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KatRina [158]
3 years ago
6

There are 500 passengers on a train.

Mathematics
2 answers:
natta225 [31]3 years ago
6 0

Answer:

There are 175 men, 200 women, and 125 children.

Step-by-step explanation:

7/20 = M/500; M = 175

0.4 x 500 = W; W = 200

175 + 200 + C = 500; C = 125

Amiraneli [1.4K]3 years ago
5 0

Answer:7/20 = 0.35 = 35%=500x.35 =175 men

40%=.40x500 =200 women

200+170=370

500-370

125 children

Step-by-step explanation:

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Answer:

b(-10) = 6

Step-by-step explanation:

Step 1: Define

b(x) = |x + 4|

b(-10) is x = -10

Step 2: Substitute and Evaluate

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Calculate the area of the equilateral triangle using the formula for area of a regular polygon, and compare it to Bianca’s answe
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The calculated areas of both are the same instead of using the different formulas that are 43. 5 squared unit.

<h3>What is the area of a regular polygon?</h3>

The area of the polygon is the product of half of the apothem of the triangle and perimeter.

The area of the polygon = \frac{1}{2} \times h \times p

The side of the triangle is 10 units.

Apothem of an equilateral triangle

s = \frac{\sqrt{3} a}{6}

s= \sqrt{3} \times 10/6

s = 2.88

Here, a is the side of the triangle.

Thus the apothem of the triangle is 2.9 units.

The perimeter of the equilateral triangle is equal to the three times the sides.

Perimeter of the equilateral triangle ,

P = 3(10)

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Therefore, the perimeter of the equilateral triangle is 30 units.

The area of the equilateral triangle-

The area of the polygon = \frac{1}{2} \times h \times p

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A = 43.5

Thus the area of the equilateral triangle is 43.5 squared units.

Therefore, The calculated areas of both are the same instead of using the different formulas.

Learn more about the apothem;

brainly.com/question/10580427

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An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberr
Sliva [168]

Answer:

A) Option B is correct.

H₀: μ₁ = μ₂

Hₐ: μ₁ - μ₂ < 0

B) t = -2.502

p-value = 0.0112

C) Option A is correct.

Reject H₀. The data suggests that cola has a higher average compression strength than the strawberry drink.

D) Option A is correct.

The distributions of compression strengths are approximately normal.

Step-by-step explanation:

The complete Question is presented in the two attached images to this answer.

A) To perform this test we first define the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, we want to test if the extra carbonation of cola results in a higher average compression strength. That is, that cola has a higher average compression strength than the strawberry drink.

Hence, the null hypothesis would be that there isn't significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

The alternative hypothesis is that there is significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

Mathematically, if the average compression strength of strawberry drink is μ₁, the average compression strength of cola is μ₂ and the difference in compression strengths is μ = μ₁ - μ₂

The null hypothesis is represented as

H₀: μ = 0 or μ₁ = μ₂

The alternative hypothesis is represented as

Hₐ: μ < 0 or μ₁ - μ₂ < 0

B) So, to perform this test, we need to compute the test statistic

Test statistic for 2 sample mean data is given as

Test statistic = (μ₁ - μ₂))/σ

σ = √[(s₂²/n₂) + (s₁²/n₁)]

μ₁ = average compression strength of strawberry drink = 537

n₁ = sample size of the sample of strawberry drink in cans surveyed = 10

s₁ = standard deviation of the compression strength of strawberry drink in cans surveyed= 22

μ₂ = average compression strength of cola = 559

n₂ = sample size of the sample of cola in cans surveyed = 10

s₂ = standard deviation of the compression strength of strawberry drink in cans surveyed = 17

σ = [(17²/10) + (22²/10)] = 77.5903160379 = 8.792

We will use the t-distribution as no information on population standard deviation is provided

t = (537 - 559) ÷ 8.792

= -2.502 = -2.50

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

Degree of freedom = df = n₁ + n₂ - 2 = 10 + 10- 2 = 18

Significance level = 0.05

The hypothesis test uses a one-tailed condition because we're testing in only one direction (whether compression strength of cola in can is greater).

p-value (for t = -2.50, at 0.05 significance level, df = 18, with a one tailed condition) = 0.011154 = 0.0112 to 4 d.p.

C) 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.

So, for this question, significance level = 0.05

p-value = 0.0112

0.0112 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis accept the alternative hypothesis & say that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

D) The necessary conditions required before a t-test is deemed valid include.

- The samples used must be a random sample of the population distribution with each variable in the sample independent of other one.

- The distribution of the population where the samples were extracted from must be normal or approximately normal to ensure some degree of normality for the samples.

Hence, the necessary assumption for this t-test among the options is that the distributions of compression strengths are approximately normal.

Hope this Helps!!!

7 0
3 years ago
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