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2 years ago

Bruelle exercises for 3/4 hours each day for 6 days in a row altogether how many hours does she exercise during the six days

Mathematics

2 answers:

Mandarinka [93]2 years ago

8 0

Answer:

18-24 hours

Step-by-step explanation:

Allushta [10]2 years ago

3 0

Answer: PRETTY SURE ITS 45?

Step-by-step explanation:

6 days a week x 3/4 hours

3/4 as a time is 45mins

45x6=270?

move the decimal over and get 27

270 divided by 6 =45

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a farm has 75 acres of wheat .the farmer can harvest the wheat from 12 acres per day. In how many days will all of the fields be

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75/12 which is about 6, or 6.25

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A gram is 300 times smaller than kilogram true or false

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False it is 1,000 times smaller

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How do you add vectors? Add the vector 3,4 to the vector that goes 7 units at an angle of 2π/3.

Gekata [30.6K]

The sum of two vectors is (- 0.5, 10.1)

<u>Explanation: </u>

To add two vectors, add the corresponding components.

Let u =⟨u1,u2⟩ and v =⟨v1,v2⟩ be two vectors.

Then, the sum of u and v is the vector

u +v =⟨u1+v1, u2+v2⟩

(b)

Two vectors = ( 3, 4 )

angle 2π/3 = 120°

In x axis, the vector is = 7 cos 120°

$= 7 X -\frac{1}{2} \\\\= -3.5$

In y axis, the vector is = 7 sin 120°

= 7 X 0.866

= 6.062

The second vectors are ( -3.5, 6.062)

Sum of two vectors = [( 3 + (-3.5) ), (4 + 6.062)]

= (- 0.5, 10.1)

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2 years ago

Is y = 5(1.04)* A. growth or B. decay? <br><br>

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growth

Step-by-step explanation:

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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

2 years ago