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3 years ago

Which choice shows how to find the greatest common factor of 18 and 72 through prime factorization

1 answer:

6 0

$\begin{array}{c|c}18&2\\9&3\\3&3\\1\end{array}\\\\18=\boxed2\cdot\boxed3\cdot\boxed3\\\\\begin{array}{c|c}72&2\\36&2\\18&2\\9&3\\3&3\\1\end{array}\\\\72=2\cdot2\cdot\boxed2\cdot\boxed3\cdot\boxed3\\\\GCF(18,\ 72)=\boxed2\cdot\boxed3\cdot\boxed3=\boxed{\boxed{18}}$

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Please answer this correctly

mars1129 [50]

Answer:

87.5%

Step-by-step explanation:

Total parts = 8

Part 6 = 1

Parts not 6 = 7

P(not 6) = 7/8

=> 87.5%

4 0

4 years ago

Read 2 more answers

Stan made 14 of 20 free throws in basketball practice. Predict the number of free throws he would make if he attempted 100 free

Scilla [17]

Answer:

okay, so you put 14/20 as a fraction, and find out the percent. that is 70%. then you find 70% of 100, which is 70.

4 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

What certain shape has a volume that is numerically equal to three times its surface area

Tanzania [10]

Answer: Need pic to help.

3 0

3 years ago

the ratio of strawberry to banana batter is 2:3. We need 30 cups of batter total. What percent of the cake is banana.

frutty [35]

70% is the correct answer to this i think

7 0

4 years ago