Find the x-coordinate of Q' after Q(-2. 5) was reflected over the line y = -x + 4.

The original selling price of the pair of socks is $8. The pair of socks is on sale for 15%. The new selling price of the pair of socks would be the original price - 15% off the original price. It becomes

8 - (15/100 × 8) = 8 - 0.15×8 = 8 - 1.2 = $6.8

if the sales tax rate is 8%, it means that the total cost of the socks would be the new selling price + 8% of the new selling price. It becomes

Total cost of the pair of socks

= 6.8 + (8/100×6.8) = 6.8 + 0.08×6.8 = 6.8 + 0.544 = $7.344

7 0

3 years ago

Byron bought 1.9 of banannas and 11/12 pounds of grapes. how many more pounds of bananas did he buy

Genrish500 [490]

Answer:

11/12

Step-by-step explanation:

I took 1.9 and turned it into a fraction and the fraction is 1 9/12 so if you have 11/12 and 1 9/12 then you bought 10/12 more pounds of banana than grapes

5 0

2 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

2 years ago