Can someone help me understand this? Because I really don’t understand this at all

A sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters. Give a point est

xxTIMURxx [149]

Answer:

Point estimate for the population variance = 3.92 * $10^{-3}$ .

Step-by-step explanation:

We are given that a sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters ;

X X - $Xbar$ $(X-Xbar)^{2}$

1.13 1.13 - 1.188 = -0.058 3.364 * $10^{-3}$

1.15 1.15 - 1.188 = -0.038 1.444 * $10^{-3}$

1.24 1.24 - 1.188 = 0.052 2.704 * $10^{-3}$

1.27 1.27 - 1.188 = 0.082 6.724 * $10^{-3}$

$\sum (X-Xbar)^{2}$ = 0.01568

Firstly, Mean of above data, $Xbar$ = $\frac{\sum X}{n}$ = $\frac{1.15+1.24+1.15+1.27+1.13}{5}$ = 1.188

Point estimate of Population Variance = Sample variance

= $\frac{\sum (X-Xbar)^{2}}{n-1}$ = $\frac{0.01568}{4}$ = 3.92 * $10^{-3}$ .

Therefore, point estimate for the population variance = 3.92 * $10^{-3}$ .

5 0

3 years ago

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In planning her retirement, Liza deposits some money at 2% interest, twice as much deposited at 2.5%. What is the amount deposit

harkovskaia [24]

She deposited two amounts, let's say "a" and "b"
"a" earning 2% interest
"b" earning 2.5% interest

the 2.5% one, is twice as much as the 2% one
so.. one can say that, whatever "a" is, "b" is twice as much,
or b = 2*a -> b = 2a

now.. .the sum of both earned interest, was $1190

so.. one can say that (2% of a) + (2.5% of b) = 1190
let' us use the decimal notation then
$\frac{2}{100}a+\frac{2.5}{100}b=1190\implies 0.02a+0.025b=1190
\\ \quad \\
\textit{however, we know "b" is 2a thus}
\\ \quad \\
0.02a+0.025\boxed{b}=1190\implies 0.02a+0.025(\boxed{2a})=1190$

solve for "a" to find what the "a" amount is,
to find "b", well, "b" is "2a", so just do a 2*a to get "b"

5 0

3 years ago

I have no idea what it is

Anuta_ua [19.1K]

Answer:

snow

Step-by-step explanation:

8 0

2 years ago

14. The diagram in Fig. 1.34 shows a uniform metre rule

kozerog [31]

Answer:

(i) The total anticlockwise moment about O, is 6000 + 50 × OD

(ii) The total clockwise moment about O, is 5000 + 50 × OC

(iii) The difference between the anticlockwise and the clockwise moment is 1000 N·cm

(iv) The distance from O where a 100 gf force should be placed to balance the meter rule is 10 cm to the right of the point O on the meter rule

Step-by-step explanation:

The given parameters are;

The mass of the uniform meter rule = 100 gf

The center of the meter rule = O

The mass of the weight at point A = 150 gf

The mass of the weight at point B = 250 gf

The distance of the point A from the point O = OA = 40 cm

The distance of the point B from the point O = OB = 20 cm

(i) Whereby the point A is on the left of the meter rule, and taking the mass of half the meter rule as acting on the center, of the midspan from the left edge to the center, O, which is the point, D we have;

The total anticlockwise moment, Mₐ = 150 × 40 + 50 × OD = 6000 + 50 × OD

(ii) The total clockwise moment about O, is given similarly as follows;

The total clockwise moment, $M_c$ = 250 × 20 + 50 × OC = 5000 + 50 × OC

(iii) The difference between the anticlockwise and the clockwise moment = Mₐ - $M_c$ = 6000 + 50 × OD - 5000 + 50 × OC

Where, OD = OC, we have;

Mₐ - $M_c$ = 6000 + 50 × OD - 5000 + 50 × OD = 1,000

Mₐ - $M_c$ = 1000 N·cm

Therefore, Mₐ > $M_c$

(iv) The distance from O where a 100 g force should be placed to balance the meter rule should give a 1000 N·cm clockwise moment as follows;

Let the 100 gf load be placed at the point E, to the right of the point O, therefore;

100 gf × OE = 1000 N·cm

OE = 1000/(100) = 10

OE = 10 cm

Therefore, the 100 gf should be placed 10 cm to the right of the point O.

3 0

3 years ago

Write a recursive definition for each sequence. 5, 22, 39, 56,...

gavmur [86]

Answer:

We can see that there is a pattern that adds 17 per number.

Step-by-step explanation:

5,22,39,56,73....

Hoped this helped.

8 0

2 years ago

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