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shepuryov [24]
3 years ago
5

Three apples and 2 oranges cost $1.25, while 2 apples and an orange were 75 cents. At the rate, find the cost of 6 apples and an

orange. ( please show steps thanks!)
Mathematics
1 answer:
Effectus [21]3 years ago
3 0
X - apple
y - orange

3x+2y=1.25\\
2x+y=0.75\\\\
3x+2y=1.25\\
-4x-2y=-1.5\\
--------\\
-x=-0.25\\
x=0.25\\\\
2\cdot0.25+y=0.75\\
0.5+y=0.75\\
y=0.25\\\\
6x+y=6\cdot 0.25+0.25=1.5+0.25=\boxed{1.75}
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The area of a playground is 50 square yards. The length of the playground is 2 times longer than its width. find the length and
Julli [10]
Length x width = area
width=x
length=2x
2x*x=50
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I need help please?!!!
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A container contains 1212 diesel engines. The company chooses 88 engines at​ random, and will not ship the container if any of t
bija089 [108]

Answer:

The probability that a container will be shipped even though it contains 2 defectives if the sample size is 88, will be P(S)=85.99\%

Step-by-step explanation:

The first step is to count the number of total possible random sets of taking a sample size of 88 engines over 1212 engines of the population, so \left[\begin{array}{ccc}1212\\88\end{array}\right] =1212C88=4.7205x10^{135}

The second step is to count the number of total possible random sets of taking a sample size of 88 engines over 1210 engines (discounting the 2 defective engines) as the possible ways to succeed, so \left[\begin{array}{ccc}1210\\88\end{array}\right] =1212C88=4.0596x10^{135}

Finally we need to compute \frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 88} =\frac{4.0596x10^{135}}{4.7205x10^{135}}=0.8599=P(S), therefore the probability that a container will be shipped is P(S)=85.99\%

7 0
3 years ago
Help please, I am not really good with this subject..
VLD [36.1K]
F(x)=(x-2)^2 +66

lemme know if u need shown work, hope this helps! can u give me brainliest??
7 0
3 years ago
An English teacher needs to pick 1111 books to put on his reading list for the next school year. He has narrowed down his choice
forsale [732]

Answer:

There are 3,659,040 ways he can choose the books to put on the list.

Step-by-step explanation:

There are

12 novels

8 plays

12 nonfiction.

He wants to include

5 novels

4 plays

2 nonfiction

The order in which the novels, plays and nonfictions are chosen is not important. So we use the combinations formula to solve this problem.

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

How many ways can he choose the books to put on the list?

Novels:

5 from a set of 12. So

C_{12,5} = \frac{12!}{5!7!} = 792

Plays:

4 from a set of 8. So

C_{8,4} = \frac{8!}{4!4!} = 70

Nonfiction:

2 from a set of 12

C_{12,2} = \frac{12!}{2!10!} = 66

Total:

Multiplication of novels, plays and nonfiction.

T = 792*70*66 = 3,659,040

There are 3,659,040 ways he can choose the books to put on the list.

4 0
3 years ago
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