Find the sample space for tossing 4 coins. Then find p( exactly 1 head)

Fit a quadratic function to these three points:<br> (-2,8), (0, -4), and (4, 68)<br> Respuesta

zloy xaker [14]

Answer:

f(x) = 4x^2 + 2x - 4.

Step-by-step explanation:

Let the quadratic function be y = f(x) = ax^2 + bx + c.

For the point (-2, 8) ( x = -2 when y = 8) we have:

a(-2)^2 + (-2)b + c = 8

4a - 2b + c = 8 For (0, -4) we have:

0 + 0 + c = -4 so c = -4. For (4, 68) we have:

16a + 4b + c = 68

So we have 2 systems of equations in a and b ( plugging in c = -4):

4a - 2b - 4 = 8

16a + 4b - 4 = 68

4a - 2b = 12

16a + 4b = 72 Multiplying 4a - 2b = 12 by 2 we get:

8a - 4b = 24

Adding the last 2 equations:

24a = 96

a = 4

Now plugging a = 4 and c = -4 in the first equation:

4(4) - 2b - 4 = 8

-2b = 8 - 16 + 4 = -4

b = 2.

3 0

3 years ago

How many x-intercepts does the graph of y=x^2 have

34kurt

Answer:

one point (on origin),

Step-by-step explanation:

graph of Y=X^2 is parabolic ,and x intecept means the point where ,Y COORDINATE (ORDINATE) becomes 0

so now after putting y=0 in eqution we got

x^2=0

i.e, x=0

that means at (0,0) ,so only one point!

✌️:)

4 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

Help me in e plz and then explain

horsena [70]

You have to take a better picture

4 0

3 years ago

8

xxTIMURxx [149]

Answer: Zoey sells over 50 boxes of cookies

Step-by-step explanation: The hypothesis is the “if” part in an if-then statement.

8 0

2 years ago