All categories

2 years ago

Brody has his computer repaired at store A. His bill was:

Mathematics

2 answers:

PIT_PIT [208]2 years ago

4 0

Initial repair cost = $1200

Gratuity = 15% of the initial repair cost

= 15% of 1200

$=\frac{15}{100} (1200)$

= 15 × 12

= 180

Hence, the gratuity for the service is $180.

7nadin3 [17]2 years ago

4 0

Answer:

180

Step-by-step explanation:

You might be interested in

Tom buys a torch and a battery.

nevsk [136]

Answer:

battery costs: 0.39...............

8 0

2 years ago

The owner of a manufacturing plant employs eighty people. As part of their personnel file, she asked each one to record to the n

wariber [46]

Answer:

Mean of the sample = 27.83

The variance of the the sample = 106.96

Standard deviation of the sample = 10.34

Step-by-step explanation:

Step(i):-

Given random sample of six employees

x 26 32 29 16 45 19

mean of the sample

$x^{-} = \frac{26+32+29+16+45+19}{6} = 27.83$

Mean of the given data = 27.83

Step(ii):-

Given data

x : 26 32 29 16 45 19

x - x⁻ : -1.83 4.17 1.17 -11.83 17.17 -8.83

(x - x⁻)² : 3.3489 17.3889 1.3689 139.9489 294.80 77.9689

∑ (x-x⁻)² = 534.8245

Given sample size 'n' =6

The variance of given data

S² = ∑(x-x⁻)² / n-1

$S^{2} = \frac{534.8245}{6-1} = 106.9649$

The variance of the given sample = 106.9649

Step(iii):-

Standard deviation of the given data

$S = \sqrt{variance} = \sqrt{106.9649} =10.3423$

Standard deviation of the sample = 10.3423

8 0

2 years ago

James and Dan are partners in a small company. From each year’s profit, James is paid a bonus of $15000 and the remainder is sha

jarptica [38.1K]

Answer:

total20000

bonus15000

remaining 20000-15000=5000

ratio 2+3=5x

Dan part =3x

Dan value=( 5000/5)*3=3000

Step-by-step explanation:

6 0

1 year ago

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

2 years ago

The Helheim glacier in Greenland enters the ocean in a 6.3 kilometer wide fjord. The terminus of the glacier retreated 7.5 kilom

vladimir1956 [14]

Answer:

7.56 km²

Step-by-step explanation:

Given data:

Width of the fjord, w = 6.3 km

Retreated terminus of the glacier between may 2001 and June 2005, d = 7.5 km

thus, the length lost , y = 7.5 - 6.3 = 1.2 km

now, the area is given as:

A = Length × width

on substituting the values, we get

A = 1.2 × 6.3

A = 7.56 km²

Hence, the surface area lost by the glacier in the fjord is 7.56 km²

7 0

2 years ago