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Zina [86]
3 years ago
15

Jonah stacks balls of yarn in a square pyramid for a store display. The number of balls of yarn, P(n), in "n" layers of a square

pyramid is given by
P(n)=P(n-1)=n^2 what number could not be the number of balls Jonah stacks.

A.14
B.24
C.5
D.30
Mathematics
2 answers:
erastovalidia [21]3 years ago
8 0

Answer: The answer is B:24. Have A Nice Day

Step-by-step explanation:

GenaCL600 [577]3 years ago
7 0
P(1) = 1\\\\P(2)=P(1)+2^2=1+2^2=1+4=5\\\\P(3)=P(2)+3^2=5+3^2=5+9=14\\\\P(4)=P(3)+4^2=14+4^2=14+16=30

Correct answer is B. 24
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Yakvenalex [24]

Answer:

<h3>∠L = 33.9°</h3><h3 />

Step-by-step explanation:

∠K = 180 - 67.8

    = 112.2°

∠J = ∠L =<u> 180 - 112.2</u>

                      2

∠J = ∠L = 33.9°

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She will have enough money in 60 months...

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4 years ago
Read 2 more answers
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
A board was 63.5 inches long. If you have 3 of them , what is the total length of all three ?
zepelin [54]
The answer is 190.5
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6 0
4 years ago
What is the equation of the line that passes through the point (5, -2) and has a<br> slope of -2/5
Alecsey [184]

Answer:

y=-2/5x

Step-by-step explanation:

y-y1=m(x-x1)

y-(-2)=-2/5(x-5)

y+2=-2/5(x-5)

y=-2/5x+10/5-2

y=-2/5x+2-2

y=-2/5x

4 0
3 years ago
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