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Ad libitum [116K]
3 years ago
12

Find the product 0.9×079=

Mathematics
2 answers:
krek1111 [17]3 years ago
6 0
The answer is 71.1. hope this helped
weqwewe [10]3 years ago
6 0

Answer is 71.1 because 0.9 x 079 in a calculator is 71.1

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Question 1(Multiple Choice Worth 5 points)
SIZIF [17.4K]

Question 1:

Substitution.

x+y=2

y=-x+2

2x+(-x+2)=4

x=2

(2)+y=2

y=0

Solution: (2,0)

There is one solution

Question 7 is too confusing, just find where the points intersect at (-5,2).

Question 9:

Elimination.

(-1)y=2x+6

y=2x+2

New: -y=-2x-6

0=-4

No Solution

Question 11:

Just look at explanations.

It's (5,17) both lines etc..

4 0
3 years ago
Q:- The rationalisation factor of √3 {<br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B3%7D%20" id="TexFormula1" title=" \sqrt
kipiarov [429]

Answer:

its the same  (root 3)

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Pleade help help please
vlada-n [284]

Answer:

A.

Step-by-step explanation:

they both have the same interquartile range so it’s A.

7 0
2 years ago
Read 2 more answers
Adding and subtracting polynomials<br><br>Subtract 2x² + 6x - 5 from 7x² + 3x + 4​
allsm [11]

Answer: 5x^2 - 3x + 9

Step-by-step explanation:

4 0
3 years ago
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In constructing a 95 percent confidence interval, if you increase n to 4n, the width of your confidence interval will (assuming
Damm [24]

Answer:

about 50 percent of its former width.

Step-by-step explanation:

Let's assume that our parameter of interest is given by \theta and in order to construct a confidence interval we can use the following formula:

\hat \theta \pm ME(\hat \theta)

Where \hat \theta is an estimator for the parameter of interest and the margin of error is defined usually if the distribution for the parameter is normal as:

ME = z_{\alpha} SE

Where z_{\alpha/2} is a quantile from the normal standard distribution that accumulates \alpha/2 of the area on each tail of the distribution. And SE represent the standard error for the parameter.

If our parameter of interest is the population proportion the standard of error is given by:

SE= \frac{\hat p (1-\hat p)}{n}

And if our parameter of interest is the sample mean the standard error is given by:

SE = \frac{s}{\sqrt{n}}

As we can see the standard error for both cases assuming that the other things remain the same are function of n the sample size and we can write this as:

SE = f(n)

And since the margin of error is a multiple of the standard error we have that ME = f(n)

Now if we find the width for a confidence interval we got this:

Width = \hat \theta + ME(\hat \theta) -[\hat \theta -ME(\hat \theta)]

Width = 2 ME (\hat \theta)

And we can express this as:

Width =2 f(n)

And we can define the function f(n) = \frac{1}{\sqrt{n}} since as we can see the margin of error and the standard error are function of the inverse square root of n. So then we have this:

Width_i= 2 \frac{1}{\sqrt{n}}

The subscript i is in order to say that is with the sample size n

If we increase the sample size from n to 4n now our width is:

Width_f = 2 \frac{1}{\sqrt{4n}} =2 \frac{1}{\sqrt{4}\sqrt{n}} =\frac{2}{2} \frac{1}{\sqrt{n}} =\frac{1}{\sqrt{n}} =\frac{1}{2} Width_i

The subscript f is in order to say that is the width for the sample size 4n.

So then as we can see the width for the sample size of 4n is the half of the wisth for the width obtained with the sample size of n. So then the best option for this case is:

about 50 percent of its former width.

7 0
4 years ago
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