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3 years ago

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°

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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?

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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.

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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