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1 year ago

What is the slope of the line that passes through the points (-20, 18) and (30, 14)?

1 answer:

7 0

Gradient =

Change in Y
———————
Change in X

Change in Y is 18 - 14 = 4
Change in X is -20 - 30 = -50

Therefore the gradient is 4/-50 which is then simplified to - (2/25)

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A teacher wants to buy supplies for two of their students, Susie and John. This teacher wants to spend at least $5 on each stude

tino4ka555 [31]

X and y represent the two students.

Teacher wants to spend at least $5 in each of x and y. That means x or y could be either equal to 5 or higher than 5.
x ≥ 5
y ≥ 5

Teacher only spends under $30. That means the sum of x and y couldn't be equal to higher than 30. It should be lower than 30.
x + y < 30

The correct answer is option D

6 0

3 years ago

The large sphere has a diameter of 12 feet.

True [87]

Answer:

2520ft

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Solve -5x2 - 20x + 35 = 30 by completing the square.

Semenov [28]

Answer:

x=−2−√5 or x=−2+√5

Step-by-step explanation:

-5x2 - 20x + 35 = 30

Answer = x=−2−√5 or x=−2+√5

5 0

2 years ago

II. Using Radians to Measure Arcs and Angles

padilas [110]

Pi radians is 180 degrees.
A
1) 180/3=60
2) 270
3) 45
4) 3240
B
1) 100/180=5/9
3) 30/180=3/18
5) 1/180
C
1) (pi) cm
2) 2pi

4 0

2 years ago

The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by WorldOne Research, included the question, "Ho

vitfil [10]

Answer:

Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The 95% interval estimate of the population mean $\mu$ is

LCL = 7.431 hours to UCL = 10.569 hours

Step-by-step explanation:

Let X be the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6.

The population standard deviation is

$\sigma = 12.6 \: hours$

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

The sample size is

$n = 250$

The sample mean is

$\bar{x} = 9 \: hours$

Since the sample size is quite large then according to the central limit theorem, the sample mean is approximately normally distributed.

The population mean would be the same as the sample mean that is

$\mu = \bar{x} = 9 \: hours$

The sample standard deviation would be

$s = {\frac{\sigma}{\sqrt{n} }$

Where is the population standard deviation and n is the sample size.

$s = {\frac{12.6}{\sqrt{250} }$

$s = 0.7969 \: hours$

Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The population mean confidence interval is given by

$\text {confidence interval} = \mu \pm MoE\\\\$

Where the margin of error is given by

$$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sampling size, s is the sample standard deviation and is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 250 - 1 = 249

From the t-table at α = 0.025 and DoF = 249

t-score = 1.9695

$MoE = t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\\\MoE = 1.9695\cdot \frac{12.6}{\sqrt{250} } \\\\MoE = 1.9695\cdot 0.7969\\\\MoE = 1.569\\\\$

So the required 95% confidence interval is

$\text {confidence interval} = \mu \pm MoE\\\\\text {confidence interval} = 9 \pm 1.569\\\\\text {LCI } = 9 - 1.569 = 7.431\\\\\text {UCI } = 9 + 1.569 = 10.569$

The 95% interval estimate of the population mean $\mu$ is

LCL = 7.431 hours to UCL = 10.569 hours

8 0

2 years ago