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3 years ago

Find the geometric mean of 2 and 98. pls explain how to solve this in steps

Mathematics

1 answer:

Nikitich [7]3 years ago

8 0

<h3>Answer: 14</h3>

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Work Shown:

$m=\text{geometric mean of x and y}\\\\m = \sqrt{x*y}\\\\m = \sqrt{2*98}\\\\m = \sqrt{196}\\\\m = \sqrt{14^2}\\\\m = 14\\\\$

The geometric mean of two numbers has us do two steps in this order

Multiply the values
Apply the square root

The geometric mean can be applied to more than two values. Regardless of how many values you have, you always multiply them out. However, the number of values will tell you what kind of root you'll take (eg: 3 values leads to a cube root; 4 values means you'll take a fourth root, etc).

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A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

3 years ago

Please help me with this quick :))))))))

vredina [299]

Answer:

129.3 in²

Step-by-step explanation:

The shaded segment is calculated as

area of sector ABC - area of triangle ABC

area of sector = area of circle × fraction of circle

A = πr² × $\frac{60}{360}$

= π × 18² × $\frac{1}{6}$

= 324π × $\frac{1}{6}$ = 54π in²

area of Δ ABC = $\frac{1}{2}$ × 18 × 18 × sin60° ≈ 40.30 in²

Then

shaded area = 54π - 40.30 ≈ 129.3 in² ( to the nearest tenth )

6 0

3 years ago

Which relation is a function?

aksik [14]

Answer:

Step-by-step explanation:

5 0

3 years ago

The area of a square concrete slab is 20 square feet. Find the side length of the square concrete slab if the area is increase b

guapka [62]

The side length of the square concrete slab if the area is increased by 25% is 5feet

The formula for calculating the area of a square is expressed as:

A = L² where:

L is the side length of the square

Given the area of the square concrete slab = 20 square feet

20 = L²

L =√20

If the area is increased by 25%, new area will be:

An = 20 + (0.25*20)

An = 20 + 5

An = 25 sq.ft

Get the new length

An = Ln²

25 = Ln²

Ln = √25

Ln = 5feet

Hence the side length of the square concrete slab if the area is increased by 25% is 5feet

Learn more here: brainly.com/question/11300671

7 0

3 years ago

What are the coordinates of the center and length of the radius of the circle whose equation is X^2+6x+4y=23?

Virty [35]

Answer:

The following equation is not a circle, but a hyperbola, because you have 4y, but no y^2

Step-by-step explanation:

Please mark for Brainliest!! :D Thanks!!

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3 0

3 years ago