If peeta scored 1060 points. how many points did vev score if her score was 30% fewer

You complete 40% of ur hw problems before dinner what fraction of the problems did u complete before dinner

yuradex [85]

Is there any answer choices? I believe it is 2/5

3 0

3 years ago

Select the equation of the line that passes through the point (3, 5) and is perpendicular to the line x = 4.

kow [346]

4 because on a graph it would be proportional 3,5 is going down in a slanted line so next would be 4,4 so answer is c

3 0

4 years ago

Cos(3pi/4)= HELLPPPP

Kitty [74]

3π/4 angle is in the second quadrant, so cos (3π/4) is going to be negative.

Reference angle of 3π/4 is (π - 3π/4) = π/4

cos(π/4) = √2/2

cos (3π/4) = - √2/2

Answer is C.

5 0

3 years ago

Pythagorean theorem<br> what is the best approximation for the length for GP

Pani-rosa [81]

Answer:

You didn't provide a clear question. Please specify your question before asking next time.

Step-by-step explanation:

Hope this helped.

7 0

3 years ago

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

C

b

C

c

C

d

A

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago

If peeta scored 1060 points. how many points did vev score if her score was 30% fewer​

If peeta scored 1060 points. how many points did vev score if her score was 30% fewer