What is the total measurement of a 30,20, degree intersecting variable of a acute triangle.

Compare the costs of bike trips from the two companies shown. For both, the cost is a linear function of the number of days.

Nutka1998 [239]

I did this quiz its Scenic Pathways has no sign up fee and a greater cost per day.

7 0

2 years ago

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Does (2, 8) make the equation y = 4x true?

guajiro [1.7K]

Answer:

Yes, it does.

Step-by-step explanation:

The equation y = 4x is the same as saying:

"The y coordinate is 4 times as much as the x coordinate"

So, with the coordinate set of (2, 8):

y = 4x

8 = 4 * 2

This is correct, meaning the coordinate does fit the equation.

Hope this helps!

6 0

2 years ago

Read 2 more answers

Adelita, Elena, Betina, and Bianca each work as a doctor, lawyer, teacher, or banker. From these clues, decide who is the doctor

frosja888 [35]

Answer: betina or adelita

Step-by-step explanation: hope it helps

7 0

2 years ago

The number of measles cases increased 13.6 % to 57 cases this year, what was the number of cases prior to the increase? (Express

ELEN [110]

Answer:

The number of cases prior to the increase is 50.

Step-by-step explanation:

It is given that the number of measles cases increased by 13.6% and the number of cases after increase is 57.

We need to find the number of cases prior to the increase.

Let x be the number of cases prior to the increase.

x + 13.6% of x = 57

$x+x\times \frac{13.6}{100}=57$

$x+0.136x=57$

$1.136x=57$

Divide both the sides by 1.136.

$\frac{1.136x}{1.136}=\frac{57}{1.136}$

$x=50.176$

$x\approx 50$

Therefore the number of cases prior to the increase is 50.

7 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

2 years ago