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aleksandrvk [35]
3 years ago
8

Mario is setting up a new tent during a camping trip. The tent came with 7 feet of rope. The instructions are to use 34.5 inches

of rope to tie a tarp on top of the tent. Then the remaining rope should be cut into 8 1/4 inches sections to tie a tent to states in the ground. Mario will use all the rope as instructed. Determine the number of 8 1/4 inch sections of rope mario can cut from the rope.
Mathematics
2 answers:
nevsk [136]3 years ago
5 0

Answer: There are 6 sections of  8\frac{1}{4}\ inches  of rope Mario can cut from the rope .

Step-by-step explanation:

Since we have given that

Length of the rope for the tent = 7 feet

As we know that

1\ feet=12\ inches\\\\7\ feet=12\times 7=84\ inches

Length of rope is used to tie a tarp on top of the tent = 34.5 inches

Remaining length of rope is given by

84-34.5=49.5\ inches

According to question, the remaining rope should be cut into

8\frac{1}{4}\ inches\\\\=\frac{33}{4}\ inches

So, Number of  8\frac{1}{4}\ inches sections of rope Mario can cut from the rope is given by

\frac{49.5}{8\frac{1}{4}}\\\\=\frac{49.5}{\frac{33}{4}}\\\\=\frac{49.5\times 4}{33}\\\\=6

Hence, there are 6 sections of 8\frac{1}{4}\ inches rope Mario can cut from the rope .

Furkat [3]3 years ago
3 0

Answer:

6

Step-by-step explanation:

Mario has 7 feet of rope.  Since there are 12 inches in every foot, this means he has

7(12) = 84 inches of rope.

34.5 inches of the rope must be used to tie the tarp to the top; this leaves Mario with

84-34.5 = 49.5

This remaining rope will be cut into 8 1/4 inch sections.  We can also write 8 1/4 as 8.25.  To find the number of sections this length he can cut, we divide:

49.5/8.25 = 6

He can cut 6 sections this length.

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2 years ago
The opponents of soccer team A are of two types: either they are a class 1 or a class 2 team. The number of goals team A scores
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Answer:

a) The expected number of goals team A will score is 5.1

b) The probability that team A will score a total of 5 goals is 0.1147

Step-by-step explanation:

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b) In order for X to be equal to 5 we have 5 possibilities

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- X1 is 1 and X2 is 4

- X1 is 2 and X2 is 3

- X1 is 3 and X2 is 2

- X1 is 4 and X2 is 1

- X1 is 5 and X2 is 0

Let T1 be a poisson distribution with mean λ = 2, then

P(T1=0) = e^{-2}

P(T1=1) = 2 * e^{-2}

P(T1=2) = 2 * e^{-2}

P(T1=3) = \frac{4}{3} \, e^{-2}

P(T1=4) = \frac{2}{3}\, e^{-2}

P(T1=5) = \frac{4}{15}\,e^{-2}

Lets do the same with a Poisson distribution T2 with mean λ = 3

P(T2=0) = e^{-3}

P(T2=1) = 3 \, e^{-3}\\P(T2=2) = \frac{9}{2} \, e^{-3}\\P(T2=3) = \frac{9}{2} \, e^{-3}\\P(T2=4) = \frac{27}{8} \, e^{-3}\\P(T2=5) = \frac{81}{40} \, e^{-3}

Now, we are ready to compute the probability that X is equal to 5.

P(X1 = 0, X2 = 5) = (0.6* e^{-2} + 0.4*e^{-3}) * (0.3*\frac{4}{15}e^{-2}  + 0.7*\frac{81}{40} e^{-3}) = 0.00823\\P(X1 = 1, X2 = 4) = (0.6* 2e^{-2} + 0.4*3 e^{-3}) * (0.3*\frac{2}{3}e^{-2}  + 0.7*\frac{27}{8} e^{-3}) = 0.03214\\P(X1 = 2, X2 = 3) = (0.6* 2e^{-2} + 0.4*\frac{9}{2}e^{-3}) * (0.3*\frac{4}{3}e^{-2}  + 0.7*\frac{9}{2} e^{-3}) = 0.05317\\P(X1 = 3, X2 = 2) = (0.6* \frac{4}{3}e^{-2} + 0.4*\frac{9}{2}*e^{-3}) * (0.3*2e^{-2}  + 0.7*\frac{9}{2} e^{-3}) = 0.0471

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