Find The Surface Area 5cm 6cm 11cm

Philipe

UkoKoshka [18]

<h3>Philipe has to make a weekly sales of $ 3000 so that he can earn $ 360</h3>

<em><u>Solution:</u></em>

Given that,

Philipe works for a computer store that pays a 12% commission and no salary

We have to find his weekly sales have to be for him to earn $360

From given, we can say,

12 % of weekly sales will earn him $ 360

12 % of weekly sales = 360

Let "x" be the weekly sales

12 % of "x" = 360

Solve the above expression

$\frac{12}{100} \times x = 360\\\\0.12x = 360\\\\x = \frac{360}{0.12}\\\\x = 3000$

Thus Philipe has to make a weekly sales of $ 3000 so that he can earn $ 360

6 0

3 years ago

A, B & C form the vertices of a triangle.

kirza4 [7]

<h3>Answer: 20.3</h3>

Work Shown:

tan(angle) = opposite/adjacent

tan(B) = AC/AB

tan(67) = x/8.6

8.6*tan(67) = x

x = 8.6*tan(67)

x = 20.2603303460843 ... make sure your calculator is in degree mode

x = 20.3 .... rounding to 3 significant figures

see diagram below

6 0

3 years ago

What is the center and radius of the circle with equation x2 + y2 - 4x + 22y + 61 = 0?

lesantik [10]

Hello,

$x^2+y^2-4x+22y+61=0\\\\
(x^2-4x+4) +(y^2+22y+121)-121-4+61=0\\\\
(x-2)^2+(y+11)^2=64


$

Center is (2,-11) and radius is 8

6 0

4 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

Solve this problem to find Y. 800 x Y = 4800

kupik [55]

Answer:

6

Step-by-step explanation:

800*Y=4800

Y=4800/800

Y=6

5 0

3 years ago

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