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3 years ago

Is 0.981 greater than less than or equal to 2.3

Mathematics

2 answers:

timofeeve [1]3 years ago

8 0

0.981 is less then 1
2.3 is greater then 1

0.981 < 2.3 (it is less then)

sattari [20]3 years ago

3 0

<h2>Answer:</h2>

0.981 is less than 2.3 i.e. mathematically it is expressed as:

0.981 < 2.3

<h2>Step-by-step explanation:</h2>

We know that while comparing two decimal numbers if the number before the decimal is greater than the number itself is greater as compared to the other number.

Here we have two decimal numbers as 0.981 and 2.3.

Now we know that: 2>0

This means that:

2.3 is greater than 0.981

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The cost of 16 television sets is $640,160. What is the cost of 1 television set? (Round off the amount to the nearest thousand

Liono4ka [1.6K]

Answer:

$40000

Step-by-step explanation:

640160≈640000

640000/16=40000

5 0

4 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).

8 0

3 years ago

Maynor rode his bike 2 miles in 8 minutes. Paul rode his bike

madam [21]

no because maynor rode 2 miles every 4 minutes and paul rode 3 miles every 4 minutes. not the same

7 0

3 years ago

Evaluate [(21 + 6) − 32] ק 9 ⋅ 2.

anzhelika [568]