What is the area? -QUICKKKKKK

X^2+14x-15=0 solve by completing square

Softa [21]

Answer:

${x}^{2} + 14x - 15 = 0 \\ {x}^{2} - x + 15x -15 = 0 \\ x(x - 1) + 15(x - 1) = 0 \\ (x - 1)(x + 15) = 0 \\ \boxed{x = 1} \\ \boxed{x = - 15}$

<h3>x=1 and x=-15 is the right answer.</h3>

7 0

3 years ago

A 2 quart container of window cleaner cost $6.96. What is the price per cup?

marissa [1.9K]

Answer:

$0.87

Step-by-step explanation:

2 quarts = 8 cups

$6.96 ÷ 8 = $0.87

4 0

3 years ago

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Data on pull-off force (pounds) for connectors used in an automobile engine application are as follows:

netineya [11]

Answer:

a. A point estimate of the mean pull-force of all connectors in the population is approximately 75.7385

The point estimate for the mean used is the sample mean

b. The point estimate of the pull force that separates the weakest 50% of the connectors from the strongest 50% is 74.3131 N

c. The point estimate of the population variance is approximately 2.8577

The point estimate for the population standard deviation is approximately 1.6905

d. The standard error of the mean is approximately 0.3315

e. The point estimate of a proportion of the connectors are;

(72.7, 0.0385) , (73.8, 0.0769) , (73.9, 0.0385) , (74, 0.0385) , (74.1, 0.0385) ,(74.2, 0.0385), (74.6, 0.0385) , (74.7, 0.0385) , (74.9, 0.0385) , (75.1, 0.0769) , (75.3, 0.0385) , (75.4, 0.0385) , (75.5, 0.0385) , (75.6, 0.0385) , (75.8, 0.0385) , (76.2, 0.0385) , (76.3, 0.0385) , (76.8, 0.0385) , (77.3, 0.0385) , (77.6, 0.0385) , (78, 0.0385) , (78.1, 0.0385) , (78.2, 0.0385) , (79.6, 0.0385)

Step-by-step explanation:

The given data for the pull-force (pounds) for connectors used in an automobile engine are presented as follows;

Pull-force (pounds); 79.6, 75.1, 78.2, 74.1, 73.9, 75.6, 77.6, 77.3, 73.8, 74.6, 75.5, 74.0, 74.7, 75.8, 72.7, 73.8, 74.2, 78.1, 75.4, 76.3, 75.3, 76.2, 74.9, 78.0, 75.1, 76.8

a. A point estimate of the mean pull-force of all connectors in the population is the sample mean given as follows;

$Mean, \ \overline x = \dfrac{\Sigma x_i}{n}$

$\Sigma x_i$ = The sum of the data = 1966.6

n = The sample size = 26

Therefore, the sample mean, $\overline x$ = 1966.6/26 ≈ 75.7385

The point estimate for the mean is approximately 75.7385

The sample mean was used as the point estimate for the mean because it is simple and representative of the sample

b. The weakest 50% of the connectors are;

72.7, 73.8, 73.8, 73.9, 74, 74.1, 74.2, 74.6, 74.7, 74.9, 75.1, 75.1, 75.3

The sum of forces that separates the weakest 50%, $\Sigma x_{i}_{50 \%}$ = 966.2

The point estimate of the pull force that separates the weakest 50% of the connectors from the strongest 50% = $\Sigma x_{i}_{50 \%}$ /13 = 966.2/13 = 74.3131 N

c. The estimate of the population variance is the sample variance, given as follows;

$s^2 =\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n - 1}}$

Where;

${\sum \left (x_i-\overline x \right )^{2} }$ ≈ 71.4415

Therefore;

$s^2 =\dfrac{71.4415 }{25}} \approx 2.8577$

The point estimate of the population variance, s², is 2.8577

The point estimate for the population standard deviation, σ, is tha sample standard deviation, 's', given as follows;

s = √s² = √2.8577 ≈ 1.6905

The point estimate for the population standard deviation ≈ 1.6905

d. The standard error of the mean is given as follows;

