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3 years ago

What is the variance of 5, 13, 32, 37, 49, 51, 63

Mathematics

2 answers:

kramer3 years ago

6 0

Answer:

s2 ≈ 438.2383

OverLord2011 [107]3 years ago

5 0

Answer:Variance: 438.2380952380951

Step-by-step explanation:First, I added up all of the numbers:5 + 13 + 32 + 37 + 49 + 51 + 63 = 250

I squared the total, and then divided the number of items in the data set250 x 250 = 62500

62500 / 7 = 8928.57142857143

I took my set of original numbers from step 1, squared them individually this time, and added them all up:(5 x 5) + (13 x 13) + (32 x 32) + (37 x 37) + (49 x 49) + (51 x 51) + (63 x 63) = 11558

I subtracted the amount in step 2 from the amount in step 3:11558 - 8928.57142857143 = 2629.4285714285706

I subtracted 1 from the number of items in my data set:7 - 1 = 6

I divided the number in step 4 by the number in step 5:2629.4285714285706 / 6 = 438.2380952380951

This is my Variance!

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A quadratic function y=f(x)y=f(x) is plotted on a graph and the vertex of the resulting parabola is (5, -6)(5,−6). What is the v

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2 years ago

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2 years ago

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A machine has four components, A, B, C, and D, set up in such a manner that all four parts must work for the machine to work pro

olchik [2.2K]

Answer:

0.756

Step-by-step explanation:

It is given that a machine has four components, A, B, C, and D.

$P(A)=P(B)=0.93, P(C)=0.95,P(D)=0.92$

If these components set up in such a manner that all four parts must work for the machine to work properly.

We need to find the probability that the machine works properly. It means we have to find the value of $P(A\cap B\cap C\cap D)$ .

If two events X and Y are independent, then

$P(X\cap Y)=P(X)\times P(Y)$

Assume the probability of one part working does not depend on the functionality of any of the other parts.

$P(A\cap B\cap C\cap D)=P(A)\times P(B)\times P(C)\times P(D)$

Substitute the given values.

$P(A\cap B\cap C\cap D)=0.93\times 0.93\times 0.95\times 0.92$

$P(A\cap B\cap C\cap D)=0.7559226$

$P(A\cap B\cap C\cap D)\approx 0.756$

Therefore, the probability that the machine works properly is 0.756.

7 0

2 years ago

Joaquín compró una docena y media de factura la docena son parte de grasa dos tercios tiene dulce de leche y el resto son medial

Juli2301 [7.4K]

Answer:

2 facturas de grasa.

12 facturas con dulce de leche

4 medialunas

Step-by-step explanation:

Una docena contiene 12 facturas y media docena contiene 6 facturas. Entonces Joaquin compro un total de 18 facturas (12+6). Si la novena parte (1/9) de las facturas son de grasa entonces solo multiplicamos esta fraccion por la totalidad de las facturas.

18 * (1/9) = 2 facturas de grasa.

La cantidad de facturas que tienen dulce de leche son dos tercios (2/3) de la totalidad. Entonces multiplicamos esta fracion for la totalidad.

18 * (2/3) = 12 facturas con dulce de leche

Por final, la cantidad de media lunas es el resto de las facturas, entonces descontamos las facturas de grasa y con dulce de leche de la totalidad para saber cuantas facturas son medialunas

18 - 2 - 12 = 4 medialunas

8 0

2 years ago