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3 years ago

Helpppppp me plzzzzzz

Mathematics

2 answers:

igomit [66]3 years ago

6 0

Answer:

D I think hmm hmm hmm hmm hmm

zhannawk [14.2K]3 years ago

4 0

C. Exponential decay

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Complete the equation so that it represents the proportional relationship<br> shown in the table.

Bumek [7]

Answer: y= 4/3x
Explanation: the table is going up by 4/3

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2 years ago

The dimensions of a room are 4.2m by 6m. On a plan, the longer side is equal to 4 cm. What is the length of the shorter side on

Fiesta28 [93]

Answer: Length of shorter side is 2.8 cm.

Explanation:

Since we have given that

Dimensions of a room are given by

$4.2\ m\times 6\ m$

On a plan, the longer side = 4 cm

As we know that the ratio of original room is equal to ratio of plan room

So,

$\frac{4.2}{6}=\frac{x}{4}\\\\4.2\times 4=6x\\\\16.8=6x\\\\\frac{16.8}{6}=x\\\\2.8=x$

So, length of shorter side is 2.8 cm.

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Need help with this question simplify -5x+14y

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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

NikAS [45]

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:

Let X be the amount of goals scored by team A in both matches. Let X1 and X2 be the total amount of goals team A scores in match 1 and 2 respectively, then X = X1+X2, and also

E(X) = E(X1+X2) = E(X1)+E(X2) = 0.6*2+0.4*3 + 0.3*2+0.7*3 = 5.1

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

- X1 is 0 and X2 is 5

- 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.

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

$P(X1 = 4, X2 = 1) = (0.6* \frac{2}{3}e^{-2} + 0.4*\frac{27}{8}e^{-3}) * (0.3*2e^{-2} + 0.7*3e^{-3}) = 0.0225\\P(X1 = 5, X2 = 0) = (0.6* \frac{4}{15}e^{-2} + 0.4*\frac{81}{40}e^{-3}) * (0.3*e^{-2} + 0.7*e^{-3}) = 0.0047$

We can conclude that

P(X = 5) = 0.00823+0.03214+0.05317+0.0471+0.0225+0.0047 = 0.1147

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

7 0

3 years ago