Answer:
1442
Step-by-step explanation:
Given,
Mean is,

Standard deviation,
,
Margin of error(E) = 0.54,
For 98% level of confidence, 
Thus, the sample size would be,
![n=[\frac{z_{\frac{\alpha}{2}}\times \sigma}{E}]^2](https://tex.z-dn.net/?f=n%3D%5B%5Cfrac%7Bz_%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ctimes%20%5Csigma%7D%7BE%7D%5D%5E2)
![=[\frac{2.33\times 8.8}{0.54}]^2](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B2.33%5Ctimes%208.8%7D%7B0.54%7D%5D%5E2)


The teacher spend some amount on buying costumes and microphones for 6 cast members.
<u>GiveN</u><u>:</u>
- Cost of each costume = $38
- Cost of each microphone = $c
- Total money spent = $354
Let's frame the equation. Total money is the total amount spent for 6 cast members.
Money spent on costumes = $38 × 6
Money spent on microphone = $c × 6
<u>Equation:</u>

Taking 6 as common,

Dividing 6 from both sides,


Solving further,


So, the cost of each microphone = <u>$21</u>
The teacher spent this much on each microphone.
#CarryOnLearning
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Answer:
40m^2 152.6m^2
Step-by-step explanation:
Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
450/3 = x.
150 = x.
So x = 150.
This means 45 is 30% of 150