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3 years ago

10.4, 10.3, 11.7, 11.1, 8.0, 4.4, 2.6, 1.8, 2.5 4.4,

Mathematics

2 answers:

Ivenika [448]3 years ago

7 0

Answer:

Mean: 7

Median: 7,65

Mode : 4.4

range: 9.9

1.8, 2.5 2.6, 4.4, 4.4, 7.3, 8.0, 9.5, 10.3, 10.4, 11.1, 11.7

Lyrx [107]3 years ago

4 0

Answer:

Step-by-step explanation:

Median: 7.65

Mean: 7

Mode: 4.4

Range: 9.9

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What is the unit rate for 14 lb for $2.99

Alexeev081 [22]

How much for 1lb
2.99/14=0.213/1=$0.21

it is 21 cents per lb

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3 years ago

Mr. Hynes buys a large variety bag of Hershey chocolates to give out for Valentine’s Day. The bag of chocolates costs $8.00 and

Tema [17]

I think the answer is 8$ but i an not sure

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3 years ago

Solve: 1.2n+1=1-n<br> HELP PLS ASAP WILL GIVE BRAINLIEST IF CORRECT

ahrayia [7]

Answer:

n = 0

Step-by-step explanation:

1.2n + 1 = 1 - n

Add n and - 1 on both sides.

1.2n + n = 1 - 1

Combine like terms.

2.2n = 0

Divide both sides by 2.2.

n = 0

5 0

2 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago

Jay organised a large wedding and 1/3 of them chose beef 5/12 of them chose Chicken and 69 of them chose vegetarian so how many

DerKrebs [107]

Answer:

276 Guests

Step-by-step explanation:

Calculation to determine how many guest were at the wedding

Guests that choose beef =1/3

Guests that choose chicken= 5/12

Guests that choose vegetarian=69

Now let calculate the how many guest were at the wedding

Let x represent the numbers of guest that were present at the wedding

Hence,

x/3 + 5x/12 + 69 = x

9x/12 + 69 = x

12x/12 - 9x/12 = 69

3x/12 = 69

x/4 = 69

x=69*4

x = 276 guests

Therefore the numbers of guests that were at the wedding is 276

7 0

3 years ago

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