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4 years ago

What is 13/99 as a repeating decimal

Mathematics

1 answer:

eduard4 years ago

8 0

Answer:

0. 13 (the 13 recurring)

Step-by-step explanation:

Simply divide 13 by 99 on a calculator, can't be demonstrated here

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How to solve and answer using elimination

grin007 [14]

Answer:

the answer is:

x=6

y=7

4 0

4 years ago

Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.

ycow [4]

This involves quite a lot of arithmetic to do manually.

The first thing you do is to make the first number in row 2 = to 0.

This is done by R2 = -3/2 R1 + R2

so the matrix becomes

( 2 1 1) ( -3 )

( 0 -13/2 3/2) (1/2 )

(5 -1 2) (-2)

Next step is to make the 5 in row 5 = 0

then the -1 must become zero

You aim for the form

( 1 0 0) (x)

(0 1 0) (y)

(0 0 1) ( z)

x , y and z will be the required solutions.

4 0

4 years ago

Read 2 more answers

At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul

vampirchik [111]

Answer:

a) Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b) The 95% confidence interval would be given by (910.05;959.95)

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) $z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750$

Since is a bilateral test the p value is given by:

$p_v =2*P(Z>2.750)=0.0059$

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b. What is the 95% confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=935$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=180$ represent the population standard deviation

n=200 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=3278.222$

The sample deviation calculated $s=97.054$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$