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kaheart [24]
2 years ago
6

One-third of the dairy cattle on a farm are in the barn. Three-fifths of the beef cattle are in the barn. There are 30 dairy cat

tle and 15 beef cattle in thr barn. How many more dairy cattle are on the farm than beef cattle?
Mathematics
1 answer:
Korolek [52]2 years ago
8 0

Let

d = the total number of dairy cattle

b = the total number of beef cattle

2/3 of the dairy cattle on a farm are in the barn, and we know that 24 dairy cattle are in the barn thus:

(2/3)d = 24

d = 36 dairy cattle

1/4 of the beef cattle are in the barn and there are 16 beef cattle in the barn thus:

(1/4)b = 16

 

b = 64 beef cattle

64 - 36 = 28

There are 28 more beef cattle on the farm than dairy cattle.

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No, Dolly won't have enough money to buy the bracelet.

Step-by-step explanation:

$30.00 - $25.00 = $5.00

$10.00 is more than $5.00

<em>hope this helps :)</em>

8 0
3 years ago
The area of the rectangle shown is 17.36 square yards. What is the perimeter?
Nostrana [21]

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3 years ago
Can someone help me with this?
larisa [96]

9514 1404 393

Answer:

  • 2b +2c
  • 35h +15m
  • ef +eg

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The outside factor multiplies each term in parentheses.

a) 2(b+c) = 2b +2c

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4 0
2 years ago
Suppose a couple planned to have three children. Let X be the number of girls the couple has.
sesenic [268]

Answer:

a) {GGG, GGB, GBG, BGG, BBG, BGB, GBB, BBB}

b) {0,1,2,3}

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P(X=2) = \dfrac{3}{8}

d)

P(\text{3 boys}) = \dfrac{1}{8}

Step-by-step explanation:

We are given the following in the question:

Suppose a couple planned to have three children. Let X be the number of girls the couple has.

a) possible arrangements of girls and boys

Sample space:

{GGG, GGB, GBG, BGG, BBG, BGB, GBB, BBB}

b) sample space for X

X is the number of girls couple has. Thus, X can take the values 0, 1, 2 and 3 that is 0 girls, 1 girl, 2 girls and three girls from three children.

Sample space: {0,1,2,3}

c) probability that X=2

P(X=2)

That is we have to compute the probability that couple has exactly two girls.

\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}

Favorable outcome: {GGB, GBG, BGG}

P(X=2) =\dfrac{3}{8}

d) probability that the couple have three boys.

Favorable outcome: {BBB}

P(BBB) = \dfrac{1}{8}

8 0
3 years ago
First five terms of n2+5
artcher [175]
a_n=n^2+5\\\\a_1=1^2+5=1+5=6\\\\a_2=2^2+5=4+5=9\\\\a_3=3^2+5=9+5=14\\\\a_4=4^2+5=16+5=21\\\\a_5=5^2+5=25+5=30
8 0
3 years ago
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