A Price of a sweater was reduced from $40 to $25. By what percentage was the price of the sweater reduced?

Please help (its a pic again)

Agata [3.3K]

Answer:

1) v= 280 in^3 sa= 262 in^2

2) v= 80 in^3 sa= 120 in^2

3) v= 36 in^3 sa= 72 in^2

4) v= 240 in^3 sa= 252 in^2

8 0

3 years ago

Please help as soon as possible PLEASE

pashok25 [27]

Area of the entire rectangle $=x^2+10x+24$

Solution:

Area of the blue box = length × width

$=x\times x$

Area of the blue box $=x^2$

Area of the pink box = length × width

$=4\times x$

Area of the pink box $=4x$

Area of the green box = length × width

$=6\times x$

Area of the green box $=6x$

Area of the orange box = length × width

$=6\times 4$

Area of the orange box = 24

Total area of the rectangle $=x^2+4x+6x+24$

$=x^2+10x+24$

Area of the entire rectangle $=x^2+10x+24$

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=2%20%5Ctimes%202" id="TexFormula1" title="2 \times 2" alt="2 \times 2" align="absmiddle" class

EastWind [94]

4 (i literally had to think about this cause i thought it was too easy )

6 0

3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago

Find the distance between the two points (-10,3) and (0,27)

crimeas [40]

Answer:

26 units

Step-by-step explanation:

$\boxed{distance \: between \: 2 \: points = \sqrt{ {(x1 - x2)}^{2} + {(y1 - y2)}^{2} } }$