought by customer i. If, for instance, X=2 customers are in line, then the total number of packages is Y=Y_1+Y_2, and

P(Y=y)=P(Y_1=y_1,Y_2=y_2)=P(Y_1=y_1)P(Y_2=y_2)

since the number of packages brought by any customer is independent from the number of packages brought by any other.

a. P(X=3,Y=3) is the probability that a total of 3 packages are waiting to get wrapped from among a total of 3 customers. This means each customer must have exactly one package. So

P(X=3,Y=3)=P(X=3)P(Y=3)^3\approx0.0416

b. P(X=4,Y=11) is the probability that a total of 11 packages are brought by 4 customers. The only way for there to be 11 packages is if 3 customers bring 3 packages and the last 1 brings 2. There are \dbinom43=4 ways this can happen, so

P(X=4,Y=11)=4P(X=4)P(Y=3)^3P(Y=2)\approx0.0002

6 0
3 years ago
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