At Basketball practice Kyla made 60 free throws this was 80% of her attempts how many free throws did kyle attempt?

2. Construct Arguments Macy says that any

melamori03 [73]

Answer:Incorrect.

For example, two equations with same y intercept.

y = 2x + 3

y = 5x + 3

This system has only one solution.

Another example,

y = x + 7

This system has infinitely many solutions.

So she is not correct because of the first example.

Step-by-step explanation:

4 0

3 years ago

The producer of a certain bottling equipment claims that the variance of all its filled bottles is .027 or less. A sample of 30

o-na [289]

Answer:

$p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459$

b. between .025 and .05

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

$\bar X$ represent the sample mean

n = 30 sample size

s= 0.2 represent the sample deviation

$\sigma_o =\sqrt{0.027}=0.164$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic

$\alpha=$ significance level

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is less than 0.027, so the system of hypothesis are:

H0: $\sigma \leq 0.027$

H1: $\sigma >0.027$

In order to check the hypothesis we need to calculate the statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(30-1) [\frac{0.2}{0.164}]^2 =42.963$

What is the approximate p-value of the test?

The degrees of freedom are given by:

$df=n-1= 30-1=29$

For this case since we have a right tailed test the p value is given by:

$p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459$

And the best option would be:

b. between .025 and .05

6 0

3 years ago

The data set of the diameters of the metal cylinders manufactured an automatic machine has sample size of n=29, mean of x= 49.98

Lena [83]

Answer:

we say for μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757 , 50.033243 )

Step-by-step explanation:

Given data

n=29

mean of x = 49.98 mm

S = 0.14 mm

μ = 50.00 mm

Cl = 95%

to find out

Can we be 95% confident that machine calibrated properly

solution

we know from t table

t at 95% and n -1 = 29-1 = 28 is 2.048

so now

Now for 95% CI for mean is

(x - 2.048 × S/√n , x + 2.048 × S/√n )

(49.98 - 2.048 × 0.14/√29 , 49.98 + 2.048 × 0.14/√29 )

( 49.926757 , 50.033243 )

hence we say for μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757 , 50.033243 )

8 0

3 years ago

I need help with #2 please I don’t understand

kipiarov [429]

1. Square
2. Circle
3. Triangle (inside of a square)
4. Pentagon
5. Rectangle

Hope this helps :)

6 0

3 years ago

Read 2 more answers

57000 x45 pls helppppppp

Varvara68 [4.7K]

Answer:

2565000

Step-by-step explanation:

Easy, multiply them in a table.

3 0

3 years ago

Read 2 more answers