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3 years ago

The larger of two numbers is five more than twice the smaller number. The sum of the two numbers is 38. Find the numbers.

Mathematics

2 answers:

Damm [24]3 years ago

8 0

Answer:

I would set this up as:

let: lesser number = x

greater number = 2x + 5

Restating the word problem:

lesser number + greater number = 38

x + 2x + 5 = 38

Solving: 3x + 5 = 38

- 5 -5

---------------

3x = 33

--- ---

3 3

x = 11

Substitution for the greater number:

2(11) + 5

22 + 5

Bezzdna [24]3 years ago

6 0

Answer:

Let the 2 no.s be x and x+15. According to the question,. x+x+15=41. 2x+15=41. 2x=41−15. 2x=26. x=262. x=13. ∴ The no.s are 13 and 13+15=28.

Step-by-step explanation:

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2 years ago

Find the quotient of -85^2-27n+76 / -17n-19

bagirrra123 [75]

Answer:
5n-4

Explanation:
The given expression is:
$\frac{-85n^2-27n+76}{-17n-19}$

1- Take the negative as a common factor from both the numerator and the denominator. This will give us:
$\frac{-(85n^2+27n-76)}{-(17n+19)}$

2- Cancel out the negative sign (common factor) from the numerator and denominator. This will give us:
$\frac{(85n^2+27n-76)}{(17n+19)}$

3- Factor the numerator. This wil give:
$\frac{(5n-4)(17n+19)}{(17n+19)}$

4- Finally, cancel out the (17n+19) which common in both numerator and denominator. This will give us the final expression:
5n-4

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2 years ago

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2 years ago

A study was conducted and two types of engines, A and B, were compared. Fifty experiments were performed using engine A and 75 u

USPshnik [31]

Answer:

a) -6 mpg.

b) 2.77 mpg

c) The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem, and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Gas mileage A: Mean 36, standard deviation 6, sample of 50:

$\mu_A = 36, s_A = \frac{6}{\sqrt{50}} = 0.8485$

Gas mileage B: Mean 42, standard deviation 8, sample of 50:

$\mu_B = 42, s_B = \frac{8}{\sqrt{50}} = 1.1314$

Distribution of the difference:

Mean:

$\mu = \mu_A - \mu_B = 36 - 42 = -6$

Standard error:

$s = \sqrt{s_A^2+s_B^2} = \sqrt{0.8485^2+1.1314^2} = 1.4142$

A. Find the point estimate.

This is the difference of means, that is, -6 mpg.

B. Find the margin of error

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = zs = 1.96*1.4142 = 2.77$

The margin of error is of 2.77 mpg

C. Construct the 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results(5 pts)

The lower end of the interval is the sample mean subtracted by M. So it is -6 - 2.77 = -8.77 mpg

The upper end of the interval is the sample mean added to M. So it is -6 + 2.77 = -3.23 mpg

The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

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2 years ago