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3 years ago

Explain how to find the distance between two integers using the difference

2 answers:

5 0

When we want to find the distance between two integers using the difference, the best way is to plot them on number line. And then count distance between them. For example -3 and -1. But distance can not be negative.

AnnZ [28]3 years ago

5 0

Answer:

Step-by-step explanation:

edge 2021

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The quadratic function f(x) has a turning point at (-3,6). the quadratic y=2/3f(x)+3 would have a turning point of

NARA [144]

Answer:

C. (-3,11)

Step-by-step explanation:

Tp is (-3,6) implies the quadratic could have been

f(x) = (x+3)²+6

(2/3)f(x) = (2/3)[(x+3)²+6]

= (2/3)(x+3)²+4

(2/3)f(x)+3 = (2/3)(x+3)²+4+3

= (2/3)(x+3)²+7

Tp at (-3,7)

Alternately,

No change in domain so x remains-3

(2/3)f(x) changes y from 6 to 4 (6×2/3)

+3 increases the y by 3

i.e 4+3 = 7

So, (-3,7)

4 0

3 years ago

What I can do after completing the course of statistics?

algol13

Answer:

You can apply for Civil services, Statistical services, and the Economic services

Step-by-step explanation:

Source: https://www.tutoreye.com/blog/what-is-the-scope-of-statistics

3 0

2 years ago

The box plots below show the distribution of salaries, in thousands, among employees of two small companies.

Shtirlitz [24]

Answer:

Mean and IQR

Step-by-step explanation:

The measure of centre gives the central or the measure which gives the best mid term of a distribution. Based in the details of the box plot, the median is the value which divides the box in the box plot.

For company A:

Range = 25 to 80 with a median value at 30 ; this means the median does not give a good centre measure of the distribution ad it is very close to the minimum value. This goes for the Company B plot too; with values ranging from (35 to 90) and the median designated at 40.

Hence, the mean will be the best measure of centre rather Than the median in this case.

For the variability, the interquartile range would best suit the distribution. With the lower quartile and upper quartile both having reasonable width to the minimum and maximum value of the distribution.

7 0

2 years ago

Read 2 more answers

Can you help me plz i need this ASAP please

astra-53 [7]

look you see that im in 5 grade in I dont konw this stuff do the anser mite be 3577643577

7 0

3 years ago

Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Consider the

Dafna1 [17]

Answer:

a) $\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2$

$s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311$

b) The 90% confidence interval would be given by (63.330;81.070)

$63.330 < \mu_{left}- \mu_{right}$

Step-by-step explanation:

1) Previous concepts and notation

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".

Let put some notation

x=left arm , y = right arm

x: 175 169 182 146 144

y: 102 101 94 79 79

The first step is define the difference $d_i=x_i-y_i$ , that is given so we have:

d: 73, 68, 88, 67, 65

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311$

2) Confidence interval

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

$\bar d \pm t_{\alpha/2}\frac{s_d}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=5-1=4$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,4)".And we see that $t_{\alpha/2}=2.13$

Now we have everything in order to replace into formula (1):

$72.2-2.13\frac{9.311}{\sqrt{5}}=63.330$

$72.2+2.13\frac{9.311}{\sqrt{5}}=81.070$

So on this case the 90% confidence interval would be given by (63.330;81.070)

$63.330 < \mu_{left}- \mu_{right}$

3 0

2 years ago