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3 years ago

Pablo bought a basketball. The price after a 7.5% tax was $34.40. What was the price before t

Mathematics

1 answer:

Alika [10]3 years ago

8 0

Answer:

$2.58

Step-by-step explanation:

7.5/100

__/34.40

This might seem confusing but basically you put a percent over 100.

That's how you would solve something like this

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3/4 ( 1/2× -12) + 4/5

Inessa [10]

Hello! Your answer would be :

-3.7

Hope this helps! And have a great day! :)

6 0

3 years ago

Read 2 more answers

Erica must tutor a minimum of 20 hours per week in order to remain in the national honor society. She has a student that she tut

irinina [24]

Answer:

3 hours every weekday

Step-by-step explanation:

20 - 5 is 15 so she must work 15 hours during the week since saturday and sunday are taken. 15 divided by 5 is 3 therefore, giving you three hours a day; Monday, Tuesday, Wednesday, Thursday, and Friday

8 0

4 years ago

ou are putting border around a poster frame that is 2 feet by 4 feet. How much border will you need to buy?

inessss [21]

well, you would use the equation P=2L+2W, or perimeter equals 2 time the length + 2 times the width. which would be 2*2+4*4= 20. you would need 20 feet of border. hope this helped ^-^

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

Percent of change from 125 to 75

grandymaker [24]

I think it would be 50% let’s hope

3 0

3 years ago