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KatRina [158]
3 years ago
9

Finn bought 2 packs of stickers. Each pack had the same number of stickers. A friend gave him 4 more stickers. Now he has 24 sti

ckers in all. How many stickers were in each pack. Explain how you solved it.
Mathematics
2 answers:
butalik [34]3 years ago
5 0

Let the number of stickers in each pack be 'x'

Finn bought 2 packs of stickers.

Number of stickers in 2 packs = 2 \times x = 2x

According to the question, a friend gave him 4 more stickers and now he has 24 stickers in all.

Therefore, 2x+4=24

2x=24-4

2x=20

x = 10

Therefore, number of stickers in each pack are 10.

frosja888 [35]3 years ago
4 0
24-4=20
20/2=10
10 stickers in each pack
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vovikov84 [41]

Answer:

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

Step-by-step explanation:

For each 3-pointer shot, there are only two possible outcomes. Either the player makes it, or the player does not. The same is valid for free throws. This means that both the number of 3-pointers and free throws made are given by binomial distributions.

Since 3-pointers and free throws are independent, first we find the probability of making exactly 3 3-pointers out of 10, then the probability of making exactly 5 free throws out of 10, and then the probability that the player will make exactly 3 3-pointers and 5-free throws is the multiplication of these probabilities.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which 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)!}

And p is the probability of X happening.

Probability of making 3 3-pointers out of 10:

The chances of a basketball player hitting a 3-pointer shot is 0.4, which means that p = 0.4. So this is P(X = 3) when n = 10.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.21499

Probability of making 5 free throws out of 10:

The probability of hitting a free-throw is 0.65, which means that p = 0.65. The probability is P(X = 5) when n = 10.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{10,5}.(0.65)^{5}.(0.35)^{5} = 0.15357

Calculate the probability that the player will make exactly 3 3-pointers and 5-free throws.

0.21499*0.15537 = 0.0334

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20x%20%7B%7D%5E%7B2%7D%20%20%2B%2012x%20%2B%2036" id="TexFormula1" title="f(x)
Mars2501 [29]

Answer:

B

Step-by-step explanation:

We are given the function:

\displaystyle f(x) = x^2 + 12x + 36

And we want to determine the value of:

\displaystyle f^{-1}(225)

Let this value equal <em>a</em>. In other words:  

\displaystyle f^{-1}(225) = a

Then by the definition of inverse functions:

\displaystyle \text{If } f^{-1}(225) = a\text{, then } f(a) = 225

Hence:

\displaystyle f(a) =225 = (a)^2 + 12(a) + 36

Solve for <em>a: </em>

\displaystyle \begin{aligned} 225 &= a^2 + 12a + 36 \\ \\ a^2 + 12a -189 &= 0 \\ \\ (a + 21)(a-9) &= 0\end{aligned}

By the Zero Product Property:

\displaystyle a + 21 = 0 \text{ or } a - 9 = 0

Hence:

\displaystyle a = -21 \text{ or } a = 9

Thus, f(9) = 225. Consequently, f⁻¹(225) = 9.

In conclusion, our answer is B.

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3 years ago
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2.105% is the answer!
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The following observations are on stopping distance (ft) of a particular truck at 20 mph under specified experi- mental conditio
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Answer:

see explaination

Step-by-step explanation:

Data : 32.1 , 30.6 , 31.4 , 30.4 , 31.0 , 31.9

Mean X-bar = 31.23

SD = 0.689

a)

Null Hypothesis : Xbar = mu

Alternate Hypothesis : Xbar > mu

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 30 ) /(0.689/sqrt(6))

= 4.372

P-value = ~0

Since P-vale < 0.01 , we will reject null hypothesis.

The data suggest that true average stopping ditance exceeds the maximum.

b)

i) SD = 0.65

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.65/sqrt(6))

= 0.867

P-value = 0.3859 Answer

ii) SD = 0.65

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.65/sqrt(6))

= -2.9

P-value = 0.0037 Answer

c)

i) SD = 0.8

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.8/sqrt(6))

= 0.704

P-value = 0.4814 Answer

ii) SD = 0.8

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.8/sqrt(6))

= -2.357

P-value = 0.0184 Answer

The probabilities obtained in part c are comparatively higher than that of part b.

d)

i) For alpha =0.01

z = (Xbar - mu) / (SD/sqrt(n))

=> -2.32 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-2.32

=> sqrt(n) = (0.65*(-2.32)) / (31.23 - 31)

=> n = 43 Answer

ii) For beta =0.10

z = (Xbar - mu) / (SD/sqrt(n))

=> -1.28 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-1.28

=> sqrt(n) = (0.65*(-1.28)) / (31.23 - 31)

=> n = 13 Answer

4 0
3 years ago
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