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rodikova [14]
3 years ago
7

Suppose that a teacher wishes to distribute 25 identical pencils to Ahmed, Barbara, Casper, and Dieter such that Ahmed and Diete

r receive at least one pencil each, Casper receives no more than five pencils, and Barbara receives at least four pencils. In how many ways can such a distribution be made?
Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
5 0

Answer:

we will change the problem in stages

pre-place 1 in x1 & x4, and 4 in x2, thus guaranteeing their minimum requirements.

now 19 pencils remain to be distributed

when there are no restrictions on distributing n identical objects to k distinct piles, the formula is (n+k-1)C(k-1) which here translates to 22C3

finally, we need to see that x3 = 5, so we place 6 in x3 (guaranteeing that it violates the condition), and subtract such violations, ie (22-6)C3 = 16C3

ans: 22C3 - 16C3 = 980

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Martha Washington walks 30 minutes each day. If she walks for 140 days, how many hours will she walk?
telo118 [61]

70

make 30 minutes into an hour then divide 140 by 2

140/2=70

70x1=70

5 0
2 years ago
A circular jogging track forms the edge of a circular lake that has a diameter of 2 miles. Johanna walked once around the track
harkovskaia [24]

Answer:

  C.  2.0 < t < 2.5

Step-by-step explanation:

  time = distance / speed

The circumference of the lake is given by ...

  C = πd = 2π miles ≈ 6.28 miles

Then Johanna's time is ...

  (6.28 mi)/(3 mi/h) ≈ 2.09 h

This time is in the interval (2, 2.5), so matches choice C.

___

<em>Alternate solution</em>

If we take pi to be 3, then this boils down to ...

  2×3/3 = 2 . . . hours

Pi is on the order of 5% more than 3, so her time will be on the order of 5% more than 2 hours, or just above 2, but not as great as 2.5 hours. This sort of estimating can get you to the correct answer without a calculator.

6 0
3 years ago
Trapezia ABCD and PQRS are similar.<br> Find the area shaded in green.
avanturin [10]

The area shaded in green is 864 cm²

<h3>Similar figures</h3>

Similar figures, corresponding angles are congruent and the sides are ratio of each other. Therefore,

AB / PQ = CD / RS

30 / 10 = 24 / RS

30RS = 240

RS = 240 / 30

RS = 8 cm

let find the height of trapezium PQRS.

AB / PQ  = 36 / h

30 / 10 = 36 / h

30h = 360

h = 360 / 30

h = 12 cm

Therefore,

area of the green portion = area of ABCD - area of PQRS

<h3>Area of a trapezium</h3>
  • area = 1 / 2(a + b)h

Therefore,

area of ABCD = 1 / 2(24 + 30)36 =  1 / 2 (54)36 = 1944 / 2 = 972 cm²

area of PQRS = 1 / 2(10 + 8)12 = 1 / 2(18)12 = 216 / 2 = 108 cm²

Area of the green portion = 972 - 108 = 864 cm²

learn more on trapezium here: brainly.com/question/11961445

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