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In-s [12.5K]
4 years ago
7

john bought a bicycle for $20 sold it for $30 rebought it for $40 and resold it for $50. How much money did he make or loose?

Mathematics
1 answer:
ValentinkaMS [17]4 years ago
7 0
He lost 50 dollars and earned 80 dollars
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1. Suppose y varies directly with x, and y = 14 when x = –4. What is the value of y when x = –6?
juin [17]
<span>1) y varies directly with x, then we have: y = kx,  k is the constant of variation 

y = kx

14 = - 4k
k = 14/- 4, or - 7/2, or - 3.5 

y = kx
y = - 3.5(- 6) 
y=21

2) </span><span>201.6 is the right answer.see the procedure used in Q1.

3) K = y/x
        = 9/12
        = 3/4</span>
7 0
4 years ago
The sum of two consecutive even integers divided by four is 189.5.
goldenfox [79]
This is not a question
6 0
4 years ago
If a person picks a pair of different Months, what is the probability that both months (different) have less than 31 days?
wlad13 [49]

Answer:

15.15% probability that both months (different) have less than 31 days.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Also, the order of the months is not important. For example, picking July and August is the same as picking August and July. So the combinations formula is used to solve this problem

Combinations formula:

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)!}

There are 12 months, of which February, April, June, September and November, that is, 5 of them, have less than 31 months.

Desired outcomes:

Picking 2 months from a set of 5(those with less than 31 days). So

D = C_{5,2} = \frac{5!}{2!(3)!} = 10

Total outcomes:

Picking 2 months from a set of 12(all the months). So

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

Probability:

P = \frac{10}{66} = 0.1515

15.15% probability that both months (different) have less than 31 days.

5 0
3 years ago
If f(x) = x^2 and g(x) = 1/2x + 3, find g(f(-1)).<br> A.) -1<br> B.) 1/25<br> C.) 1/5<br> D.) 1
Veseljchak [2.6K]
F(x)=x^2
x=-1→f(-1)=(-1)^2→f(-1)=1

g(x)=1/(2x+3)
g(f(-1))=g(1)→x=1→g(1)=1/[2(1)+3]=1/(2+3)→g(f(-1))=1/5

Answer: Option C.) 1/5
5 0
4 years ago
Find the Sum of the Odd Integers between 140 and 204
Scorpion4ik [409]

Answer: 5504

Step-by-step explanation:

This involves us finding the sum of all odd integers from 141 to 203.

We can interpret this as an arithmetic sequence with first term 141 and common difference 2.

This sequence has \frac{203-141}{2}+1=32 terms.

So, the sum is:

32\left(\frac{141+203}{2} \right)=\boxed{5504}

8 0
2 years ago
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