When comparing data quickly, it’s best to use a

Ada charges a flat rate of $56.00 for painting houses plus an additional $8.00 for each hour she works. Which of the following e

Evgen [1.6K]

Answer:

cost = 56 + 8t, where t is the number of hours worked for.

for 4 hours of work, cost = 56 + 8 x 4 = $88

Step-by-step explanation:

5 0

3 years ago

Which of the following has the most inertia, four kilograms of iron or four kilograms of cork? ...?

evablogger [386]

<span>Inertia is defined as "the tendency for a body to resist acceleration; the tendency of a body at rest to remain at rest or of a body in motion to stay in motion in a straight line unless disturbed by an external force" Therefore, both will have the same inertia since they have the same mass.</span>

5 0

3 years ago

Read 2 more answers

What your expect to gen.math subject

sveta [45]

Can you elaborate on ur question

6 0

3 years ago

Read 2 more answers

In early spring, you are wanting to purchase 6 potted rose plants. The plants to choose from come in 8-inch pots that sell for $

topjm [15]

Things we know:
Six total pots = # of 10" + # of 8"
$78 total = cost of 10" + cost of 8"

So:
6 = x +y
and
78 = x*(cost of x) + y * (cost of y)
78 = x*12 + y*15

We need:
x+y=6 AND 12x+15y=78

So our answer can only be B.

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