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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

I don't know what to do

True [87]

Your answer is 1/4, -7/3,-4

8 0

3 years ago

The deer population in a national wildlife refuge has been decreasing by 5% each year.

mestny [16]

Answer:

750 $(1-\frac{5}{100})^{n}$

Step-by-step explanation:

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3 years ago

Somebody help me !!!! HELPP please!!!

stellarik [79]

Step 1: Read and understand the problem statement.

You are given (time, depth) pairs of (20 s, 8 cm) and (40 s, 0 cm) and asked to write an equation that describes the relationship of depth (y) to time (x).

The rate of change is (0 cm -8 cm)/(40 s -20 s) = -8 cm/(20 s) = -2/5 cm/s. Then in point-slope form using the second point, the linear function rule is
y = (-2/5)(x -40) +0
You can expand this to
y = (-2/5)x +16
y = -0.4x +16 . . . . . . using a decimal number for the slope

_____
If the bathtubs in your "draining race" start with the same level, the one with the steepest slope (-0.5 cm/s) will win.

8 0

3 years ago

A geometric sequence is defined by the general term tn = 75(5n), where n ∈N and n ≥ 1. What is the recursive formula of the sequ

andreyandreev [35.5K]

The correct answer is C) t₁ = 375, $t_n=5t_{n-1}$ .

From the general form,
$t_n=75(5)^n$ , we must work backward to find t₁.

The general form is derived from the explicit form, which is
$t_n=t_1(r)^{n-1}$ . We can see that r = 5; 5 has the exponent, so that is what is multiplied by every time. This gives us

$t_n=t_1(5)^{n-1}$

Using the products of exponents, we can "split up" the exponent:
$t_n=t_1(5)^n(5)^{-1}$

We know that 5⁻¹ = 1/5, so this gives us
$t_n=t_1(\frac{1}{5})(5)^n
\\
\\=\frac{t_1}{5}(5)^n$

Comparing this to our general form, we see that
$\frac{t_1}{5}=75$

Multiplying by 5 on both sides, we get that
t₁ = 75*5 = 375

The recursive formula for a geometric sequence is given by
$t_n=t_{n-1}(r)$ , while we must state what t₁ is; this gives us

$t_1=375; t_n=t_{n-1}(5)$

3 0

3 years ago