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netineya [11]
3 years ago
12

A mechanic purchased two motor bikes for a total of 15,000 and made a profit of 10% when he sold both. He bought the first one f

or $5000 and sold that for a profit of 20%. How much did he lose or gain on the second one as a percentage? ( with explanation )
Mathematics
1 answer:
Sloan [31]3 years ago
4 0

Answer:

16

Step-by-step explanation:

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12 feet of flooring cost $51

Step-by-step explanation:

68/16 = 4.25

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Suppose you have a set of data points with a correlation of 0.85. What is the variance? A. Variance = √ r = 0.92195 B. Variance
TiliK225 [7]

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B. Variance = r2 = 0.7225

Step-by-step explanation:

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WILL GIVE BRAINLIEST! HELP QUICK
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Step-by-step explanation:

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3 years ago
Can somebody help me? -2-7p<26
LekaFEV [45]

Answer: p>−4

Step-by-step explanation:

Let's solve your inequality step-by-step.

−2−7p<26

Step 1: Simplify both sides of the inequality.

−7p−2<26

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4 0
3 years ago
A colony of cats were introduced into a reserve. The number of cats present at timet(measured in years since the colony was intr
RUDIKE [14]

ANSWER:

a) 12 cats

b) 232

c) 246

d) 14

EXPLANATION:

Given the expression for number of cats C present at time t:

C\text{ = }\frac{12.31}{0.05+0.56^t^{}}

a) To find the number of cats initially on the reserve, let t = 0

Therefore, substitute 0 for t in the equation

\begin{gathered} C\text{ = }\frac{12.31}{0.05+0.56^0} \\ \text{   = }\frac{12.31}{0.05\text{ + 1}} \\ \text{   = }\frac{12.31}{1.05} \\ =\text{ }11.72 \end{gathered}

Number of cats initially on the reserve are approximately 12 cats

b) C(10):

\begin{gathered} C(10)\text{ = }\frac{12.31}{0.05+0.56^{10}}\text{ = 232.12} \\  \end{gathered}

C(10) = 232

Here, C(10) means that C is a function of 10. This means at time = 10 years

C)Using function notation to express the number of cats present after 17 years, we have:

C(17)\text{ = }\frac{12.31}{0.05+0.56^{17}}

C(17)\text{ = }\frac{12.31}{0.05+0.56^{17}}\text{ = }245.94

Therefore, number of cats present after 17 years are approximately 246 cats

C(17) = 246 cats

d) In this case, first find the number of cats present in the 10th year and subtract from the number of cats present in the 17th year.

C(10)\text{ = }\frac{12.31}{0.05+0.56^{10}}=\text{ }232.12

From question C above, we know C(17) = 246

Therefore, the increase in cat population to be expected from the 10th year to the 17th year is:

C(17) - C(10) = 246 - 232 = 14 cats

7 0
1 year ago
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