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3 years ago

Michael Bought 0.44 pounds of sliced turkey. What is the value of the digit in the hundredths place?

Mathematics

2 answers:

Art [367]3 years ago

3 0

.04 hundredths. I cant believe i am still doing this in middle school.<span />

RoseWind [281]3 years ago

3 0

.04 is in the hundreths place i hope this helps.

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Wayne Gretsky scored a Poisson mean six number of points per game. sixty percent of these were goals and forty percent were assi

BartSMP [9]

Answer:

a) The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

b) 0.05 = 5% probability that he has four goals and two assists in one game

Step-by-step explanation:

In hockey, a point is counted for each goal or assist of the player.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval. The standard deviation is the square root of the mean.

(a) Find the mean and standard deviation for the total revenue he earns per game.

60% of six are goals, which means that 60% of the time he earned 3K.

40% of six are goals, which means that 40% of the time he earned 1K.

The mean is:

$\mu = 6*0.6*3 + 6*0.4*1 = 13.2$

The standard deviation is:

$\sigma = \sqrt{\mu} = \sqrt{13.2} = 3.63$

The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

(b) What is the probability that he has four goals and two assists in one game

Goals and assists are independent of each other, which means that we find the probability P(A) of scoring four goals, the probability P(B) of getting two assists, and multiply them.

Probability of four goals:

60% of 6 are goals, which means that:

$\mu = 6*0.6 = 3.6$

The probability of scoring four goals is:

$P(A) = P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.19122$

Probability of two assists:

40% of 2 are assists, which means that:

$\mu = 6*0.4 = 2.4$

The probability of getting two assists is:

$P(B) = P(X = 2) = \frac{e^{-2.4}*(2.4)^{2}}{(2)!} = 0.26127$

Probability of four goals and two assists:

$P(A \cap B) = P(A)*P(B) = 0.19122*0.26127 = 0.05$

0.05 = 5% probability that he has four goals and two assists in one game

5 0

2 years ago

Can someone fr help me with this like right now i’ll appreciate it sm

rjkz [21]

Step-by-step explanation:

Since

$\frac{108}{72} = \frac{3}{2}$

$\frac{72}{48} = \frac{3}{2}$

$\frac{48}{32} = \frac{3}{2}$

The function has a constant factor so we have a exponential function.

The equation of a exponential function is

$y = a(b) {}^{x}$

Where a is the initial value and b is the growth factor

The growth factor is 3/2,

The initial value in 32

$y = 32 ( \frac{3}{2} ) {}^{x}$

Plug in 4 for x.

You will get 162

Plug in 5 for x you will get

243.

2013- 162

2014- 243

8 0

2 years ago

An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

andrey2020 [161]

Answer:

The probability that A selects the first red ball is 0.5833.

Step-by-step explanation:

Given : An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

To find : What is the probability that A selects the first red ball?

Solution :

A wins if the first red ball is drawn 1st,3rd,5th or 7th.

A red ball drawn first, there are $E(1)= ^9C_2$ places in which the other 2 red balls can be placed.

A red ball drawn third, there are $E(3)= ^7C_2$ places in which the other 2 red balls can be placed.

A red ball drawn fifth, there are $E(5)= ^5C_2$ places in which the other 2 red balls can be placed.

A red ball drawn seventh, there are $E(7)= ^3C_2$ places in which the other 2 red balls can be placed.

The total number of total event is $S= ^{10}C_3$

The probability that A selects the first red ball is

$P(A \text{wins})=\frac{(^9C_2)+(^7C_2)+(^5C_2)+(^3C_2)}{^{10}C_3}$

$P(A \text{wins})=\frac{36+21+10+3}{120}$

$P(A \text{wins})=\frac{70}{120}$

$P(A \text{wins})=0.5833$

6 0

3 years ago

How do I solve this problem below!

mojhsa [17]

Answer:

$5.44

Step-by-step explanation:

$1.75 + $1.26 + $1.08 + $0.52 = $4.56

$10.00 - $4.56 = $5.44

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3 years ago

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Describe a model that represents 3/3 times 4/4

galben [10]

They are both fractions, but 3/3 is 1 whole and 4/4 is 1 whole also. so 1 whole times 1 whole equals 1 number form of my explanation: 3/3=1 4/4=1 1x1=1 Describe complete!

6 0

3 years ago

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