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VLD [36.1K]
2 years ago
13

help plzz) A teacher made a copy of a map. To make the map easier to see, the teacher enlarged the area of the map by 38%. Let d

represent the area of the original map. The expression d +0.38d is one way to represent the area of the new map. Write two expressions that represent the area of the new map.​
Mathematics
1 answer:
Stels [109]2 years ago
3 0

Answer:

first expressions = 1.38d

second expressions = (1 + (38/100)d

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x=2; the extraneous root x=42.

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01: Which gives the correct order of steps needed to multiply the problem below? 83 X 39 O Multiply the ones. 8 times 39 is 312.
Burka [1]

Answer:

Multiply the ones. 9 times 83 is 747. Multiply the tens. 30 times 83 is 2,490. Add the products to get 3,237.

The rest of the question:

Which gives the correct order of steps needed to multiply the problem below?   "83x39 "

The steps to multiply  83x39

1) multiply the ones 9 times 83 = 9 times 3 + 9 times 80 = 27 + 720 = 747

2) multiply the tens 30 times 83 = 30 times 3 + 30 times 80 = 90 + 2400 = 2490

3) add 747 and 2490 = 747 + 2490 = 3,237

According to the previous steps, the answer is option 3

Multiply the ones. 9 times 83 is 747. Multiply the tens. 30 times 83 is 2,490. Add the products to get 3,237.

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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
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