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3 years ago

Help me out question #9

Mathematics

2 answers:

-Dominant- [34]3 years ago

8 0

The answer is
20 + 10h

plant grows 2h inches each week for 4 weeks
(4 * 2h) = 8h

every 2 weeks thereafter another 2h
6 wks totaling 10h

add to original height of the 20, which will give you (20 + 10h)

hope that help

Alex777 [14]3 years ago

4 0

The original height of the plant is 20 inches, and each week for 4 weeks its height increases by 2h inches, so at the end of 4 weeks its height would be:
20 + 4(2h) = 20 + 8h
Now for the next two weeks the plant grows in height by h inches each week, so we need to add a further 2h inches to the equation. Thus the expression that shows the height of the plant after 6 weeks is:
20 + 8h + 2h = 20 + 10h, therefor A is the correct answer

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3 years ago

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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

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4 years ago

Answer this questions please!!!

Svetradugi [14.3K]

Answer:

sin^2 x

Step-by-step explanation:

By the formula,

sin^2 x + cos^2x = 1

so, 1 - cos^ 2 x = sin ^2

<h2>HOPE IT HELPS U!!!! </h2>

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