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il63 [147K]
3 years ago
7

Chips Ahoy wants to perform a study to determine the number of chocolate chips in their cookies. To that end, they collected a s

ample of 40 cookies. The mean of this sample is 23.95 chocolate chips. From past studies, we know that the standard deviation is 2.55 chocolate chips. Construct a 99% confidence interval of the mean of chocolate chips in all such cookies.
Mathematics
1 answer:
Dahasolnce [82]3 years ago
5 0
The formula for the confidence interval is given by

Sample mean + z*[σ/√n], and
Sample mean - z*[σ/√n]

We have:
Sample mean = 23.95
n = 40 
σ = 2.55
z* for 99% confidence = 2.58

Substitute these values into the formula, we have

23.95 + (2.58)(2.55÷√40) = 24.99
23.95 - (2.58)(2.55÷√40) = 22.91

So the lower interval is 22.91 and the highest interval is 24.99
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3 years ago
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Fittoniya [83]
<h3><u>Answer:</u></h3>

\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}

<h3><u>Step-by-step explanation:</u></h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2  \\\\\bf\implies (2y+3)^2 \leq 1  \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1  \\\\\bf\implies 4y^2+9+12y - 1 \leq 0  \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0  \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup   \\\\\bf\implies y^2y+2y+y+2 \leq 0  \\\\\bf\implies y(y+2)+1(y+2)\leq 0  \\\\\bf\implies ( y+2)(y+1)\leq 0  \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}

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vesna_86 [32]
Parallel = same slope
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3 years ago
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Mrac [35]
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3 years ago
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