A rectangular prism is made up of 320 unit cubes. The prism is 8 unit cubes wide.

Graph y<1−3x. .......................... . .

Mkey [24]

Answer:

algobn tu jale

Step-by-step explanation:

yo digo q C es la megor opceon

4 0

3 years ago

Select all the examples of categorical data. colors number of siblings favorite pet profits genre of music

bulgar [2K]

Answer:

Color, favorite pet, genre of music

Step-by-step explanation:

Categorical variables are simply statistical variables which are non - numeric, usually employed in characterization and groupings based on a certain number of fixed attributes, categories. In the options atated above, the categorical variables fall under a heading with a limited and fixed attributes such as color, pet and genre of music. However, options such as number of siblings and profit are purely numeric variables which can take up any numeric digits and allow for direct numeric computation. They are called quantitative variables.

3 0

3 years ago

The functions f(x) and g(x) are described below: f(x) = 7x + 9 g(x) = 7x −4 The graph of g(x) is obtained by shifting down the g

Slav-nsk [51]

Correct Answer:
Option A

Solution:

f(x) = 7x + 9
g(x) = 7x - 4

We can re-write g(x) as:

g(x) = 7x + 9 - 13

Since 7x+9 is equal to f(x), we can write:

g(x) = f(x) - 13

This relation shows that g(x) is obtained by shifting f(x) 13 units vertically down. So the correct answer is option A.

6 0

3 years ago

Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor

castortr0y [4]

Answer:

A) 95% confidence interval for the population mean PEF for children in biomass households = (3.214, 3.386)

95% confidence interval for the population mean PEF for children in LPG households

= (4.125, 4.375)

Simultaneous confidence interval for both = (3.214, 4.375)

B) The result of the hypothesis test is significant, hence, the true average PEF is lower for children in biomass households than it is for children in LPG households.

C) 95% confidence interval for the population mean FEY for children in biomass households = (2.264, 2.336)

Simultaneous confidence interval for both = (2.264, 4.375)

This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).

Step-by-step explanation:

A) Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the z-distribution. This is because although, there is no information provided for the population standard deviation, the sample sizes are large enough for the sample properties to approximate the population properties.

Finding the critical value from the z-tables,

Significance level for 95% confidence interval

= (100% - 95%)/2 = 2.5% = 0.025

z (0.025) = 1.960 (from the z-tables)

For the children in the biomass households

Sample mean = 3.30

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation of the sample = 1.20

N = sample size = 755

σₓ = (1.20/√755) = 0.0436724715 = 0.04367

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 3.30 ± (1.960 × 0.04367)

CI = 3.30 ± 0.085598

95% CI = (3.214402, 3.385598)

95% Confidence interval = (3.214, 3.386)

For the children in the LPG households

Sample mean = 4.25

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation of the sample = 1.75

N = sample size = 750

σₓ = (1.75/√750) = 0.063900965 = 0.063901

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 4.25 ± (1.960 × 0.063901)

CI = 4.25 ± 0.125246

95% CI = (4.12475404, 4.37524596)

95% Confidence interval = (4.125, 4.375)

Simultaneous confidence interval for both = (3.214, 4.375)

B) The null hypothesis usually goes against the claim we are trying to test and would be that the true average PEF for children in biomass households is not lower than that of children in LPG households.

The alternative hypothesis confirms the claim we are testing and is that the true average PEF is lower for children in biomass households than it is for children in LPG households.

Mathematically, if the true average PEF for children in biomass households is μ₁, the true average PEF for children in LPG households is μ₂ and the difference is μ = μ₁ - μ₂

The null hypothesis is

H₀: μ ≥ 0 or μ₁ ≥ μ₂

The alternative hypothesis is

Hₐ: μ < 0 or μ₁ < μ₂

Test statistic for 2 sample mean data is given as

Test statistic = (μ₂ - μ₁)/σ

σ = √[(s₂²/n₂) + (s₁²/n₁)]

μ₁ = 3.30

n₁ = 755

s₁ = 1.20

μ₂ = 4.25

n₂ = 750

s₂ = 1.75

σ = √[(1.20²/755) + (1.75²/750)] = 0.07740

z = (3.30 - 4.25) ÷ 0.07740 = -12.27

checking the tables for the p-value of this z-statistic

Significance level = 0.01

The hypothesis test uses a one-tailed condition because we're testing in only one direction.

p-value (for z = -12.27, at 0.01 significance level, with a one tailed condition) = < 0.000000001

The interpretation of p-values is that

When the p-value > significance level, we fail to reject the null hypothesis and when the p-value < significance level, we reject the null hypothesis and accept the alternative hypothesis.

Significance level = 0.01

p-value = 0.000000001

0.000000001 < 0.01

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that true average PEF is lower for children in biomass households than it is for children in LPG households.

C) For FEY for biomass households,

Sample mean = 2.3 L/s

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation = 0.5

N = sample size = 755

σₓ = (0.5/√755) = 0.0182

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 2.30 ± (1.960 × 0.0182)

CI = 2.30 ± 0.03567

95% CI = (2.264, 2.336)

Simultaneous confidence interval for both = (2.264, 4.375)

This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).

Hope this Helps!!!

6 0

2 years ago

The statistics of nequals22 and sequals14.3 result in this 95% confidence interval estimate of sigma: 11.0less thansigmaless t

dimulka [17.4K]

Answer: no, the confidence interval for the standard deviation σ cannot be expressed as 15.7 $\pm$ 4.7

There are three ways in which you can possibly express a confidence interval:

1) inequality
The two extremities of the interval will be each on one side of the "less then" symbol (the smallest on the left, the biggest on the right) and the symbol for the standard deviation (σ) will be in the middle:
11.0 < σ < 20.4
This is the first interval given in the question and it means that the standard deviation can vary between 11.0 and 20.4

2) interval
The two extremities will be inside a couple of round parenthesis, separated by a comma, always the smallest on the left and the biggest on the right:
(11.0, 20.4)
This is the second interval given in the question.

3) point estimate $\pm$ margin of error
This is the most common way to write a confidence interval because it shows straightforwardly some important information.
However, this way can be used only for the confidence interval of the mean or of the popuation, not for he confidence interval of the variance or of the standard deviation.

This is due to the fact that in order to calculate the confidence interval of the standard variation (and similarly of the variance), you need to apply the formula:
$\sqrt{ \frac{(n-1) s^{2} }{\chi^{2}_{\alpha / 2} } } \leq \sigma \leq \sqrt{ \frac{(n-1) s^{2} }{\chi^{2}_{1 - \alpha / 2} } }$

which involves a χ² distribution, which is not a symmetric function. For this reason, the confidence interval we obtain is not symmetric around the point estimate and the third option cannot be used to express it.

4 0

2 years ago