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3 years ago

Since you're lucky and found this question, here's some fr.ee points :V

Mathematics

2 answers:

Daniel [21]3 years ago

8 0

Answer:

Thank you :):):)

Step-by-step explanation:

allsm [11]3 years ago

3 0

Answer:

THANK YOUU :))))

Step-by-step explanation:

:)))

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I don’t get this and solve it well I know the answer but I testing this app

a_sh-v [17]

Answer:x=0.089

Step-by-step explanation:

5 0

4 years ago

Can some one help me with this

pychu [463]

Answer:

Actually i don't know what is the wanted but now

i will find the total of the teenagers.

Step-by-step explanation:

$\frac{4000}{45\%} = \frac{x}{100\%} \\ x = \frac{4000 \times 100}{45} \\ x = 8889 \: \: teenagers$

I hope that is useful for you :)

6 0

3 years ago

Read 2 more answers

Troy assembles wooden toya at Tindall's Toy Shoppe. He is paid $1.46 for each snake, $1.11 ab each fan, and $1.74 for each teddy

Ilia_Sergeevich [38]

Answer:

$3005.98

Step-by-step explanation:

did the math with calculator

3 0

3 years ago

A book claims that more hockey players are born in January through March than in October through December. The following data sh

astra-53 [7]

Answer:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference of birthdates distributed throughout the year

H1: There is a difference between birthdates distributed throughout the year

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total}{4}$

And replacing we got:

$E_{1} =\frac{67+56+30+37}{4}=47.5$

And now we can calculate the statistic:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

3 0

4 years ago

2) Radius = 1.5 cm Height = 6 cm

tino4ka555 [31]

Answer:

42.41 cm^3

Step-by-step explanation:

4 0

3 years ago