$SE_{\mu_x} = \dfrac{s}{\sqrt{n} }$

Therefore, we have;

$SE_{\mu_x}$ = 1.6905/√(26) ≈ 0.3315

The standard error indicates the likely hood of the difference between the sample mean and the population mean

e. The point estimate of a proportion of the connectors are;

(Number of sample with a given pull-force value)/(Sample size (26))

Therefore, using Microsoft Excel, we have

(72.7, 0.0385) , (73.8, 0.0769) , (73.9, 0.0385) , (74, 0.0385) , (74.1, 0.0385) ,(74.2, 0.0385), (74.6, 0.0385) , (74.7, 0.0385) , (74.9, 0.0385) , (75.1, 0.0769) , (75.3, 0.0385) , (75.4, 0.0385) , (75.5, 0.0385) , (75.6, 0.0385) , (75.8, 0.0385) , (76.2, 0.0385) , (76.3, 0.0385) , (76.8, 0.0385) , (77.3, 0.0385) , (77.6, 0.0385) , (78, 0.0385) , (78.1, 0.0385) , (78.2, 0.0385) , (79.6, 0.0385)

8 0

3 years ago

1) How do you find the sum of the exterior angle measures of any convex polygon?

skelet666 [1.2K]

Answer:

1) sum of all convex polygons' EXTERIOR ANGLES IS ALWAYS 360 DEGREES. If it is a triangle or even if it is a 25- gon.

2) if the polygon is regular then this means all of the sides are equal and all of the angles are equal to each other.

for example: Find the measure of an exterior angle of a REGULAR HEXAGON. so a regular hexagon has 6 sides all the same length and 6 exterior angles all the same measures. so to find the measure of 1 exterior angle you do 360 degrees /6 = 60 degrees. so an exterior angle of a regular hexagon is 60 degrees.

8 0

2 years ago

A research team conducted a study showing that approximately 25% of all businessmen who wear ties wear them so tightly that they

Sonbull [250]

Answer:A) 0.9866

Step-by-step explanation:

Probability of at least one tie is too tight is

Probability of tight tie 25/100

Probability of not a tight tie is 1-25/100=3/4

=1-(3/4)^15=0.9866

B)0.764

C)0.013

D)0.236

Method 2

This is binomial probability.

n=15 (number of businessmen

p=.25 (probability that a businessman wears the tie so tightly

A)

x = number of businessmen out of 15 who wears the tie so tightly

P( at least 1) = P( x >= 1) = 1-P(x=0)

P(x=0) = 15C0 (.25)^0 (.75)^(15-0) = (0.75)^15 = 0.013363

P( at least 1) = 1- 0.013363 = 0.9866

B)

P(more than 2) = P( x > 2) = 1-P( x=0)-P(x=1)-P(x=2)

P(x=0) = 15C0 (.25)^0 (.75)^(15-0) = (0.75)^15 = 0.013363

P(x=1) = 15C1 (.25)^1 (.75)^14 = 0.066817

P(x=2) = 15C2 (.25)^2 (.75)^13 = 0.155907

1-P( x=0)-P(x=1)-P(x=2) = 1 - 0.013363 - 0.066817 - 0.155907 = 0.763912

C)

P(x=0) = 15C0 (.25)^0 (.75)^(15-0) = (0.75)^15 = 0.013363

D)

P( at least 13) = P(x=13)+P(x=14)+P(x=15)

P(x=13)= 15C13 (.25)^13 (.75)^2 = 0.000001

P(x=14)= 15C14 (.25)^14 (.75)^1 = 0.000000

P(x=15)= 15C15 (.25)^15 (.75)^0 = 0.000000

P(x=13)+P(x=14)+P(x=15) = .000001+.000000+.000000 = 0.000001

4 0

4 years ago

Read 2 more answers