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My name is Ann [436]
3 years ago
10

Find the selling price of a $270 bike with a 24% markup

Mathematics
1 answer:
ipn [44]3 years ago
7 0
Try $64.80 + the $270
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Plz help with numbers 3 and 4 !!
goblinko [34]

Answer:

3. opposite

30°

4.  50° + 30° + 2x = 180°

80° + 2x = 180°

2x = 100°

4 0
3 years ago
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The following information matrices shows how many of each vehicle type sold and the bonus amount
Ber [7]

Do you know the answers to 8.4 quiz?

4 0
3 years ago
How to convert a repeating decimal to a fraction
egoroff_w [7]
Always place it out of 9. For example, 0.333= 3/9, or if simplified, 1/3.

0.5555? this will be 5/9.

for the other way around, make it a repeating decimal, such as 4/9, make that 0.444

If it is 33/99, or even 333/999, the rule should be the same.
8 0
3 years ago
It took Sharon 105 minutes to wash three cars and put everything away. If she spent 10
shtirl [24]
Well that would be 105 - 10. If you spent 105 minutes in total and 10 of that was putting your stuff away, take that 10 minutes away and you’d get 95. So, she spent 95 minutes washing 3 cars in total. If you divide that 95 into 3s, you get rounded 31 minutes each.
3 0
3 years ago
40% of Oatypop cereal boxes contain a prize. Hannah plans to keep buying cereal until she gets a prize. What is the probability
aleksley [76]

Answer:

(0.6)^{n}+n(0.4)(0.6)^{n-1}+\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}

Step-by-step explanation:

We will use the binomial distribution. Let X be the random variable representing the no. of boxes Hannah buys before betting a prize.

Our success is winning the prize, p =40/100 = 0.4

Then failure q = 1-0.4  = 0.6

Hannah keeps buying cereal boxes until she gets a prize. Then n be no. times she buys the boxes.

P(X ≤ 3) = P(X=0) +P(X=1)+P(X=2)+P(X=3)

             = \binom{n}{0}p^{0}q^{n-0} +  \binom{n}{1}p^{1}q^{n-1}+ \binom{n}{2}p^{2}q^{n-2}+ \binom{n}{3}p^{3}q^{n-3}


             =  q^{n}+npq^{n-1}+\frac{n(n-1)(p)^{2}q^{n-2}}{2}+\frac{n(n-1)(n-2)p^{3}q^{n-3}}{6}

             = (0.6)^{n}+n(0.4)(0.6)^{n-1}+[tex]\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}


7 0
3 years ago
Read 2 more answers
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