25 times what is fifty?

Svetllana [295]

Answer:

B= $\frac{2A}{H}$

Step-by-step explanation:

If you noticed, A= $\frac{1}{2}$ BH is the formula for finding the area for a triangle. Your goal is to get B by it's self. Your first step will be to clear of the fraction first, so you will multiply both sides by 2. 2(A)=2( $\frac{1}{2}$ BH). On the left, you have 2×A= On the right side, you have 2( $\frac{1}{2}$ BH), but since you have a number in the equation, you will only use 2× $\frac{1}{2}$ . To solve 2× $\frac{1}{2}$ , you will cnacel out both 2's and you have 1. 1×BH will still equal BH, so you are now left will B×H.

(Your new equation looks like this by the way). 2A=BH

Since you need to get B by its self, the way to clear the H away from the B is by dividing. You will now divide the B and H aswell as 2A and H. (It will look like this) $\frac{2A}{H} = \frac{BxH}{H}$ . (Again when you have the same number or letter, you cross it out. When you divide, you won't change anything on the left side, and all you have to do on the right id to cross out the H next to the B and cross out the H on the bottom of the equation). You should be left with $\frac{2A}{H}$ = B. Now you can turn it around for your final answer. B= $\frac{2A}{H}$ .

Please let me know if i helped, how I did, and if you have any questions.

3 0

3 years ago

please solve the question and show your work find the missing number in 34x =0 show how u got your answer and write full explana

Anastaziya [24]

34x=0
So we need to isolate x. Since 34 is multiplied by x, we need to isolate x by moving 34 over the equal sign. Whenever you move something over the equal sign, the opposite is done. For 34, instead of being multiplied, it will be divided.
34x=0
X=0/34
Then you solve for x
X=0
Hope this helps!

8 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

3 years ago

PLEASE HELP‼️‼️ILL MARK BRAINLYEST ‼️‼️I MEAN IT PLEASE ‼️‼️‼️‼️‼️‼️‼️15 POINTS PLEASE ‼️‼️

adell [148]

Answer:

1. Yes; 2. Yes ; 3. No; 4. Yes

Step-by-step explanation:

The proportion will have to be read the same way. For example top Δ bottom to the lower Δ bottom (locate in the same place)

14/21 = 6 /9 Yes because the top Δ bottom to lower Δ bottom = top side to

lower same side

13.5/21 = 9/ 14 Yes because the same two sides to the same two sides in

the lower Δ

9/13.5 = 6 21 No because top Δ side to lower Δ same location but the

second one top Δ side to the lower bottom

14 /6 = 21/6 Yes because top Δ bottom to top side = lower Δ bottom to

lower side (in the same place)

8 0

2 years ago

The amounts of time employees at a large corporation work each day are normally distributed, with a mean of 7.5 hours and a stan

Ber [7]

Answer:

When sample size increases from n = 23 to n = 35, the mean of the distribution of sample means remains same but the standard deviation of the distribution of sample means decreases from 0.073 to 0.059.

Step-by-step explanation:

We are given the amounts of time employees at a large corporation work each day are normally distributed, with a mean of 7.5 hours and a standard deviation of 0.35 hour.

Random samples of size 23 and 35 are drawn from the population and the mean of each sample is determined.

Here, $\mu$ = population mean = 7.5 hours

$\sigma$ = population standard deviation = 0.35 hour

Let $\bar X$ = sample mean

(a) The random samples of size of n = 23 is drawn from the population;

So, mean of the distribution of sample means = $\mu$ = 7.5 hours

Standard deviation for the distribution of sample means is given by;

s = $\frac{\sigma}{\sqrt{n} }$ = $\frac{0.35}{23}$ = 0.073

(b) The random samples of size of n = 35 is drawn from the population;

So, mean of the distribution of sample means = $\mu$ = 7.5 hours

Standard deviation for the distribution of sample means is given by;

s = $\frac{\sigma}{\sqrt{n} }$ = $} \frac{0.35}{\sqrt{35} }$ = 0.059

(c) So, as we can see that when sample size increases from n = 23 to n = 35, the mean of the distribution of sample means remains same but the standard deviation of the distribution of sample means decreases from 0.073 to 0.059.

4 0

3 years ago