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netineya [11]
3 years ago
14

How do you write the number for 3,152,308

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
8 0

If you mean in word form, then it should be as follows;

Three million one hundred and fifty-two thousand three hundred and eight.

If you don't mean word form, you will have to be a little more specific.

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Line A: y=-1/2x+3
sashaice [31]
A) For the equation y=- \frac{1}{2}x+3, the slope is - \frac{1}{2}.
Slope-intercept form is y = mx + b, where m is slope and b is the y-intercept. This means that if m=- \frac{1}{2}, the slope is  - \frac{1}{2}.

B) Since this equation is in the same form, you just find what m is equal to.
Since m=2, that's the slope.

For the solution, set y=- \frac{1}{2}x+3 equal to y=2x-4, so it would be put together like this:
y=[tex]- \frac{1}{2}x+3=2x-4\\ -\frac{1}{2}=2x-7\\ 1\frac{1}{2}x=-7\\ x=-4.66667
So your answer is -4.667.
8 0
3 years ago
Written in simplest form explain how you found the answer
Anettt [7]
The answer is dividing stuff
8 0
3 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:

So

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

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

So

\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
2 years ago
The math test has 10 multiple choice questions each with 4 answer for Betty guesses correctly on the first two questions what is
lord [1]
If she already gets the first two, thats a given 100% situation. then it is just 1/4
3 0
3 years ago
Read 2 more answers
List the 2 other things ASAP (DONT BE GIVING ME LINKS JUST ANSWER PLZ)
stepladder [879]

Answer:

Step-by-step explanation:

it can e the median of AC

and the angle bisector of B

3 0
3 years ago
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